Existence criteria of solutions for a fractional nonlocal boundary value problem and degeneration to corresponding integer-order case

Abstract

In this paper, we mainly discuss the existence and uniqueness results of solutions to fractional differential equations with multi-strip boundary conditions. When the fractional order α becomes integer, the existence theorem of positive solutions can be established by a monotone iterative technique. Also, some examples are presented to illustrate the main results.

1 Introduction

Differential equations attract many scholars’ interest since they can succinctly establish the relationship between variables and their derivatives. And fractional order calculus has been used as an important tool to improve mathematical modeling of many complex problems, such as in fluid mechanics, rheology, fractional model of nerve and fractional regression model; see [1–5], for instance.

In the last decades, fractional order boundary value problems have also received plenty of attention from many researchers. There are many achievements derived from some fractional equations with various boundary conditions, some recent contribution can be found in [6–13].

For example, in [14], authors considered a discrete multi-point boundary value problem such as

$$ \textstyle\begin{cases} ^{c}D^{\alpha }u(t)+f(t,u(t))=0,\quad t\in (0,1), \\ u(0)=0,\qquad D_{0+}^{\beta }u(0)=0, \\ D_{0+}^{\beta }u(1)=\sum_{i=1}^{\infty }\xi_{i}D_{0+}^{\beta }u(\eta_{i}), \end{cases} $$

where $2<\alpha \leq 3$, $1\leq \beta \leq 2$, $\alpha -\beta \geq 1$ and $0<\xi_{i},\eta_{i}<1$ with $\sum_{i=1}^{\infty }\xi_{i} \eta_{i}^{\alpha -\beta -1}<1$.

In [15], the authors considered the following equation with integral boundary conditions:

$$ \textstyle\begin{cases} {}^{c}D^{q}u(t)+f(t,u(t))=0,\quad t\in [0,1], \\ \alpha u(0)-\beta u'(0)=\int_{0}^{1}g(s)u(s)\,ds, \\ \gamma u(1)+\delta u'(1)=\int_{0}^{1}h(s)u(s)\,ds, \end{cases} $$

where $q\in (1,2]$, α, β, γ and δ are nonnegative constants, and $^{c}D^{q}$ is the standard Caputo fractional derivative of fractional order q.

Different from [14] and [15], some work focused on the solvability of the fractional differential equations with both multi-point and integral boundary conditions. In [16], Ahmad et al. were concerned with the following problem:

$$ \textstyle\begin{cases} (^{c}D^{q}+k^{c}D^{q-1})x(t)=f(t,x(t),^{c}D^{\beta }x(t),I^{\gamma }(t)),\quad t\in [0,1], \\ x(0)=0,\qquad x'(0)=0,\qquad \sum_{i=1}^{m}x(\zeta_{i})= \lambda \int_{0}^{\eta } \frac{(\eta -s)^{\delta -1}}{\Gamma (\delta )}x(s)\,ds, \end{cases} $$

where $2< q\leq 3$, $0<\beta ,\gamma <1$, $k>0$, $\delta <1$, $0< \eta <\zeta_{1}<\zeta_{2}<\cdots <\zeta_{m}<1$, and λ, $a_{i}$, $i=1,2,\dots , m$ are real constants.

Motivated by the above works, in this paper, we first deal with the following fractional order differential equation with multi-point and multi-strip boundary conditions:

$$ \textstyle\begin{cases} D_{0+}^{\alpha }u(t)+q(t)f(t,u(t), D_{0+}^{\beta }u(t), D_{0+}^{ \gamma }u(t))=0,\quad t\in (0,1), \\ u(0)=D_{0+}^{\gamma }u(0)=0, \\ u(1)+\sum_{i=1} ^{m}a_{i}D_{0+}^{\beta }u(\xi_{i})=\sum_{i=1}^{m}b_{i}\int_{0} ^{\xi_{i}}u(s)\,ds, \end{cases} $$

(1.1)

where $D_{0+}^{\alpha }$ is the Riemann–Liouville fractional derivative of order α, $f:[0,1]\times \mathbb{R}^{3}\rightarrow \mathbb{R}$ is a continuous function, $2<\alpha \leq 3$, $0<\beta \leq 1<\gamma <\alpha -1$, $0<\xi_{i}\leq 1$, $a_{i}$, $b_{i}$ are nonnegative constants satisfying $a_{i}\geq \frac{\Gamma (\alpha - \beta )}{\Gamma (\alpha +1)}b_{i}$, for $i=1,2,\dots ,m$.

It is worth mentioning that the nonlinear term of BVP (1.1) depends on all the lower fractional order derivatives of the unknown function, which implies more complete consideration from the practical application problems’ point of view. Although the complexity of the nonlinearity of BVP (1.1) is increased, we still get three Green’s functions with concise forms and satisfactory properties. Meanwhile, boundary conditions of (1.1) include both multiple discrete points and multiple band-like integrals, which is a broad generalization of most models in [17–24]. By using the Leray–Schauder alternative theorem and the Banach’s contraction mapping principle, existence and uniqueness theorems of solutions to BVP (1.1) are proved.

However, it is known that sometimes only positive solutions are significant in the real world. For this reason, in the second part we degenerate the fractional order model and choose $\alpha =3$, $\gamma =2$, $\beta =1$. Hence, the following integer-order differential equation is discussed:

$$ \textstyle\begin{cases} u{'''}(t)+q(t)f(t,u(t),u'(t),u{''}(t))=0,\quad t\in (0,1), \\ u(0)=u{''}(0)=0,\qquad u(1)+\sum_{i=1}^{m}a_{i}u'(\xi_{i})=\sum_{i=1}^{m}b_{i}\int_{0}^{\xi_{i}}u(s)\,ds, \end{cases} $$

(1.2)

where $f:[0,1]\times [0,\infty ) \times \mathbb{R} \times (-\infty ,0] \rightarrow [0,\infty )$ is continuous, $a_{i}$, $b_{i}$ are nonnegative constants satisfying $\frac{1}{6}b_{i}\leq a_{i} <\frac{\xi_{i}^{2}b _{i}}{2}$ with $\frac{\sqrt{3}}{3}<\xi_{i}\leq 1$, for $i=1,2, \dots ,m$.

The arguments for BVP (1.2) are based on a monotone iterative technique. It is important that the Green’s function associated with BVP (1.2) is nonnegative, which is different from that of BVP (1.1). In this part, not only the existence results of positive solutions are obtained, but also the approximate solutions of BVP (1.2) can be presented.

2 Fractional order differential equation

In this section, we consider the fractional order BVP (1.1) and establish the existence and uniqueness criteria of solutions. We put forward some indispensable definitions and theorems in advance.

Definition 2.1

([25])

The Riemann–Liouville fractional integral of order $\alpha >0$ of a function $f:(0,\infty )\rightarrow \mathbb{R}$ is given by

$$ I_{0+}^{\alpha }f(t)=\frac{1}{\Gamma (\alpha )} \int_{0}^{t}(t-s)^{ \alpha -1}f(s)\,ds, $$

provided the right-hand side is pointwise defined on $(0,\infty )$, where $\Gamma (\alpha )$ is the Euler gamma function defined by $\Gamma (\alpha )=\int_{0}^{\infty }t^{\alpha -1}e^{-t}\,dt$, for $\alpha >0$.

Definition 2.2

([25])

The Riemann–Liouville fractional derivative of order $\alpha >0$ for a function $f:(0,\infty )\rightarrow \mathbb{R}$ is given on $[0,\infty )$ by

$$ D_{0+}^{\alpha }f(t)=\frac{1}{\Gamma (n-\alpha )} \biggl( \frac{d}{dt} \biggr)^{n} \int_{0}^{t}(t-s)^{n-\alpha -1}f(s)\,ds, $$

where $n=[\alpha ]+1$ and $[\alpha ]$ stands for the largest integer not greater than α.

From the definitions of Riemann–Liouville’s derivative, the following lemmas can be obtained.

Lemma 2.1

([25])

For $\alpha >0$, if we assume that $u\in C[0,\infty ) \cap L ^{1}(0,1)$, then

$$ I_{0+}^{\alpha } \bigl(D_{0+}^{\alpha }u(t) \bigr)=u(t)+m_{1}t^{\alpha -1}+m_{2}t ^{\alpha -2}+ \cdots +m_{n}t^{\alpha -n}, $$

for some $m_{i}\in \mathbb{R},i=1,2,\dots ,n$, where n is the smallest integer greater than or equal to α.

Remark 2.1

([20])

The following properties are useful for our discussion:

• (i) As a basic example, we quote, for $\lambda > -1$,

  $$\begin{aligned} D_{0+}^{\alpha }t^{\lambda }=\frac{\Gamma (\lambda +1)}{\Gamma ( \lambda -\alpha +1)}t^{\lambda -\alpha }; \end{aligned}$$

• (ii) $D_{0+}^{\alpha }I_{0+}^{\alpha }u(t)=u(t)$, for $u(t)\in L^{1}(0,1)$.

Before presenting the main results, we give the following assumptions:

• (F1) $2<\alpha \leq 3$, $0<\beta \leq 1<\gamma <\alpha -1$, $0<\xi_{i}< 1$, for $i=1,2,\dots ,m$;

• (F2) $a_{i}$, $b_{i}$ are nonnegative constants and $a_{i}\geq \frac{\Gamma (\alpha -\beta )}{\Gamma (\alpha +1)}b_{i}$, for $i=1,2,\dots ,m$;

• (F3) $q\in L^{1}[0,1]$ is nonnegative, and $f\in C([0,1] \times \mathbb{R}^{3}, \mathbb{R})$.

For convenience, denote

$$ \begin{aligned} &A=\frac{\Gamma (\alpha )}{\Gamma (\alpha -\beta )}\sum _{i=1}^{m}a _{i}\xi_{i}^{\alpha -\beta -1}, \qquad B=\frac{\Gamma (\alpha )}{\Gamma ( \alpha +1)}\sum_{i=1}^{m}b_{i} \xi_{i}^{\alpha }, \\ &\Delta_{1}=1+A-B,\qquad \varphi (t)=\frac{t^{\alpha -1}}{\Delta_{1}}. \end{aligned} $$

(2.1)

In view of (F1) and (F2), it is obvious that $A>B\geq 0$, as well as $\Delta_{1}> 1$ and $\varphi (t)\geq 0$ for $t\in [0,1]$.

Lemma 2.2

For $h(t)\in C(0,1)\cap L^{1}(0,1)$, the boundary value problem

$$ \textstyle\begin{cases} D_{0+}^{\alpha }u(t)+h(t)=0,\quad t\in (0,1), \\ u(0)=D_{0+}^{\gamma }u(0)=0, \\ u(1)+\sum_{i=1} ^{m}a_{i}D_{0+}^{\beta }u(\xi_{i})=\sum_{i=1}^{m}b_{i} \int_{0} ^{\xi_{i}}u(s)\,ds \end{cases} $$

(2.2)

has a unique solution

$$ u(t)= \int_{0}^{1}H_{0}(t,s)h(s)\,ds+P(h) \varphi(t), $$

(2.3)

where

$$\begin{aligned}& H_{0}(t,s)=\frac{1}{\Gamma (\alpha )} \textstyle\begin{cases} \varphi (t)(1-s)^{\alpha -1}-(t-s)^{\alpha -1},& 0\leq s\leq t\leq 1, \\ \varphi (t)(1-s)^{\alpha -1},& 0\leq t\leq s\leq 1, \end{cases}\displaystyle \end{aligned}$$

(2.4)

$$\begin{aligned}& P(h)=\sum_{i=1}^{m}a_{i}I_{0+}^{\alpha -\beta }h( \xi_{i})-\sum_{i=1} ^{m}b_{i}I_{0+}^{\alpha +1}h( \xi_{i}). \end{aligned}$$

(2.5)

Proof

From Lemma 2.1, we can reduce $D_{0+}^{\alpha }u(t)+h(t)=0$ to the following equivalent equation:

$$ u(t) =-\frac{1}{\Gamma (\alpha )} \int_{0}^{t}(t-s)^{\alpha -1}h(s)\,ds+m _{1}t^{\alpha -1}+m_{2}t^{\alpha -2}+m_{3}t^{\alpha -3}, $$

(2.6)

where $m_{1}$, $m_{2}$ and $m_{3}$ are arbitrary real constants.

Since $u(0)=D_{0+}^{\gamma }u(0)=0$, we have $m_{2}=m_{3}=0$. Integrating (2.6) from 0 to $\xi_{i}$, for $i=1,\dots ,m$, we get

$$ \int_{0}^{\xi_{i}}u(t)\,dt = \int_{0}^{\xi_{i}} \bigl[-I_{0+}^{\alpha }h(s)+m _{1}s^{\alpha -1} \bigr]\,ds =-I_{0+}^{\alpha +1}h( \xi_{i})+\frac{\Gamma (\alpha )}{\Gamma ( \alpha +1)}m_{1}\xi_{i}^{\alpha }. $$

(2.7)

By Remark 2.1, we have

$$ D_{0+}^{\beta }u(t)=-\frac{1}{\Gamma (\alpha -\beta )} \int_{0}^{t}(t-s)^{ \alpha -\beta -1}h(s)\,ds+ \frac{\Gamma (\alpha )}{\Gamma (\alpha - \beta )}m_{1}t^{\alpha -\beta -1}. $$

(2.8)

Thus, together with (2.6), (2.7) and (2.8), it can be seen that

$$ m_{1}=\frac{1}{\Delta_{1}} \bigl[I_{0+}^{\alpha }h(1)+P(h) \bigr], $$

(2.9)

where $\Delta_{1}$ is defined by (2.1), $P(h)$ is given by (2.5). Hence, the solution of problem (2.2) can be expressed as

$$\begin{aligned} u(t) &= -I_{0+}^{\alpha }h(t)+\frac{t^{\alpha -1}}{\Delta_{1}}I_{0+} ^{\alpha }h(1)+\frac{P(h)}{\Delta_{1}}t^{\alpha -1} \\ &=-\frac{1}{\Gamma (\alpha )} \int_{0}^{t}(t-s)^{\alpha -1}h(s)\,ds+ \frac{ \varphi (t)}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}h(s)\,ds+ \varphi (t)P(h) \\ &= \int_{0}^{1}H_{0}(t,s)h(s)\,ds+\varphi (t)P(h), \end{aligned}$$

where $H_{0}(t,s)$ is given by (2.4) and $\varphi (t)$ is introduced by (2.1).

This completes the proof of the lemma. □

After replacing $m_{1}$ in (2.8), we get

$$ D_{0+}^{\beta }u(t) = \int_{0}^{1}H_{\beta }(t,s)h(s)\,ds+ \frac{\Gamma (\alpha )}{\Delta_{1}\Gamma (\alpha -\beta )}P(h)t^{\alpha -\beta -1}, $$

(2.10)

where

$$\begin{aligned} H_{\beta }(t,s)=\frac{1}{\Delta_{1}\Gamma (\alpha -\beta )} \textstyle\begin{cases} t^{\alpha -\beta -1}(1-s)^{\alpha -1}-\Delta_{1}(t-s)^{\alpha - \beta -1},& 0\leq s \leq t \leq 1, \\ t^{\alpha -\beta -1}(1-s)^{\alpha -1},&0\leq t\leq s\leq 1. \end{cases}\displaystyle \end{aligned}$$

Similarly, we have

$$ D_{0+}^{\gamma }u(t) = \int_{0}^{1}H_{\gamma }(t,s)h(s)\,ds+ \frac{ \Gamma (\alpha )}{\Delta_{1}\Gamma (\alpha -\gamma )}P(h)t^{\alpha - \gamma -1}, $$

(2.11)

where

$$\begin{aligned} H_{\gamma }(t,s)=\frac{1}{\Delta_{1}\Gamma (\alpha -\gamma )} \textstyle\begin{cases} t^{\alpha -\gamma -1}(1-s)^{\alpha -1}-\Delta_{1}(t-s)^{\alpha - \gamma -1},& 0\leq s \leq t \leq 1, \\ t^{\alpha -\gamma -1}(1-s)^{\alpha -1},&0\leq t\leq s\leq 1. \end{cases}\displaystyle \end{aligned}$$

Next, we present some properties of Green’s functions and $P(h)$.

Lemma 2.3

For $t,s\in [0,1]$, the Green functions $H_{0}(t,s),H_{\beta }(t,s),H _{\gamma }(t,s)$ and $P(h)$ satisfy the following properties:

• (a) $\vert H_{0}(t,s) \vert \leq H(s)$, where $H(s)=\frac{(1-s)^{ \alpha -1}}{\Delta_{1}\Gamma (\alpha )}(1+\Delta_{1})$;

• (b) $\vert P(h) \vert \leq \overline{P}(h)$, for $h(t)\geq 0$, where

  $$\begin{aligned} \overline{P}(h)=\sum_{i=1}^{m} \frac{a_{i}}{\Gamma (\alpha -\beta )} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{\alpha -\beta -1} \bigl[1+(\xi_{i}-s)^{ \beta +1} \bigr]h(s)\,ds; \end{aligned}$$

• (c) $\vert H_{\beta }(t,s) \vert \leq \Lambda_{\beta }(s)$, $\vert H_{\gamma }(t,s) \vert \leq \Lambda_{\gamma }(s)$, where

  $$\begin{aligned} &\Lambda_{\beta }(s)=\frac{1}{\Gamma (\alpha -\beta )}(1-s)^{\alpha - \beta -1} \biggl( \frac{(1-s)^{\beta }}{\Delta_{1}}+1 \biggr), \\ &\Lambda_{\gamma }(s)=\frac{1}{\Gamma (\alpha -\gamma )}(1-s)^{ \alpha -\gamma -1} \biggl( \frac{(1-s)^{\gamma }}{\Delta_{1}}+1 \biggr). \end{aligned}$$

Proof

(a) For $0\leq s\leq t\leq 1$, we have

$$\begin{aligned} \bigl\vert H_{0}(t,s) \bigr\vert &=\frac{1}{\Gamma (\alpha )} \bigl\vert \varphi(t) (1-s)^{\alpha -1}-(t-s)^{\alpha -1} \bigr\vert \\ &\leq \frac{1}{\Gamma (\alpha )} \biggl( \frac{t^{\alpha -1}}{\Delta _{1}}(1-s)^{\alpha -1}+(1-s)^{\alpha -1} \biggr) \\ &\leq \frac{(1-s)^{\alpha -1}}{\Delta_{1}\Gamma (\alpha )}(1+\Delta _{1}) =H(s); \end{aligned}$$

while, for $0\leq t\leq s \leq 1$, we have

$$\begin{aligned} \bigl\vert H_{0}(t,s) \bigr\vert &=\frac{1}{\Gamma (\alpha )} \bigl\vert \varphi(t) (1-s)^{\alpha -1} \bigr\vert \leq \frac{(1-s)^{\alpha -1}}{\Delta _{1}\Gamma (\alpha )}(1+ \Delta_{1}). \end{aligned}$$

(b) According to (2.5), (F1) and (F2), for $h(t)\geq 0$, we have

$$\begin{aligned} \bigl\vert P(h) \bigr\vert &= \Biggl\vert \sum _{i=1}^{m}\frac{a_{i}}{\Gamma (\alpha -\beta )} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{\alpha -\beta -1}h(s) \,ds- \sum_{i=1}^{m}\frac{b_{i}}{\Gamma (\alpha +1)} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{\alpha }h(s) \,ds \Biggr\vert \\ &\leq \sum_{i=1}^{m} \Biggl\vert \frac{a_{i}}{\Gamma (\alpha -\beta )} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{\alpha -\beta -1}h(s) \,ds \Biggl\vert +\sum_{i=1}^{m} \Biggr\vert \frac{b_{i}}{\Gamma (\alpha +1)} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{\alpha }h(s) \,ds \Biggr\vert \\ &\leq \sum_{i=1}^{m}\frac{a_{i}}{\Gamma (\alpha -\beta )} \int_{0} ^{\xi_{i}}(\xi_{i}-s)^{\alpha -\beta -1}h(s) \,ds+ \sum_{i=1}^{m}\frac{a _{i}}{\Gamma (\alpha -\beta )} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{\alpha }h(s) \,ds \\ &=\sum_{i=1}^{m}\frac{a_{i}}{\Gamma (\alpha -\beta )} \int_{0}^{\xi _{i}}(\xi_{i}-s)^{\alpha -\beta -1} \bigl[1+(\xi_{i}-s)^{\beta +1} \bigr]h(s)\,ds = \overline{P}(h). \end{aligned}$$

(c) For $0\leq s \leq t \leq 1$, we have

$$\begin{aligned} \bigl\vert H_{\beta }(t,s) \bigr\vert &= \frac{1}{\Delta_{1}\Gamma (\alpha - \beta )} \bigl\vert t^{\alpha -\beta -1}(1-s)^{\alpha -1}-\Delta_{1}(t-s)^{\alpha -\beta -1} \bigr\vert \\ &\leq \frac{1}{\Delta_{1}\Gamma (\alpha -\beta )} \bigl( t^{\alpha - \beta -1}(1-s)^{\alpha -1}+ \Delta_{1}(t-ts)^{\alpha -\beta -1} \bigr) \\ &= \frac{t^{\alpha -\beta -1}}{\Gamma (\alpha -\beta )}(1-s)^{\alpha -\beta -1} \biggl(\frac{(1-s)^{\beta }}{\Delta_{1}}+1 \biggr) \\ &\leq \frac{1}{\Gamma (\alpha -\beta )}(1-s)^{\alpha -\beta -1} \biggl(\frac{(1-s)^{\beta }}{\Delta_{1}}+1 \biggr) = \Lambda_{\beta }(s); \end{aligned}$$

while, for $0\leq t \leq s \leq 1$, we have

$$\begin{aligned} \bigl\vert H_{\beta }(t,s) \bigr\vert &= \frac{1}{\Delta_{1}\Gamma (\alpha - \beta )} \bigl\vert t^{\alpha -\beta -1}(1-s)^{\alpha -1} \bigr\vert \\ &\leq \frac{1}{\Gamma (\alpha -\beta )}(1-s)^{\alpha -\beta -1} \cdot \frac{(1-s)^{\beta }}{\Delta_{1}} \\ &\leq \Lambda_{\beta }(s). \end{aligned}$$

Similarly, we can have

$$\begin{aligned} \bigl\vert H_{\gamma }(t,s) \bigr\vert &\leq \frac{1}{\Gamma (\alpha -\gamma )}(1-s)^{ \alpha -\gamma -1} \biggl(\frac{(1-s)^{\gamma }}{\Delta_{1}}+1 \biggr) = \Lambda_{\gamma }(s). \end{aligned}$$

Then the proof is completed. □

Let $E_{1}=\{u(t)\mid u(t)\in C[0,1] \text{ and } D_{0+}^{\beta }u(t), D_{0+}^{\gamma }u(t) \in C[0,1]\}$ be endowed with the norm

$$\begin{aligned} \Vert u \Vert = \max \bigl\{ \Vert u \Vert _{0}, \bigl\Vert D_{0+}^{\beta }u \bigr\Vert _{0}, \bigl\Vert D_{0+}^{\gamma }u \bigr\Vert _{0} \bigr\} , \end{aligned}$$

where $\Vert u \Vert _{0} = \max_{t\in [0,1]} \vert u(t) \vert$. In order to ensure the feasibility of the conclusion, we should prove the following lemmas.

Lemma 2.4

$(E_{1},\Vert \cdot \Vert )$ is a Banach space.

Proof

Let $\{u_{n}\}_{n=1}^{\infty }$ be a Cauchy sequence in the space $(E_{1},\Vert \cdot \Vert )$. It is clear that $\{u_{n}\}_{n=1}^{ \infty }$, $\{D_{0+}^{\beta }u_{n}\}_{n=1}^{\infty }$ and $\{D_{0+} ^{\gamma }u_{n}\}_{n=1}^{\infty }$ are Cauchy sequences in the space $C[0,1]$. Accordingly, $\{u_{n}\}_{n=1}^{\infty }$, $\{D_{0+}^{\beta }u_{n}\}_{n=1}^{\infty }$ and $\{D_{0+}^{\gamma }u_{n}\}_{n=1}^{ \infty }$ uniformly converge to some u, v and w on $[0,1]$ while $u, v, w \in C[0,1]$. Now we should prove that $v=D_{0+}^{\beta }u$ and $w=D_{0+}^{\gamma }u$.

For $t\in [0,1]$, we notice that

$$\begin{aligned} \bigl\vert I_{0+}^{\beta }D_{0+}^{\beta }u_{n}(t)-I_{0+}^{\beta }v(t) \bigr\vert &\leq \frac{1}{\Gamma (\beta )} \int_{0}^{t}(t-s)^{\beta -1} \bigl\vert D_{0+}^{\beta }u_{n}(s)-v(s) \bigr\vert \,ds \\ &\leq \frac{1}{\Gamma (\beta +1)}\max_{s\in [0,1]} \bigl\vert D_{0+}^{\beta }u_{n}(s)-v(s) \bigr\vert . \end{aligned}$$

Due to the convergence of $\{D_{0+}^{\beta }u_{n}\}_{n=1}^{\infty }$, we obtain that $\lim_{n \to \infty }I_{0+}^{\beta }D_{0+}^{ \beta }u_{n}(t)= I_{0+}^{\beta }v(t)$ uniformly for $t\in [0,1]$. Meanwhile, by Lemma 2.1, we get $I_{0+}^{\beta }D_{0+}^{\beta }u_{n}(t)=u _{n}(t)+m_{1}t^{\beta -1}$, for $t\in [0,1]$ and some $m_{1}\in \mathbb{R}$. These two facts yield

$$\begin{aligned} \lim_{n\to \infty }{I_{0+}^{\beta }D_{0+}^{\beta }u_{n}(t)}= \lim_{n\to \infty }{u_{n}(t)+m_{1}t^{\beta -1}}=I_{0+}^{\beta }v(t), \quad \text{for } t\in [0,1]. \end{aligned}$$

Together with $\lim_{n \to \infty }u_{n}(t)= u(t)$ for $t\in [0,1]$, we have

$$ u(t)+m_{1}t^{\beta -1}=I_{0+}^{\beta }v(t), \quad \text{for } t\in [0,1]. $$

(2.12)

Taking the derivative of order β of both sides of Eq. (2.12), as a result we have

$$\begin{aligned} D_{0+}^{\beta }I_{0+}^{\beta }v(t)=D_{0+}^{\beta } \bigl(u(t)+m_{1}t^{ \beta -1} \bigr)=D_{0+}^{\beta }u(t), \quad \text{for } t\in [0,1]. \end{aligned}$$

From Remark 2.1, it is easy to see that

$$\begin{aligned} v(t)=D_{0+}^{\beta }u(t),\quad \text{for } t \in [0,1]. \end{aligned}$$

Also, for $t\in [0,1]$, $\omega (t)=D_{0+}^{\gamma }u(t)$ can be proved using similar steps.

The proof of this lemma is completed. □

For $u\in E_{1}$, we define an operator $T_{1}$ as follows:

$$\begin{aligned} (T_{1}u) (t)= \int_{0}^{1}H_{0}(t,s)q(s)f_{u}(s) \,ds+ \varphi (t)P \bigl[q(s)f _{u}(s) \bigr], \end{aligned}$$

where

$$\begin{aligned} &P \bigl[q(s)f_{u}(s) \bigr]= \sum_{i=1}^{m}a_{i}I_{0+}^{\alpha -\beta }q( \xi_{i})f _{u}(\xi_{i})-\sum _{i=1}^{m}b_{i}I_{0+}^{\alpha +1}q( \xi_{i})f_{u}(\xi _{i}), \end{aligned}$$

and $f_{u}(s)=f(s,u(s),D_{0+}^{\beta }u(s),D_{0+}^{\gamma }u(s))$. Also, from (2.10) and (2.11), we have

$$\begin{aligned} &D_{0+}^{\beta }(T_{1}u) (t) = \int_{0}^{1}H_{\beta }(t,s)q(s)f_{u}(s) \,ds+ \frac{ \Gamma (\alpha )}{\Delta_{1}\Gamma (\alpha -\beta )}P \bigl[q(s)f_{u}(s) \bigr]t ^{\alpha -\beta -1}, \\ &D_{0+}^{\gamma }(T_{1}u) (t) = \int_{0}^{1}H_{\gamma }(t,s)q(s)f_{u}(s) \,ds+ \frac{ \Gamma (\alpha )}{\Delta_{1}\Gamma (\alpha -\gamma )}P \bigl[q(s)f_{u}(s) \bigr]t ^{\alpha -\gamma -1}. \end{aligned}$$

Lemma 2.5

$T_{1}:E_{1}\rightarrow E_{1}$ is completely continuous.

Proof

By the continuity of Green’s function $H_{0}(t,s)$ and $f(t,u(t),D _{0+}^{\beta }u(t),D_{0+}^{\gamma }u(t))$, $T_{1}$ is continuous.

Then, we show $T_{1}$ is uniformly bounded. Let $\Omega \subset E_{1}$ be bounded. We set $\Vert u \Vert \leq K$, with $K >0$, for all $u\in \Omega$. Let $F=\max \{\vert f_{u}(t) \vert \mid 0\leq t \leq 1, -K\leq u(t) \leq K, -K\leq D_{0+}^{\beta }u(t) \leq K, -K \leq D_{0+}^{\gamma }u(t)\leq K\}$. Then from Lemma 2.3, we have

$$\begin{aligned} \bigl\vert (T_{1}u) (t) \bigr\vert &\leq \int_{0}^{1}H_{0}(t,s)q(s) \bigl\vert f_{u}(s) \bigr\vert \,ds+\frac{1}{\Delta_{1}}P \bigl[q(s) \bigl\vert f_{u}(s) \bigr\vert \bigr] \\ &\leq F \biggl( \int_{0}^{1}H(s)q(s)\,ds+\frac{1}{\Delta_{1}} \overline{P} \bigl[q(s) \bigr] \biggr) \end{aligned}$$

and

$$\begin{aligned} \bigl\vert D_{0+}^{\beta }(T_{1}u) (t) \bigr\vert & \leq \int_{0}^{1}\Lambda_{ \beta }(s)q(s) \bigl\vert f_{u}(s) \bigr\vert \,ds+\frac{\Gamma (\alpha )}{\Delta _{1}\Gamma (\alpha -\beta )}t^{\alpha -\beta -1}P \bigl[q(s) \bigl\vert f_{u}(s) \bigr\vert \bigr] \\ &\leq F \biggl( \int_{0}^{1}\Lambda_{\beta }(s)q(s)\,ds+ \frac{\Gamma ( \alpha )}{\Delta_{1}\Gamma (\alpha -\beta )}\overline{P} \bigl[q(s) \bigr] \biggr), \\ \bigl\vert D_{0+}^{\gamma }(T_{1}u) (t) \bigr\vert &\leq \int_{0}^{1} \Lambda_{\gamma }(s)q(s) \bigl\vert f_{u}(s) \bigr\vert \,ds+\frac{\Gamma (\alpha )}{\Delta_{1}\Gamma (\alpha -\gamma )}t^{\alpha -\gamma -1}P \bigl[q(s) \bigl\vert f_{u}(s) \bigr\vert \bigr] \\ &\leq F \biggl( \int_{0}^{1}\Lambda_{\gamma }(s)q(s)\,ds+ \frac{\Gamma ( \alpha )}{\Delta_{1}\Gamma (\alpha -\gamma )}\overline{P} \bigl[q(s) \bigr] \biggr). \end{aligned}$$

Hence, we can find upper bounds of $\vert (T_{1}u)(t) \vert$, $\vert D_{0+}^{\beta }(T_{1}u)(t) \vert$ and $\vert D_{0+}^{\gamma }(T_{1}u)(t) \vert$, for $t\in [0,1]$. Thus, $\Vert T_{1}u \Vert$ is bounded, which implies that the operator $T_{1}$ is uniformly bounded.

Finally, we show $T_{1}$ is equicontinuous. Indeed, for any $u\in \Omega$, $t_{1},t_{2}\in [0,1]$, with $t_{1}< t_{2}$, we can infer that

$$\begin{aligned} & \bigl\vert (T_{1}u) (t_{2})-(T_{1}u) (t_{1}) \bigr\vert \\ & \quad \leq \biggl\vert \int_{0}^{1} H_{0}(t_{2},s)q(s)f_{u}(s) \,ds- \int_{0}^{1} H_{0}(t_{1},s)q(s)f_{u}(s) \,ds \biggr\vert + \bigl\vert \varphi(t_{2})-\varphi (t_{1}) \bigr\vert P \bigl[q(s) \bigl\vert f_{u}(s) \bigr\vert \bigr] \\ &\quad \leq \frac{F}{\Delta_{1}\Gamma (\alpha )} \int_{0}^{1} \bigl( \bigl\vert \bigl(t_{2}^{\alpha -1}-t_{1}^{\alpha -1} \bigr) (1-s)^{\alpha -1} \bigr\vert +\Delta _{1} \bigl\vert (t_{2}-s)^{\alpha -1}-(t_{1}-s)^{\alpha -1} \bigr\vert \bigr) q(s)\,ds \\ & \quad \quad {}+ \frac{F}{\Delta_{1}} \bigl\vert t_{2}^{\alpha -1}-t_{1}^{\alpha -1} \bigr\vert \overline{P} \bigl[q(s) \bigr]. \end{aligned}$$

Applying the mean value theorem, the following inequalities hold:

$$\begin{aligned} &t_{2}^{\alpha -1}-t_{1}^{\alpha -1}\leq (\alpha -1) (t_{2}-t_{1}), \\ &(t_{2}-s)^{\alpha -1}-(t_{1}-s)^{\alpha -1}\leq ( \alpha -1) (t_{2}-t _{1}), \end{aligned}$$

from which we can deduce that

$$\begin{aligned} & \bigl\vert (T_{1}u) (t_{2})-(T_{1}u) (t_{1}) \bigr\vert \\ &\quad \leq \frac{(\alpha -1)F}{\Delta_{1}\Gamma (\alpha )} \biggl\{ \int_{0} ^{1} \bigl[(1-s)^{\alpha -1}+ \Delta_{1} \bigr]q(s)\,ds+\Gamma (\alpha ) \overline{P} \bigl[q(s) \bigr] \biggr\} (t_{2}-t_{1}) \\ &\quad \rightarrow 0 \quad \text{as } t_{2}\to t_{1}. \end{aligned}$$

In addition,

$$\begin{aligned} & \bigl\vert D_{0+}^{\beta }(T_{1}u) (t_{2})-D_{0+}^{\beta }(T_{1}u) (t_{1}) \bigr\vert \\ & \quad \leq F \int_{0}^{1} \biggl\vert \frac{(t_{2}^{\alpha -\beta -1}-t_{1}^{\alpha -\beta -1})}{\Delta_{1}\Gamma (\alpha -\beta )}(1-s)^{\alpha -1}- \frac{(t_{2}-s)^{\alpha -\beta -1}-(t_{1}-s)^{\alpha -\beta -1}}{\Gamma (\alpha -\beta )} \biggr\vert q(s)\,ds \\ &\quad \quad {} +F\frac{\Gamma (\alpha )}{\Delta_{1}\Gamma (\alpha -\beta )} \bigl\vert t_{2}^{\alpha -\beta -1}-t_{1}^{\alpha -\beta -1} \bigr\vert \overline{P} \bigl[q(s) \bigr] \\ &\quad \leq \frac{F}{\Delta_{1}\Gamma (\alpha -\beta )} \biggl[ \int_{0} ^{1} \biggl\vert \bigl(t_{2}^{\alpha -\beta -1}-t_{1}^{\alpha -\beta -1} \bigr) (1-s)^{\alpha -1} \biggl\vert q(s)\,ds \\ &\quad \quad {}+\Delta_{1} \int_{0}^{1} \biggr\vert (t_{2}-s)^{\alpha -\beta -1}-(t_{1}-s)^{\alpha -\beta -1} \biggr\vert q(s)\,ds \biggr] \\ &\quad \quad {} +\frac{\Gamma (\alpha )F}{\Delta_{1}\Gamma (\alpha -\beta )} \bigl\vert t_{2}^{\alpha -\beta -1}-t_{1}^{\alpha -\beta -1} \bigr\vert \overline{P} \bigl[q(s) \bigr] \\ & \quad \leq \frac{(\alpha -\beta -1)F}{\Delta_{1}\Gamma (\alpha -\beta )} \biggl\{ \int_{0}^{1} \bigl[(1-s)^{\alpha -1}+ \Delta_{1} \bigr]q(s)\,ds+\Gamma (\alpha ) \overline{P} \bigl[q(s) \bigr] \biggr\} (t_{2}-t_{1}) \\ &\quad \rightarrow 0 \quad \text{as } t_{2}\to t_{1}. \end{aligned}$$

This also leads to $\vert D_{0+}^{\gamma }(T_{1}u)(t_{2})-D_{0+}^{\gamma }(T_{1}u)(t_{1}) \vert \rightarrow 0$, as $t_{2}\to t_{1}$.

Therefore, $T_{1}$ is equicontinuous for all $u\in \Omega$. Thus, by means of the Arzelà–Ascoli Theorem, we obtain that $T_{1}:E_{1} \rightarrow E_{1}$ is completely continuous. □

Lemma 2.6

(Leray–Schauder alternative theorem)

Let $T:E\rightarrow E$ be a completely continuous operator (i.e., a map that restricted to any bounded set in E is compact). Let

$$\begin{aligned} &\varepsilon (T)= \bigl\{ u\in E:u=\lambda T(u), 0< \lambda < 1 \bigr\} . \end{aligned}$$

Then, either the set is unbounded, or T has at least one fixed point.

From the above facts, if operator $T_{1}$ has fixed points, we can observe that BVP (1.1) has solutions.

The next stage is devoted to obtaining the existence result.

For convenience, set

$$ \begin{aligned} &L_{P}=\sum _{i=1}^{m}\frac{a_{i}}{\Gamma (\alpha -\beta )} \int_{0} ^{\xi_{i}}(\xi_{i}-s)^{\alpha -\beta -1} \bigl[1+(\xi_{i} -s)^{\beta +1} \bigr]q(s)\,ds, \\ &L_{\beta }= \int_{0}^{1}(1-s)^{\alpha -\beta -1} \biggl( \frac{(1-s)^{ \beta }}{\Delta_{1}}+1 \biggr)q(s)\,ds, \\ &L_{\gamma }= \int_{0}^{1}(1-s)^{\alpha -\gamma -1} \biggl( \frac{(1-s)^{ \gamma }}{\Delta_{1}}+1 \biggr)q(s)\,ds, \\ &L_{H}= \int_{0}^{1} \biggl(\frac{1}{\Delta_{1}}+1 \biggr) (1-s)^{\alpha -1}q(s)\,ds, \\ &M_{1}=\max \biggl\{ \frac{L_{H}}{\Gamma (\alpha )}+\frac{L_{P}}{\Delta _{1}}, \frac{L_{\beta }}{\Gamma (\alpha -\beta )}+\frac{\Gamma ( \alpha )L_{P}}{\Delta_{1}\Gamma (\alpha -\beta )}, \frac{L_{\gamma }}{ \Gamma (\alpha -\gamma )}+\frac{\Gamma (\alpha )L_{P}}{\Delta_{1} \Gamma (\alpha -\gamma )} \biggr\} . \end{aligned} $$

(2.13)

Theorem 2.7

Assume there exist real constants $\mu_{i}\geq 0$ ($i=1,2,3$) and $\mu_{0}>0$ such that for $t\in [0,1]$, $u,v,w \in \mathbb{R}$, we have

$$\begin{aligned} \bigl\vert f(t,u,v,w) \bigr\vert \leq \mu_{0}+\mu_{1} \vert u \vert +\mu_{2} \vert v \vert +\mu_{3}\vert w \vert . \end{aligned}$$

If $(\mu_{1}+\mu_{2}+\mu_{3})M_{1}<1$, then BVP (1.1) has at least one solution.

Proof

In order to verify that problem (1.1) has at least one solution by Lemma 2.6, we should prove that the set $\varepsilon =\{ u\in E_{1} \mid u= \lambda T_{1}(u),0\leq \lambda \leq 1\}$ is bounded.

Let $u\in \varepsilon$, for any $t\in [0,1]$, we have

$$\begin{aligned} \bigl\vert u(t) \bigr\vert &= \bigl\vert \lambda (T_{1}u) (t) \bigr\vert \leq \bigl\vert (T_{1}u) (t) \bigr\vert \\ &\leq \biggl\vert \int_{0}^{1}H_{0}(t,s)q(s)f_{u}(s) \,ds \biggr\vert + \bigl\vert \varphi (t)P \bigl[q(s)f_{u}(s) \bigr] \bigr\vert \\ &\leq \bigl(\mu_{0}+\mu_{1}\vert u \vert + \mu_{2}\vert v \vert +\mu _{3}\vert w \vert \bigr) \int_{0}^{1}H(s)q(s)\,ds \\ &\quad {}+ \bigl(\mu_{0}+ \mu_{1} \vert u \vert + \mu_{2}\vert v \vert +\mu_{3}\vert w \vert \bigr) \frac{ \overline{P}[q(s)]}{\Delta_{1}} \\ &\leq \bigl(\mu_{0}+\mu_{1}\Vert u \Vert _{0}+ \mu_{2}\Vert v \Vert _{0}+ \mu_{3}\Vert w \Vert _{0} \bigr) \biggl[ \frac{L_{H}}{\Gamma (\alpha )}+\frac{L _{P}}{\Delta_{1}} \biggr]. \end{aligned}$$

Also, we have

$$\begin{aligned} \bigl\vert D_{0+}^{\beta }u(t) \bigr\vert = & \bigl\vert D_{0+}^{\beta }\lambda (T_{1}u) (t) \bigr\vert \leq \bigl\vert D_{0+}^{\beta }(T_{1}u) (t) \bigr\vert \\ \leq & \biggl\vert \int_{0}^{1}H_{\beta }(t,s)q(s)f_{u}(s) \,ds \biggr\vert + \biggl\vert \frac{\Gamma (\alpha )}{\Delta_{1}\Gamma (\alpha -\beta )}P \bigl[q(s)f_{u}(s) \bigr]t^{\alpha -\beta -1} \biggr\vert \\ \leq & \bigl(\mu_{0}+\mu_{1}\vert u \vert + \mu_{2}\vert v \vert +\mu _{3}\vert w \vert \bigr) \biggl[ \int_{0}^{1}\Lambda_{\beta }(s)q(s)\,ds+ \frac{ \Gamma (\alpha )}{\Delta_{1}\Gamma (\alpha -\beta )}\overline{P} \bigl[q(s) \bigr] \biggr] \\ \leq & \bigl(\mu_{0}+\mu_{1}\Vert u \Vert _{0}+\mu_{2}\Vert v \Vert _{0}+ \mu_{3}\Vert w \Vert _{0} \bigr) \biggl[ \frac{L_{\beta }}{\Gamma ( \alpha -\beta )}+\frac{\Gamma (\alpha )L_{P}}{\Delta_{1}\Gamma ( \alpha -\beta )} \biggr] \end{aligned}$$

and

$$\begin{aligned} \bigl\vert D_{0+}^{\gamma }u(t) \bigr\vert \leq \bigl( \mu_{0}+\mu_{1}\Vert u \Vert _{0}+ \mu_{2}\Vert v \Vert _{0}+\mu_{3}\Vert w \Vert _{0} \bigr) \biggl[\frac{L_{\gamma }}{\Gamma (\alpha -\gamma )}+\frac{\Gamma ( \alpha )L_{P}}{\Delta_{1}\Gamma (\alpha -\gamma )} \biggr]. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} &\Vert u \Vert _{0} \leq \bigl(\mu_{0}+( \mu_{1}+\mu_{2}+\mu_{3})\Vert u \Vert \bigr) \biggl[ \frac{L_{H}}{\Gamma (\alpha )}+\frac{L_{P}}{\Delta _{1}} \biggr], \\ & \bigl\Vert D_{0+}^{\beta }u \bigr\Vert _{0} \leq \bigl(\mu_{0}+(\mu_{1}+\mu_{2}+ \mu_{3})\Vert u \Vert \bigr) \biggl[ \frac{L_{\beta }}{\Gamma (\alpha - \beta )}+ \frac{\Gamma (\alpha )L_{P}}{\Delta_{1}\Gamma (\alpha - \beta )} \biggr], \\ & \bigl\Vert D_{0+}^{\gamma }u \bigr\Vert _{0} \leq \bigl(\mu_{0}+(\mu_{1}+\mu _{2}+ \mu_{3})\Vert u \Vert \bigr) \biggl[ \frac{L_{\gamma }}{\Gamma ( \alpha -\gamma )}+ \frac{\Gamma (\alpha )L_{P}}{\Delta_{1}\Gamma ( \alpha -\gamma )} \biggr]. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert u \Vert \leq \frac{\mu_{0}M_{1}}{1- ( \mu_{1}+\mu_{2}+\mu _{3} )M_{1}}, \end{aligned}$$

where $M_{1}$ has been given in (2.13), which proves that $\Vert u \Vert$ is bounded. Thus, operator $T_{1}$ has at least one fixed point and, consequently, we can derive that BVP (1.1) has at least one solution. □

In the following, we should verify the uniqueness of the solution to BVP (1.1) by Banach’s contraction mapping principle.

Theorem 2.8

Let $f:[0,1]\times \mathbb{R}^{3}\rightarrow \mathbb{R}$ be a continuous function. For all $t\in [0,1]$, $u_{i},v_{i},w_{i}\in \mathbb{R}$ ($i=1,2$), we have

$$\begin{aligned} \bigl\vert f(t,u_{2},v_{2},w_{2})-f(t,u_{1},v_{1},w_{1}) \bigr\vert \leq M \bigl(\vert u_{2}-u_{1} \vert + \vert v_{2}-v_{1} \vert +\vert w_{2}-w_{1} \vert \bigr), \end{aligned}$$

where M is the Lipschitz constant. If $3MM_{1} \leq 1$, BVP (1.1) has a unique solution, where $M_{1}$ is given by (2.13).

Proof

Denote $M_{0}=\sup_{t\in [0,1]}\vert f(t,0,0,0) \vert$. Set $r \geq \frac{M_{0}M_{1}}{1-3MM_{1}}$ subject to the above mentioned M, $M_{1}$ and $M_{0}$. The set $B_{r}\subset E_{1}$ is defined by $B_{r}=\{u \in E_{1}\mid \Vert u \Vert \leq r \}$, and we will show that $T_{1}B_{r} \subset B_{r}$. For $u\in B_{r}$, we obtain

$$\begin{aligned} & \bigl\vert f \bigl(t,u(t), D_{0+}^{\beta }u(t),D_{0+}^{\gamma }u(t) \bigr) \bigr\vert \\ &\quad \leq \bigl\vert f \bigl(t,u(t),D_{0+}^{\beta }u(t),D_{0+}^{\gamma }u(t) \bigr)-f(t,0,0,0) \bigr\vert + \bigl\vert f(t,0,0,0) \bigr\vert \\ &\quad \leq M \bigl(\Vert u \Vert _{0}+ \bigl\Vert D_{0+}^{\beta }u \bigr\Vert _{0}+ \bigl\Vert D_{0+}^{\gamma }u \bigr\Vert _{0} \bigr)+M_{0} \\ &\quad \leq 3Mr+M_{0}. \end{aligned}$$

(2.14)

For $u\in B_{r}$, from (2.14) and similar to the proof of Theorem 2.7, we have

$$\begin{aligned}& \bigl\vert (T_{1}u) (t) \bigr\vert = \biggl\vert \int_{0}^{1}H_{0}(t,s)q(s)f_{u}(s) \,ds+ \varphi (t)P \bigl[q(s)f_{u}(s) \bigr] \biggr\vert \\& \hphantom{ \vert (T_{1}u) (t) \vert }\leq (3Mr+M_{0}) \biggl[ \frac{L_{H}}{\Gamma (\alpha )}+ \frac{L_{P}}{ \Delta_{1}} \biggr], \\& \bigl\vert D_{0+}^{\beta }(T_{1}u) (t) \bigr\vert = \biggl\vert \int_{0}^{1}H_{\beta }(t,s)q(s)f_{u}(s) \,ds+ \frac{\Gamma (\alpha )}{\Delta_{1}\Gamma(\alpha -\beta )}P \bigl[q(s)f_{u}(s) \bigr]t^{\alpha -\beta -1} \biggr\vert \\& \hphantom{ \vert D_{0+}^{\beta }(T_{1}u) (t) \vert }\leq (3Mr+M_{0}) \biggl[\frac{L_{\beta }}{\Gamma (\alpha -\beta )}+ \frac{ \Gamma (\alpha )L_{P}}{\Delta_{1}\Gamma (\alpha -\beta )} \biggr], \\& \bigl\vert D_{0+}^{\gamma }(T_{1}u) (t) \bigr\vert \leq (3Mr+M_{0}) \biggl[\frac{L _{\gamma }}{\Gamma (\alpha -\gamma )}+\frac{\Gamma (\alpha )L_{P}}{ \Delta_{1}\Gamma (\alpha -\gamma )} \biggr], \end{aligned}$$

which yields $\Vert T_{1}u \Vert \leq (3Mr+M_{0})M_{1}< r$. This shows that $T_{1}$ maps $B_{r}$ into itself. Now, setting $u_{1},u_{2} \in E_{1}$, for each $t\in [0,1]$, we get

$$\begin{aligned} & \bigl\vert (T_{1}u_{2}) (t)-(T_{1}u_{1}) (t) \bigr\vert \\ &\quad = \biggl\vert \int_{0}^{1}H_{0}(t,s)q(s) \bigl(f_{u_{2}}(s)-f_{u_{1}}(s) \bigr)\,ds+\varphi (t) \bigl(P \bigl[q(s)f_{u_{2}}(s) \bigr]-P \bigl[q(s)f_{u_{1}}(s) \bigr] \bigr) \biggr\vert \\ & \quad \leq M \bigl(\Vert u_{2}-u_{1} \Vert _{0}+ \bigl\Vert D_{0+}^{\beta }u_{2}-D_{0+}^{\beta }u_{1} \bigr\Vert _{0}+ \bigl\Vert D_{0+}^{\gamma }u_{2}-D_{0+}^{\gamma }u_{1} \bigr\Vert _{0} \bigr) \biggl( \frac{L_{H}}{\Gamma (\alpha )}+\frac{L_{P}}{\Delta_{1}} \biggr) \\ & \quad \leq 3M\cdot \biggl( \frac{L_{H}}{\Gamma (\alpha )}+\frac{L_{P}}{ \Delta_{1}} \biggr) \Vert u_{2}-u_{1} \Vert \\ &\quad \leq 3MM_{1}\Vert u_{2}-u_{1} \Vert . \end{aligned}$$

Additionally, we obtain

$$\begin{aligned} & \bigl\vert D_{0+}^{\beta }(T_{1}u_{2}) (t)-D_{0+}^{\beta }(T_{1}u_{1}) (t) \bigr\vert \\ &\quad = \biggl\vert \int_{0}^{1}H_{\beta }(t,s)q(s) \bigl(f_{u_{2}}(s)-f_{u_{1}}(s) \bigr)\,ds+\frac{t^{\alpha -\beta -1}\Gamma (\alpha )}{\Delta_{1}\Gamma (\alpha -\beta )} \bigl(P \bigl[q(s)f_{u_{2}}(s) \bigr]-P \bigl[q(s)f_{u_{1}}(s) \bigr] \bigr) \biggr\vert \\ &\quad \leq M \bigl(\Vert u_{2}-u_{1} \Vert _{0}+ \bigl\Vert D_{0+}^{\beta }u_{2}-D_{0+}^{\beta }u_{1} \bigr\Vert _{0} \\ &\quad \quad {}+ \bigl\Vert D_{0+}^{\gamma }u_{2}-D_{0+}^{\gamma }u_{1} \bigr\Vert _{0} \bigr) \biggl[ \frac{L_{\beta }}{\Gamma (\alpha - \beta )}+\frac{\Gamma (\alpha )L_{P}}{\Delta_{1}\Gamma (\alpha - \beta )} \biggr] \\ &\quad \leq 3M\cdot \biggl[ \frac{L_{\beta }}{\Gamma (\alpha -\beta )}+\frac{ \Gamma (\alpha )L_{P}}{\Delta_{1}\Gamma (\alpha -\beta )} \biggr] \bigl\Vert u_{2}-u_{1} \Vert \\ &\quad \leq 3MM_{1} \Vert u_{2}-u_{1} \bigr\Vert \end{aligned}$$

and

$$\begin{aligned} \bigl\vert D_{0+}^{\gamma }(T_{1}u_{2}) (t)-D_{0+}^{\gamma }(T_{1}u_{1}) (t) \bigr\vert &\leq 3M\cdot \biggl[ \frac{L_{\gamma }}{\Gamma (\alpha -\gamma )}+\frac{\Gamma (\alpha )L _{P}}{\Delta_{1}\Gamma (\alpha -\gamma )} \biggr]\Vert u_{2}-u_{1} \Vert \\ &\leq 3MM_{1}\Vert u_{2}-u_{1} \Vert . \end{aligned}$$

From the above, we have $\Vert T_{1}u_{2}-T_{1}u_{1} \Vert \leq 3MM _{1}\Vert u_{2}-u_{1} \Vert$. In view of $3MM_{1}<1$, the operator $T_{1}$ is a contraction. Thus, the uniqueness of solution to BVP (1.1) follows from Banach’s contraction mapping principle. □

Example 2.1

Consider the nonlinear fractional differential equation

$$ \textstyle\begin{cases} D_{0+}^{2.2}u(t)+ [\frac{2}{5}t+\frac{1}{48} (2u(t)+ \frac{8}{5}D_{0+}^{0.7}u(t)+ \frac{4}{3}D_{0+}^{1.1}u(t) ) ]=0,\quad t\in (0,1), \\ u(0)=D_{0+}^{1.1}u(0)=0, \\ u(1)+\sum_{i=1}^{m}a _{i}D_{0+}^{0.7}u( \xi_{i})=\sum_{i=1}^{m}b_{i} \int_{0}^{\xi _{i}}u(s)\,ds. \end{cases} $$

(2.15)

In this model, we set

$$\begin{aligned}& m=2,\qquad q(t)=1,\qquad \xi_{1}=\frac{2}{3},\qquad \xi_{2}=\frac{5}{6}, \\& a_{1}=\frac{3}{5},\qquad a_{2}=\frac{2}{5}, \qquad b_{1}=\frac{3}{2}, \qquad b_{2}=1. \end{aligned}$$

It is easy to verify that (F1)–(F3) hold. By calculation, we have

$$\begin{aligned} &A=1.063,\qquad B=0.5838,\qquad \Delta_{1}=1.4793, \\ &L_{P}=0.6111,\qquad L_{\beta }=0.9739,\qquad L_{\gamma }=1.2164,\qquad L_{H}=0.7618, \\ &M_{1}=\max \biggl\{ \frac{L_{H}}{\Gamma (\alpha )}+\frac{L_{P}}{\Delta_{1}}, \frac{L_{\beta }}{\Gamma (\alpha -\beta )}+\frac{\Gamma (\alpha )L_{P}}{\Delta_{1}\Gamma (\alpha -\beta )}, \frac{L_{\gamma }}{ \Gamma (\alpha -\gamma )}+\frac{\Gamma (\alpha )L_{P}}{\Delta_{1} \Gamma (\alpha -\gamma )} \biggr\} =1.757. \end{aligned}$$

Meanwhile, we see

$$\begin{aligned}& \bigl\vert f \bigl(t,u,D_{0+}^{0.7}u(t),D_{0+}^{1.1}u(t) \bigr) \bigr\vert \\& \quad \leq \frac{2}{5}+\frac{1}{24}\vert u \vert + \frac{1}{30} \bigl\vert D_{0+}^{0.7}u(t) \bigr\vert + \frac{1}{36} \bigl\vert D_{0+}^{1.1}u(t) \bigr\vert , \\& \bigl\vert f \bigl(t,u_{2},D_{0+}^{0.7}u_{2}(t),D_{0+}^{1.1}u_{2}(t)\bigr)-f \bigl(t,u_{1},D_{0+}^{0.7}u_{1}(t),D_{0+}^{1.1}u_{1}(t) \bigr) \bigr\vert \\& \quad \leq \frac{1}{9} \bigl(\Vert u_{2}-u_{1} \Vert _{0}+ \bigl\Vert D_{0+}^{0.7}u_{2}-D_{0+}^{0.7}u_{1}\bigr\Vert _{0}+ \bigl\Vert D_{0+}^{1.1}u_{2}-D_{0+}^{1.1}u_{1} \bigr\Vert _{0} \bigr). \end{aligned}$$

Therefore, $(\mu_{1}+\mu_{2}+\mu_{3} )M_{1}<1$ and $3MM_{1}<1$.

Thus, all the conditions of the above theorems are satisfied. Hence, by Theorem 2.7 problem (2.15) has at least one solution, and by Theorem 2.8 it has a unique solution.

3 Integer-order differential equation

In this section, in order to establish the existence results of positive solutions, we try to degenerate the fractional order problem into a corresponding integer-order differential model.

Necessarily, we give the following assumptions:

• (H1) $a_{i}$, $b_{i}$ are nonnegative constants satisfying $\frac{1}{6}b_{i}\leq a_{i} <\frac{\xi_{i}^{2}b_{i}}{2}$, with $\frac{\sqrt{3}}{3}<\xi_{i}\leq 1$ for $i=1,2,\dots ,m$;

• (H2) $q\in L^{1}[0,1]$ is nonnegative and $f\in C([0,1] \times [0,\infty )\times \mathbb{R}\times (-\infty ,0], [0,\infty ))$.

For convenience, we denote

$$ \begin{aligned} &E=\sum_{i=1}^{m}a_{i}, \qquad F=\sum_{i=1}^{m}\frac{b_{i}\xi _{i}^{2}}{2}, \\ &\Delta_{2}=1+E-F,\qquad \psi (t)=\frac{t}{\Delta_{2}}. \end{aligned} $$

(3.1)

From (H1), it is easy to see that $F>E\geq 0$. In the following, we always assume that $0< F-E <1$. Hence, we have $0<\Delta_{2}< 1$, and $\psi (t)\geq 0$ for $t\in [0,1]$.

Lemma 3.1

Let $h(t)\in C[0,1]\cap L^{1}(0,1)$. Then the boundary value problem

$$ \textstyle\begin{cases} u{'''}(t)+h(t)=0, \quad t\in (0,1), \\ u(0)=u{''}(0)=0, \\ u(1)+\sum_{i=1}^{m}a_{i}u'(\xi_{i})=\sum_{i=1}^{m}b_{i}\int_{0}^{\xi_{i}}u(s)\,ds \end{cases} $$

(3.2)

has an integral representation

$$ u(t)= \int_{0}^{1}G_{0}(t,s)h(s)\,ds+Q(h)\psi (t), $$

(3.3)

where

$$\begin{aligned}& G_{0}(t,s)=\frac{1}{2} \textstyle\begin{cases} \psi (t) (1-s)^{2}-(t-s)^{2},&0\leq s\leq t\leq 1, \\ \psi (t) (1-s)^{2},&0\leq t\leq s\leq 1, \end{cases}\displaystyle \end{aligned}$$

(3.4)

$$\begin{aligned}& Q(h)=\sum_{i=1}^{m}a_{i} \int_{0}^{\xi_{i}}(\xi_{i}-s)h(s)\,ds-\sum _{i=1} ^{m}\frac{b_{i}}{6} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{3}h(s) \,ds. \end{aligned}$$

(3.5)

The proof is similar to that of Lemma 2.2, so we omit it. Moreover, one has

$$ u'(t)= \int_{0}^{1}g(t,s)h(s)\,ds+\frac{Q(h)}{\Delta_{2}}, $$

(3.6)

where

$$ g(t,s)= \textstyle\begin{cases} \frac{1}{2\Delta_{2}}(1-s)^{2}-(t-s),&0\leq s\leq t\leq 1, \\ \frac{1}{2\Delta_{2}}(1-s)^{2},&0\leq t\leq s\leq 1. \end{cases} $$

(3.7)

Now, we will provide some properties of $G_{0}(t,s)$, $Q(h)$ and $g(t,s)$.

Lemma 3.2

For $(t,s)\in [0,1]\times [0,1]$, the functions $G_{0}(t,s)$, $Q(h)$ and $g(t,s)$ satisfy the following properties:

• (a) $0\leq \frac{t}{2\Delta_{2}}(1-s)^{2}(1-\Delta_{2}t) \leq G_{0}(t,s)\leq \frac{t}{2\Delta_{2}}(1-s)^{2}+t^{2}(1-s)$;

• (b) $0\leq \underline{Q}(h)\leq Q(h)\leq \overline{Q}(h)$, for $h(t)\geq 0$, where

  $$\begin{aligned}& \underline{Q}(h)=\sum_{i=1}^{m} \frac{b_{i}}{6} \int_{0}^{\xi_{i}}(\xi _{i}-s) \bigl[1-( \xi_{i}-s)^{2} \bigr]h(s)\,ds, \\& \overline{Q}(h)=\sum_{i=1}^{m}a_{i} \int_{0}^{\xi_{i}}(\xi_{i}-s) \bigl[1+( \xi_{i}-s)^{2} \bigr]h(s)\,ds; \end{aligned}$$

• (c) $\vert g(t,s) \vert \leq \frac{(1-s)^{2}}{2\Delta_{2}}+(1-s)$.

Proof

(a) For $0\leq s\leq t\leq 1$, we have

$$\begin{aligned} G_{0}(t,s) &=\frac{1}{2} \bigl[\psi (t) (1-s)^{2}-(t-s)^{2} \bigr] \\ &\geq \frac{1}{2} \biggl[\frac{t}{\Delta_{2}}(1-s)^{2}-(t-ts)^{2} \biggr] \\ &=\frac{t}{2\Delta_{2}}(1-s)^{2}(1-\Delta_{2}t) \geq 0, \end{aligned}$$

while, for $0\leq t\leq s\leq 1$, we have

$$\begin{aligned} G_{0}(t,s) &=\frac{1}{2}\psi (t) (1-s)^{2} \\ &\geq \frac{1}{2} \biggl[\frac{t}{\Delta_{2}}(1-s)^{2}-(t-ts)^{2} \biggr] \\ &=\frac{t}{2\Delta_{2}}(1-s)^{2}(1-\Delta_{2}t). \end{aligned}$$

On the other hand, for $0\leq s\leq t\leq 1$, we have

$$\begin{aligned} G_{0}(t,s) &=\frac{1}{2} \bigl[\psi (t) (1-s)^{2}-(t-s)^{2} \bigr] \\ &\leq \frac{1}{2} \biggl[\frac{t}{\Delta_{2}}(1-s)^{2}+t^{2}(1-s) \biggr] \\ &\leq \frac{t}{2\Delta_{2}}(1-s)^{2}+t^{2}(1-s), \end{aligned}$$

while, for $0\leq t\leq s\leq 1$, we have

$$\begin{aligned} G_{0}(t,s) &=\frac{1}{2}\psi (t) (1-s)^{2}\leq \frac{t}{2\Delta_{2}}(1-s)^{2}+t ^{2}(1-s). \end{aligned}$$

(b) From (3.5) and (H1), for $h(t)\geq 0$, we have

$$\begin{aligned} Q(h) &=\sum_{i=1}^{m}a_{i} \int_{0}^{\xi_{i}}(\xi_{i}-s)h(s)\,ds-\sum _{i=1}^{m}\frac{b_{i}}{6} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{3}h(s) \,ds \\ &\geq \sum_{i=1}^{m}\frac{b_{i}}{6} \int_{0}^{\xi_{i}}(\xi_{i}-s)h(s)\,ds- \sum _{i=1}^{m}\frac{b_{i}}{6} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{3}h(s) \,ds \\ &=\sum_{i=1}^{m}\frac{b_{i}}{6} \int_{0}^{\xi_{i}}(\xi_{i}-s) \bigl[1-(\xi _{i}-s)^{2} \bigr]h(s)\,ds \\ &=\underline{Q}(h)\geq 0. \end{aligned}$$

For $h(t)\geq 0$, we also have

$$\begin{aligned} Q(h) &=\sum_{i=1}^{m}a_{i} \int_{0}^{\xi_{i}}(\xi_{i}-s)h(s)\,ds-\sum _{i=1}^{m}\frac{b_{i}}{6} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{3}h(s) \,ds \\ &\leq \sum_{i=1}^{m}a_{i} \int_{0}^{\xi_{i}}(\xi_{i}-s)h(s)\,ds+\sum _{i=1}^{m}a_{i} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{3}h(s) \,ds \\ &=\sum_{i=1}^{m}a_{i} \int_{0}^{\xi_{i}}(\xi_{i}-s) \bigl[1+( \xi_{i}-s)^{2} \bigr]h(s)\,ds \\ &=\overline{Q}(h). \end{aligned}$$

(c) From (3.7), for $0\leq s\leq t\leq 1$, we have

$$\begin{aligned} \bigl\vert g(t,s) \bigr\vert &= \biggl\vert \frac{1}{2\Delta_{2}}(1-s)^{2}-(t-s) \biggr\vert \leq \frac{(1-s)^{2}}{2\Delta_{2}}+(1-s), \end{aligned}$$

and, for $0\leq t\leq s\leq 1$, we have

$$\begin{aligned} \bigl\vert g(t,s) \bigr\vert &= \biggl\vert \frac{1}{2\Delta_{2}}(1-s)^{2} \biggr\vert \leq \frac{(1-s)^{2}}{2\Delta_{2}}+(1-s). \end{aligned}$$

This completes the proof of the lemma. □

In this section, we introduce the Banach space $E_{2} = C^{2}[0,1]$ equipped with the norm

$$\begin{aligned} \Vert u \Vert :=\max \Bigl\{ \max_{0\leq t \leq 1} \bigl\vert u(t) \bigr\vert , \max_{0\leq t \leq 1} \bigl\vert u'(t) \bigr\vert , \max_{0\leq t \leq 1} \bigl\vert u{''}(t) \bigr\vert \Bigr\} \end{aligned}$$

and define a cone $P\subset E_{2}$ by $P = \{u\in E_{2}: u(t)\geq 0, u{''}(t)\leq 0 \}$. Then, for all $u\in E_{2}$, we define an integral operator $T_{2}:P \rightarrow E_{2}$ by

$$\begin{aligned} (T_{2}u) (t)= \int_{0}^{1}G_{0}(t,s)q(s)f_{u}(s) \,ds+Q \bigl[q(s)f_{u}(s) \bigr] \psi (t), \end{aligned}$$

where $f_{u}(s)=f(s,u(s),u'(s),u{''}(s))$ and

$$\begin{aligned} Q \bigl[q(s)f_{u}(s) \bigr]=\sum_{i=1}^{m}a_{i} \int_{0}^{\xi_{i}}(\xi_{i}-s)q(s)f _{u}(s)\,ds-\sum_{i=1}^{m} \frac{b_{i}}{6} \int_{0}^{\xi_{i}}(\xi_{i}-s)^{3}q(s)f _{u}(s)\,ds. \end{aligned}$$

Lemma 3.3

If (H1) and (H2) are satisfied, $T_{2}:P \rightarrow P$ is completely continuous.

Proof

After introducing the operator $T_{2}$, for $u\in P$, we can get that

$$ \begin{aligned} &(T_{2}u)'(t)= \int_{0}^{1}g(t,s)q(s)f_{u}(s)\,ds+ \frac{Q[q(s)f_{u}(s)]}{ \Delta_{2}}, \qquad (T_{2}u)''(t) \leq 0, \\ &(T_{2}u) (0)=0, \qquad (T_{2}u) (1)= \int_{0}^{1}G_{0}(1,s)q(s)f_{u}(s) \,ds+ \frac{Q[q(s)f _{u}(s)]}{\Delta_{2}} \geq 0. \end{aligned} $$

(3.8)

Thus, $(T_{2}u)(t)$ is concave and $(T_{2}u)(t)\geq 0$, for $0\leq t \leq 1$, which implies that operator $T_{2}$ maps P into P.

It is obvious that $T_{2}$ is continuous, but we need to prove that $T_{2}$ is also compact. Let $\Omega \subset P$ be a bounded set. Similar to Lemma 2.5, we can easily prove that $T_{2}(\Omega )$ is bounded and equicontinuous. Thus, by the Arzelà–Ascoli Theorem, $T_{2}(\Omega )$ is relatively compact, which implies $T_{2}$ is compact. Consequently, we get that $T_{2}:P \rightarrow P$ is completely continuous. □

For convenience, we denote

$$ \begin{aligned} &L_{1}= \int_{0}^{1}(1-s)^{2}q(s)\,ds,\qquad L_{2}= \int_{0}^{1}(1-s)q(s)\,ds,\qquad L _{3}= \int_{0}^{1}q(s)\,ds, \\ &L_{Q}=\sum_{i=1}^{m}a_{i} \int_{0}^{\xi_{i}}(\xi_{i}-s) \bigl[1+( \xi_{i}-s)^{2} \bigr]q(s)\,ds. \end{aligned} $$

(3.9)

According to (H1) and (H2), it is obviously that $L_{1}$, $L_{2}$, $L _{3}$ and $L_{Q}$ are nonnegative.

Now, based on Lemmas 3.2 and 3.3, in what follows, we show that there exist positive extremal solutions for BVP (1.2) by a monotone iterative method.

Theorem 3.4

Assume that (H1) and (H2) hold, let $l_{1}$ and l be two positive numbers, satisfying $l=\max \{\frac{1}{2\Delta_{2}}L_{1}+L_{2}+\frac{1}{ \Delta_{2}}L_{Q}+L_{3}, \frac{L_{3}}{\Delta_{2}} \}l_{1}$, and

• (S1) $f(t,u_{1},v_{1},w_{1})\leq f(t,u_{2},v_{2},w_{2})$, for $0\leq t \leq 1, 0\leq u_{1} \leq u_{2} \leq l, 0\leq \vert v_{1} \vert \leq \vert v_{2} \vert \leq l, -l\leq w_{2} \leq w_{1} \leq 0$;

• (S2) $\max_{0\leq t \leq 1}f(t,l,l,-l)\leq l_{1}$;

• (S3) $f(t,0,0,0)\not \equiv 0$, for $0\leq t \leq 1$.

Then BVP (1.2) has concave positive solutions $v^{*}$ and $\omega^{*}$, which satisfy

$$\begin{aligned} &0\leq \bigl\Vert v^{*} \bigr\Vert \leq l,\qquad 0 \leq \bigl\Vert \omega^{*} \bigr\Vert \leq l, \\ &v_{0}(t)=0,\qquad \omega_{0}(t)= \biggl[ \biggl( \frac{1}{2\Delta_{2}}L _{1}+L_{2}+\frac{L_{Q}}{\Delta_{2}} \biggr)t+ \biggl(t-\frac{t^{2}}{2} \biggr)L_{3} \biggr]l_{1}, \\ &v_{n} = T_{2}v_{n-1},\qquad\lim _{n\rightarrow \infty }v_{n}=v^{*}, \qquad \omega_{n} = T_{2}\omega_{n-1},\qquad\lim _{n\rightarrow \infty } \omega_{n}=\omega^{*},\quad n=1,2, \dots , \\ &v_{n}' = (T_{2}v_{n-1})', \qquad\lim_{n\rightarrow \infty }(v_{n})'= \bigl(v ^{*} \bigr)', \qquad \omega_{n}' = (T_{2}\omega_{n-1})',\qquad\lim _{n\rightarrow \infty }(\omega_{n})'= \bigl( \omega^{*} \bigr)', \\ &v_{n}'' = (T_{2}v_{n-1})'', \qquad\lim_{n\rightarrow \infty }(v_{n})''= \bigl(v ^{*} \bigr)'', \qquad \omega_{n}'' = (T_{2} \omega_{n-1})'',\qquad\lim_{n\rightarrow \infty }( \omega_{n})'= \bigl(\omega^{*} \bigr)''. \end{aligned}$$

Proof

Denote $P_{l}= \{u\in P\mid \Vert u \Vert \leq l \}$. In the following, we first prove that $T_{2}:P_{l}\rightarrow P_{l}$. Let $u\in P_{l}$. Then for $t\in [0,1]$, we have

$$\begin{aligned} &0\leq u(t) \leq \Vert u \Vert \leq l,\qquad0\leq \bigl\vert u'(t) \bigr\vert \leq \Vert u \Vert \leq l,\qquad-l \leq - \Vert u \Vert \leq u{''}(t) \leq 0. \end{aligned}$$

So, for $0\leq t \leq 1$, by (S1) and (S2), we get

$$\begin{aligned} 0\leq f \bigl(t, u(t), u'(t), u{''}(t) \bigr) \leq \max _{0\leq t \leq 1}f(t,l,l,-l) \leq l_{1}. \end{aligned}$$

Consequently, for $t\in [0,1]$, we have

$$\begin{aligned}& \bigl\vert (T_{2}u) (t) \bigr\vert = \biggl\vert \int_{0}^{1}G_{0}(t,s)q(s)f_{u}(s) \,ds+ \frac{t}{\Delta_{2}}Q \bigl[q(s)f_{u}(s) \bigr] \biggr\vert \\& \hphantom{ \vert (T_{2}u) (t) \vert }\leq \int_{0}^{1} \biggl[\frac{t}{2\Delta_{2}}(1-s)^{2}+t^{2}(1-s) \biggr]q(s)f_{u}(s)\,ds+\frac{\overline{Q}[q(s)f_{u}(s)]}{\Delta_{2}} \\& \hphantom{ \vert (T_{2}u)(t) \vert } \leq \biggl( \frac{1}{2\Delta_{2}}L_{1}+L_{2}+ \frac{L_{Q}}{\Delta_{2}} \biggr)l_{1} \leq l, \\& \bigl\vert (T_{2}u)'(t) \bigr\vert = \biggl\vert \int_{0}^{1}g(t,s)q(s)f_{u}(s)\,ds+ \frac{Q[q(s)f_{u}(s)]}{\Delta_{2}} \biggr\vert \\& \hphantom{ \vert (T_{2}u)'(t) \vert } \leq \int_{0}^{1}\frac{1}{2\Delta_{2}}(1-s)^{2}q(s)f_{u}(s) \,ds+ \int _{0}^{1}(1-s)q(s)f_{u}(s)\,ds+ \frac{\overline{Q}[q(s)f_{u}(s)]}{\Delta _{2}} \\& \hphantom{ \vert (T_{2}u)'(t) \vert }\leq \biggl( \frac{1}{2\Delta_{2}}L_{1}+L_{2}+ \frac{L_{Q}}{\Delta_{2}} \biggr)l_{1} \leq l, \\& \bigl\vert (T_{2}u){''}(t) \bigr\vert \leq \biggl\vert \int_{0}^{1}q(s)f_{u}(s)\,ds \biggr\vert \leq L _{3}l_{1} \leq l. \end{aligned}$$

To sum up, we obtain

$$\begin{aligned} \Vert T_{2}u \Vert &= \Bigl\{ \max_{0\leq t \leq 1} \bigl\vert (T_{2}u) (t) \bigr\vert , \max_{0\leq t \leq 1} \bigl\vert (T_{2}u)'(t) \bigr\vert , \max _{0\leq t \leq 1} \bigl\vert (T_{2}u){''}(t) \bigr\vert \Bigr\} \leq l \end{aligned}$$

and $T_{2}:P_{l}\rightarrow P_{l}$.

Set $\omega_{0}= [ \frac{t}{\Delta_{2}}(\frac{1}{2}L_{1}+L_{Q}+ \Delta_{2}L_{2})+(t-\frac{t^{2}}{2})L_{3} ]l_{1}$ and $v_{0}=0$. Obviously, $\omega_{0}$, $v_{0}\in P_{l}$. By using the completely continuous operator $T_{2}$, we define the sequences $\{\omega_{n}\}$ and $\{v_{n}\}$ as $\omega_{n}=T_{2}\omega_{n-1}$, $v_{n}=T_{2}v_{n-1}$, for $n=1,2,\dots$ . Since $T_{2}:P_{l}\rightarrow P_{l}$, we get that $\omega_{n}, v_{n} \in P_{l}$, $n=1,2,\dots$ . Also we assert that $\{\omega_{n}\}$ and $\{v_{n}\}$ have relatively compact subsequences, for $n=0,1,2,\dots$ . Hence, we prove that there exist $\omega^{*}$, $v ^{*}$, satisfying $\lim_{n\rightarrow \infty }\omega_{n}=\omega^{*}$ and $\lim_{n\rightarrow \infty }v_{n}=v^{*}$, which are monotone positive solutions of problem (1.2).

For $t\in [0,1]$, according to the definition of the iterative scheme, we have

$$\begin{aligned} \omega_{1}(t)&= T_{2}\omega_{0}(t) \\ &= \int_{0}^{1}G_{0}(t,s)q(s)f_{ \omega_{0}}(s) \,ds+ \frac{Q[q(s)f_{\omega_{0}}(s)]t}{\Delta_{2}} \\ &\leq \int_{0}^{1} \biggl[ \frac{t}{2\Delta_{2}}(1-s)^{2}+t^{2}(1-s) \biggr]q(s)f_{\omega_{0}}(s)\,ds+\frac{\overline{Q}[q(s)f_{\omega_{0}}(s)]t}{ \Delta_{2}} \\ &\leq \biggl[ \biggl(\frac{1}{2\Delta_{2}}L_{1}+tL_{2} \biggr)t+\frac{L_{Q}t}{ \Delta_{2}} \biggr]l_{1} \\ &\leq \biggl[ \biggl(\frac{1}{2\Delta_{2}}L_{1}+L_{2}+ \frac{L_{Q}}{\Delta _{2}} \biggr)t+ \biggl(t-\frac{t^{2}}{2} \biggr)L_{3} \biggr]l_{1} = \omega_{0}(t); \end{aligned}$$

$$\begin{aligned}& \bigl\vert \omega_{1}'(t) \bigr\vert = \bigl\vert (T_{2}\omega_{0})'(t) \bigr\vert = \biggl\vert \int_{0}^{1}g(t,s)q(s)f_{\omega_{0}}(s)\,ds+ \frac{Q[q(s)f_{\omega_{0}}(s)]}{\Delta_{2}} \biggr\vert \\& \hphantom{ \vert \omega_{1}'(t) \vert } \leq \int_{0}^{1} \biggl[ \frac{1}{2\Delta_{2}}(1-s)^{2}+(1-s) \biggr]q(s)f _{\omega_{0}}(s)\,ds+\frac{\overline{Q}[q(s)f_{\omega_{0}}(s)]}{\Delta _{2}} \\& \hphantom{ \vert \omega_{1}'(t) \vert }\leq \biggl(\frac{1}{2\Delta_{2}}L_{1}+L_{2}+ \frac{L_{Q}}{\Delta_{2}} \biggr)l_{1} \\& \hphantom{ \vert \omega_{1}'(t) \vert } \leq \biggl[ \biggl(\frac{1}{2\Delta_{2}}L_{1}+L_{2}+ \frac{L_{Q}}{ \Delta_{2}} \biggr)+(1-t)L_{3} \biggr]l_{1} = \bigl\vert \omega_{0}'(t) \bigr\vert ; \\& \omega_{1}{''}(t)= (T_{2}\omega_{0}){''}(t) =- \int_{0}^{t}q(s)f _{\omega_{0}}(s)\,ds \geq -L_{3}l_{1}= \omega_{0}{''}(t). \end{aligned}$$

Thus, for $0\leq t \leq 1$, we have

$$\begin{aligned}& \omega_{2}(t) = T_{2}\omega_{1}(t)\leq T_{2}\omega_{0}(t)=\omega_{1}(t), \\& \bigl\vert \omega_{2}'(t) \bigr\vert = \bigl\vert (T_{2}\omega_{1})'(t) \bigr\vert \leq \bigl\vert (T_{2}\omega_{0})'(t) \bigr\vert = \bigl\vert \omega_{1}'(t) \bigr\vert , \\& \omega_{2}{''}(t)= (T_{2}\omega_{1}){''}(t) \geq (T_{2}\omega_{0}){''}(t)= \omega_{1}{''}(t). \end{aligned}$$

By induction, one has

$$\begin{aligned} &\omega_{n+1}(t)\leq \omega_{n}(t),\qquad \bigl\vert \omega ' _{n+1}(t) \bigr\vert \leq \bigl\vert \omega '_{n}(t) \bigr\vert , \\ &\omega_{n+1}{''}(t)\geq \omega_{n}{''}(t),\quad 0\leq t \leq 1, n=0,1,2,\dots\,. \end{aligned}$$

So, we assert that $\omega_{n}\rightarrow \omega^{*}$ and $T_{2}\omega ^{*}=\omega^{*}$, since $T_{2}$ is completely continuous and $\omega_{n+1}=T_{2}\omega_{n}$.

For the sequence $\{v_{n} \}_{n=1}^{\infty }$, we apply a similar argument. For $t\in [0,1]$, we have

$$\begin{aligned}& v_{1}(t)= T_{2}v_{0}(t) = \int_{0}^{1}G_{0}(t,s)q(s)f_{v_{0}}(s) \,ds+\frac{Q[q(s)f_{v_{0}}(s)]t}{\Delta_{2}} \\& \hphantom{v_{1}(t)}\geq \int_{0}^{1} \biggl[ \frac{t}{2}(1-s)^{2} \biggl( \frac{1}{\Delta_{2}}-t \biggr) \biggr]q(s)f_{v_{0}}(s)\,ds+ \frac{ \underline{Q}[q(s)f_{v_{0}}(s)]t}{\Delta_{2}} \\& \hphantom{v_{1}(t)} \geq 0 = v_{0}(t); \\& \bigl\vert v_{1}'(t) \bigr\vert = \bigl\vert (T_{2}v_{0})'(t) \bigr\vert = \biggl\vert \int_{0}^{1}g(t,s)q(s)f_{v_{0}}(s)\,ds+ \frac{Q[q(s)f_{v_{0}}(s)]}{\Delta_{2}} \biggr\vert \geq 0 = \bigl\vert v_{0}'(t)\bigr\vert ; \\& v_{1}{''}(t)= (T_{2}v_{0}){''}(t)=-\int_{0}^{1}q(s)f_{v_{0}}(s)\,ds \leq 0=v_{0}{''}(t). \end{aligned}$$

So, for $0\leq t \leq 1$, we have

$$\begin{aligned} &v_{2}(t) = T_{2}v_{1}(t)\geq T_{2}v_{0}(t)=v_{1}(t),\quad \text{while } v_{1}(t) = T_{2}v_{0}(t)\geq 0, \\ & \bigl\vert v_{2}'(t) \bigr\vert = \bigl\vert (T_{2}v_{1})'(t) \bigr\vert \geq \bigl\vert (T_{2}v_{0})'(t)\bigr\vert = \bigl\vert v_{1}'(t) \bigr\vert ,\quad \text{while } \bigl\vert v_{1}'(t) \bigr\vert = \bigl\vert (T_{2}v_{0})'(t)\bigr\vert \geq 0, \\ &v_{2}{''}(t)= (T_{2}v_{1}){''}(t)\leq (T_{2}v_{0}){''}(t)=v_{1}{''}(t),\quad \text{while } v_{1}{''}(t)=(T_{2}v_{0}){''}(t) \leq 0. \end{aligned}$$

Similarly, one has

$$\begin{aligned} v_{n+1}(t)\geq v_{n}(t) , \qquad \bigl\vert v'_{n+1}(t) \bigr\vert \geq \bigl\vert v'_{n}(t) \bigr\vert , \qquad v_{n+1}{''}(t) \leq v_{n}{''}(t) , \quad 0\leq t \leq 1 , n=0,1,2,\dots \end{aligned}$$

And, we assert that $v_{n}\rightarrow v^{*}$ and $T_{2}v^{*}=v^{*}$, since $T_{2}$ is completely continuous and $v_{n+1}=T_{2}v_{n}$.

Consequently, there exist $\omega^{*}$ and $v^{*}$ in $P_{l}$, which are nonnegative solutions of BVP (1.2). From (S3), it is obvious that $\omega^{*}(t)>0, v^{*}(t)>0$, for $t\in [0,1]$, since zero is not a solution of Eq. (1.2). The proof is completed. □

Example 3.1

Consider the following third-order boundary value problem:

$$ \textstyle\begin{cases} u{'''}(t)+ (\frac{1}{2}t^{2}+\frac{1}{10}u(t)+\frac{1}{50}u^{\prime\,2}(t)+ \sin \frac{2}{u''(t)-2}+\frac{11}{10} )=0, \quad t\in (0,1), \\ u(0)=u{''}(0)=0, \qquad u(1)+\sum_{i=1}^{m}a_{i}u'( \xi_{i})=\sum_{i=1}^{m}b_{i} \int_{0}^{\xi_{i}}u(s)\,ds. \end{cases} $$

(3.10)

In this model, we set

$$\begin{aligned}& m=2,\qquad q(s)=1,\qquad \xi_{1}=\frac{3}{5},\qquad \xi_{2}=\frac{4}{5}, \\& a_{1}=\frac{3}{7},\qquad a_{2}=\frac{2}{7}, \quad b_{1}=\frac{5}{2}, \qquad b_{2}= \frac{5}{3}. \end{aligned}$$

It is obvious that (H1) and (H2) hold. By calculation, we get

$$\begin{aligned} &E=\frac{5}{7},\qquad F=\frac{295}{300},\qquad \Delta_{2}=0.731, \\ &L_{1}=\frac{1}{3},\qquad L_{2}= \frac{1}{2},\qquad L_{3}=1,\qquad L_{Q}=0.2117, \end{aligned}$$

and $l=\max \{\frac{1}{2\Delta_{2}}L_{1}+L_{2}+\frac{1}{\Delta _{2}}L_{Q}+L_{3}, \frac{L_{3}}{\Delta_{2}} \}l_{1} \approx 2.0176l _{1}$. Set $l_{1}=5$. Then all the hypotheses of Theorem 3.4 are satisfied with $l=10$. Hence, BVP (3.10) has monotone positive solutions $v^{*}$ and $\omega^{*}$, which satisfy, for $t\in [0,1]$,

$$\begin{aligned} & v_{0}(t)=0\quad \text{and} \quad \omega_{0}(t)= \biggl[ \biggl( \frac{1}{2\Delta _{2}}L_{1}+L_{2}+\frac{L_{Q}}{\Delta_{2}} \biggr)t+ \biggl(t- \frac{t^{2}}{2} \biggr)L_{3} \biggr]l_{1}=- \frac{5}{2}t^{2}+10t. \end{aligned}$$

For $n=1,2,\dots$, the two iterative schemes are

$$\begin{aligned}& \omega_{0}(t) =-\frac{5}{2}t^{2}+10t, \quad \quad \omega_{1}(t)=-\frac{1}{80}t^{5}+ \frac{1}{24}t^{4}-\frac{4697}{10\text{,}000}t ^{3}+ \frac{383}{500}t,\qquad \dots , \\& \omega_{n+1}(t) = \int_{0}^{t}G_{0}(t,s)f_{\omega_{n}}(s) \,ds+ \frac{Q[f _{\omega_{n}(s)}]t}{\Delta_{2}} \\& \hphantom{\omega_{n+1}(t)} =-\frac{1}{2} \int_{0}^{t}(t-s)^{2} \biggl( \frac{1}{2}s^{2}+ \frac{1}{10}\omega_{n}(s)+ \frac{1}{50}\omega_{n}^{\prime\,2}(s)+\sin \frac{2}{ \omega_{n}''(s)-2}+\frac{11}{10} \biggr)\,ds \\& \hphantom{\omega_{n+1}(t)=}{}+\frac{t}{2\Delta_{2}} \int_{0}^{1}(1-s)^{2} \biggl( \frac{1}{2}s^{2}+ \frac{1}{10}\omega_{n}(s)+ \frac{1}{50}\omega_{n}^{\prime\,2}(s)+\sin \frac{2}{ \omega_{n}''(s)-2}+\frac{11}{10} \biggr)\,ds \\& \hphantom{\omega_{n+1}(t)=} {}+\frac{t}{\Delta_{2}} \biggl\{ \int_{0}^{\frac{3}{5}} \biggl( \frac{5}{12}s^{3}- \frac{3}{4}s^{2}+\frac{3}{140}s+\frac{117}{700} \biggr) \biggl( \frac{1}{2}s^{2}+\frac{1}{10} \omega_{n}(s)+\frac{1}{50} \omega_{n}^{\prime\,2}(s) \\& \hphantom{\omega_{n+1}(t)=}{}+\sin \frac{2}{\omega_{n}''(s)-2}+\frac{11}{10} \biggr)\,ds \\& \hphantom{\omega_{n+1}(t)=}{}+ \int_{0}^{\frac{4}{5}} \biggl(\frac{5}{18}s^{3}- \frac{2}{3}s^{2}+ \frac{26}{105}s+\frac{136}{1575} \biggr) \biggl( \frac{1}{2}s^{2}+ \frac{1}{10} \omega_{n}(s)+\frac{1}{50}\omega_{1}^{\prime\,2}(s) \\& \hphantom{\omega_{n+1}(t)=}{}+ \sin \frac{2}{ \omega_{n}''(s)-2}+\frac{11}{10} \biggr)\,ds \biggr\} \end{aligned}$$

and

$$\begin{aligned}& v_{0}(t) =0, \quad \quad v_{1}(t)=-\frac{1}{120}t^{5}- \frac{53}{1000}t^{3}+\frac{9}{100}t, \qquad \dots , \\& v_{n+1}(t) = \int_{0}^{t}G_{0}(t,s)f_{v_{n}}(s) \,ds+ \frac{Q[f_{v_{n}(s)}]t}{ \Delta_{2}} \\& \hphantom{v_{n+1}(t)}=-\frac{1}{2} \int_{0}^{t}(t-s)^{2} \biggl( \frac{1}{2}s^{2}+ \frac{1}{10}v_{n}(s)+ \frac{1}{50}v_{n}^{\prime\,2}(s)+\sin \frac{2}{v_{n}''(s)-2}+ \frac{11}{10} \biggr)\,ds \\& \hphantom{v_{n+1}(t)=}{}+\frac{t}{2\Delta_{2}} \int_{0}^{1}(1-s)^{2} \biggl( \frac{1}{2}s^{2}+ \frac{1}{10}v_{n}(s)+ \frac{1}{50}v_{n}^{\prime\,2}(s)+\sin \frac{2}{v_{n}''(s)-2}+ \frac{11}{10} \biggr)\,ds \\& \hphantom{v_{n+1}(t)=}{}+\frac{t}{\Delta_{2}} \biggl\{ \int_{0}^{\frac{3}{5}} \biggl( \frac{5}{12}s^{3}- \frac{3}{4}s^{2}+\frac{3}{140}s+\frac{117}{700} \biggr) \biggl( \frac{1}{2}s^{2}+\frac{1}{10}v_{n}(s) \\& \hphantom{v_{n+1}(t)=} +\frac{1}{50}v_{n}^{\prime\,2}(s)+ \sin \frac{2}{v_{n}''(s)-2}+ \frac{11}{10} \biggr)\,ds \\& \hphantom{v_{n+1}(t)=}{}+ \int_{0}^{\frac{4}{5}} \biggl(\frac{5}{18}s^{3}- \frac{2}{3}s^{2}+ \frac{26}{105}s+\frac{136}{1575} \biggr) \biggl( \frac{1}{2}s^{2}+ \frac{1}{10}v_{n}(s) \\& \hphantom{v_{n+1}(t)=}{}+\frac{1}{50}v_{n}^{\prime\,2}(s)+\sin \frac{2}{v_{n}''(s)-2}+ \frac{11}{10} \biggr)\,ds \biggr\} . \end{aligned}$$