Positive solutions for integral boundary value problems of fractional differential equations with delay

Abstract

In this article, a class of integral boundary value problems of fractional delayed differential equations is discussed. Based on the Guo–Krasnoselskii theorem, some existence results on the positive solutions are derived. Two simple examples are given to show the validity of the conditions of our main theorems.

1 Introduction

Differential equation models have been widely used in control system, aerodynamics, fluid flows and many other branches of engineering [1–9]. Recently, fractional calculus has attracted great interest. There are several kinds of fractional operators have been proposed so far, among which we have the well-known Grünwald–Letnikov, Riemann–Liouville, Caputo derivative.

In the past several decades, fractional boundary value problems have obtained abundant theoretical achievements. In [10, 11], the nonexistence of positive solutions for differential equations of fractional order is analyzed with the help of reduction to absurdity. In [12], the authors study a class of Riemann–Liouville fractional derivative equations and present sufficient condition on the unique positive solution by employing a $u_{0}$-positive operator. There are many articles devoted to the existence and multiplicity of positive solutions for the fractional boundary value problems, the approaches mainly include Leray–Schauder degree theory [13, 14], the monotone iterative method [15, 16], the Leggett–Williams theorem [17, 18], the fixed point theorem on cones [17–19]. Especially, compared with the previous results, papers [11–13, 17, 20] contain integral boundary conditions. In [21], the authors first introduced a new method, called Avery–Peterson theory, which illustrates the existence of at least three positive solutions. Since then, more and more attention [22–24] was paid to this method, and many questions were solved.

At present, fractional delayed equations have aroused the extensive attention of many scholars. They dealt with the existence of solutions under various boundary conditions by different methods. For details, one can refer to [20, 25–27] and the references therein.

In [28], Cabada and Wang considered a class of nonlinear fractional differential equations with integral boundary value conditions:

$$\begin{aligned} \textstyle\begin{cases} {}^{c}D^{\alpha }u(t)+f(t,u(t))=0, \quad 0< t< 1, \\ u(0)=u^{\prime \prime }(0)=0, \quad \quad u(1)=\lambda \int _{0}^{1}u(s)\,ds, \end{cases}\displaystyle \end{aligned}$$

where $2<\alpha <3$, $0<\lambda <2$, $f:[0,1]\times [0,\infty )\rightarrow [0,\infty )$ is a continuous function. The existence of at least one positive solution is obtained by using the Guo–Krasnoselskii fixed point theorem.

Enlightened by the above literature, we discuss the following equation:

$$\begin{aligned} \textstyle\begin{cases} {}^{c}D^{\beta }z(t)+g(t,z_{t})=0, \quad t\in [0,1], \\ z(t)=\phi (t),\quad t\in [-\tau ,0], \\ z(0)=z^{\prime \prime }(0)=z^{\prime \prime \prime }(0)=0, \quad\quad z(1)=k\int _{0}^{1}z(\theta )\,d\theta , \end{cases}\displaystyle \end{aligned}$$

(1)

where $3<\beta \leq 4$, $0< k<2$, ${}^{c}D^{\beta }$ is the Caputo fractional derivative, $g:[0,1]\times C_{\tau }\longrightarrow [0,+\infty )$ is a continuous function, $z_{t}(s)=z(t+s)$, for $t\in [0,1]$, $s\in [-\tau ,0]$. $\phi \in C_{\tau }(:=C[-\tau ,0])$, $C_{\tau }$ is a Banach space with $\Vert \phi \Vert _{[-\tau ,0]}=\max_{s \in [-\tau ,0]} \vert \phi (s) \vert$ and let $C_{\tau }^{+}=\{z\in C[-\tau ,0]\vert z(t)\geq 0, t\in [-\tau ,0]\}$.

In this paper, the sufficient conditions are obtained for the existence of at least two positive solutions for a class of integral boundary value problems of fractional differential equations with delay. The integral boundary value condition and the time delay make the results significant.

2 Preliminaries

This part introduce some useful definitions and important lemmas.

Definition 2.1

([1])

The β order fractional integral for a function $g(t)$ is defined as follows:

$$ I^{\beta }g(t)=\frac{1}{\varGamma (\beta )} \int ^{t}_{0}(t-\theta )^{ \beta -1}g(\theta )\,d \theta ,\quad \beta >0. $$

Definition 2.2

([1])

The β order Caputo fractional derivative for a function $g(t)$ is defined as follows:

$$ {}^{c}D^{\beta }g(t)=\frac{1}{\varGamma (n-\beta )} \int ^{t}_{0}(t- \theta )^{n-\beta -1}g^{(n)}( \theta )\,d\theta ,\quad n-1< \beta \leq n. $$

Definition 2.3

([2])

Let $P \subseteq X$ be a nonempty, convex closed set and X a real Banach space. The P is called a cone in X provided that

• (i) $\mu z\in P$, for all $z\in P$ and $\mu \geq 0$;

• (ii) $z, -z\in P$ imply $z=0$.

Definition 2.4

([2])

Let P is a cone in real Banach space X. If map $\psi : P\rightarrow [0,\infty )$ is continuous and satisfies

$$ \psi \bigl(tz_{1}+(1-t)z_{2}\bigr)\geq t\psi (z_{1})+(1-t)\psi (z_{2}),\quad z_{1},z_{2} \in P, t\in [0,1], $$

then ψ is called a nonnegative continuous concave functional on P.

Lemma 2.1

([1])

Assume that$\beta >0$and$n=[\beta ]+1$. If the function$g\in L[0,1]\cap C[0,1]$, then there exists$c_{i}\in R$, $i=1,2,\ldots,n$, such that

$$ I^{\beta c}D^{\beta }g(t)=g(t)-c_{1}-c_{2}t- \cdots -c_{n}t^{n-1}. $$

Lemma 2.2

Equation (1) has a unique solution as follows:

$$\begin{aligned} z(t)= \textstyle\begin{cases} \int _{0}^{1}G(t,\theta )g(\theta ,z_{\theta })\,d\theta , &t\in [0,1], \\ \phi (t), &t\in [-\tau ,0], \end{cases}\displaystyle \end{aligned}$$

(2)

where

$$\begin{aligned} G(t,\theta )= \textstyle\begin{cases} \frac{2t(1-\theta )^{\beta -1}(\beta -k+k\theta )-(2-k)\beta (t-\theta )^{\beta -1}}{(2-k)\varGamma (\beta +1)}, &0\leq \theta \leq t\leq 1, \\ \frac{2t(1-\theta )^{\beta -1}(\beta -k+k\theta )}{(2-k)\varGamma (\beta +1)}, &0\leq t\leq \theta \leq 1. \end{cases}\displaystyle \end{aligned}$$

(3)

Proof

From Lemma 2.1, we have

$$\begin{aligned} z(t)=-\frac{1}{\varGamma (\beta )} \int ^{t}_{0}(t-\theta )^{\beta -1}g( \theta ,z_{\theta })\,d\theta +c_{0}+c_{1}t+c_{2}t^{2}+c_{3}t^{3}. \end{aligned}$$

According to $z(0)=z^{\prime \prime }(0)=z^{\prime \prime \prime }(0)=0$, $z(1)=k\int _{0}^{1}z(\theta )\,d\theta$, we know that

$$\begin{aligned} c_{0}=c_{2}=c_{3}=0,\quad \quad c_{1}=\frac{1}{\varGamma (\beta )} \int ^{1}_{0}(1- \theta )^{\beta -1}g(\theta ,z_{\theta })\,d\theta +k \int _{0}^{1}z( \theta )\,d\theta . \end{aligned}$$

Thus, Eq. (1) has a unique solution

$$ \begin{aligned}[b] z(t)&=-\frac{1}{\varGamma (\beta )} \int ^{t}_{0}(t-\theta )^{\beta -1}g( \theta ,z_{\theta })\,d\theta \\ &\quad{} +\biggl(\frac{1}{\varGamma (\beta )} \int ^{1}_{0}(1- \theta )^{\beta -1}g(\theta ,z_{\theta })\,d\theta +k \int _{0}^{1}z( \theta )\,d\theta \biggr)t. \end{aligned} $$

(4)

Letting $J=\int _{0}^{1}z(\theta )\,d\theta$, from (4) we get

$$\begin{aligned} J =& \int ^{1}_{0}z(t)\,dt \\ =&- \int _{0}^{1} \int ^{t}_{0} \frac{(t-\theta )^{\beta -1}}{\varGamma (\beta )}g(\theta ,z_{\theta })\,d \theta \,dt \\ &{}+ \int _{0}^{1} \int ^{1}_{0} \frac{t(1-\theta )^{\beta -1}}{\varGamma (\beta )}g(\theta ,z_{\theta })\,d \theta \,dt+ \int ^{1}_{0} kJt \,dt \\ =& - \int ^{1}_{0}\frac{(1-\theta )^{\beta }}{\beta \varGamma (\beta )}g( \theta ,z_{\theta })\,d\theta +\frac{1}{2} \int ^{1}_{0} \frac{(1-\theta )^{\beta -1}}{\varGamma (\beta )}g(\theta ,z_{\theta })\,d \theta +\frac{1}{2}k J. \end{aligned}$$

It follows that

$$ J=-\frac{2}{2-k} \int ^{1}_{0} \frac{(1-\theta )^{\beta }}{\beta \varGamma (\beta )}g(\theta ,z_{ \theta })\,d\theta +\frac{1}{2-k} \int ^{1}_{0} \frac{(1-\theta )^{\beta -1}}{\varGamma (\beta )}g(\theta ,z_{\theta })\,d \theta . $$

(5)

Substituting (5) into (4), we derive

$$\begin{aligned} z(t) =& - \int ^{t}_{0}\frac{(t-\theta )^{\beta -1}}{\varGamma (\beta )}g( \theta ,z_{\theta })\,d\theta + \int _{0}^{1} \frac{t(1-\theta )^{\beta -1}}{\varGamma (\beta )}g(\theta ,z_{\theta })\,d \theta \\ & {} - \frac{2k}{2-k} \int ^{1}_{0} \frac{t(1-\theta )^{\beta }}{\beta \varGamma (\beta )}g(\theta ,z_{ \theta })\,d\theta +\frac{k}{2-k} \int ^{1}_{0} \frac{t(1-\theta )^{\beta -1}}{\varGamma (\beta )}g(\theta ,z_{\theta })\,d \theta \\ =&- \int ^{t}_{0}\frac{(t-\theta )^{\beta -1}}{\varGamma (\beta )}g( \theta ,z_{\theta })\,d\theta + \int _{0}^{1} \frac{2t(1-\theta )^{\beta -1}(\beta -k+k \theta )}{(2-k)\varGamma (\beta +1)}g( \theta ,z_{\theta })\,d\theta \\ =& \int ^{1}_{0}G(t,\theta )g(\theta ,z_{\theta }) \,d\theta . \end{aligned}$$

The conclusion have been proved. □

Lemma 2.3

([28])

The function $G(t,\theta )$ satisfies

• (1) $0< G(t,\theta )\leq \frac{2}{(2-k)\varGamma (\beta )}$for$t,\theta \in (0,1)$if and only if$0< k<2$;

• (2) $tG(1,\theta )\leq G(t,\theta )\leq MG(1,\theta )$, $M= \frac{2\beta }{k(\beta -2)}$for all$t,\theta \in (0,1)$, $3<\beta \leq 4$and$0< k<2$.

Lemma 2.4

([2])

Suppose thatPis a cone in Banach space X. If$\varOmega _{1}$, $\varOmega _{2}$are bounded open sets inXsuch that$0\in \varOmega _{1}$, $\overline{\varOmega _{1}}\subset \varOmega _{2}$and operator$T:P\cap (\overline{\varOmega _{2}}\setminus \varOmega _{1}) \longrightarrow P$is completely continuous satisfying

• (i) $\Vert Tz \Vert \geq \Vert z \Vert$, $z\in P\cap \partial \varOmega _{1}$and$\Vert Tz \Vert \leq \Vert z \Vert$, $z\in P\cap \partial \varOmega _{2}$; or

• (ii) $\Vert Tz \Vert \leq \Vert z \Vert$, $z\in P\cap \partial \varOmega _{1}$and$\Vert Tz \Vert \geq \Vert z \Vert$, $z\in P\cap \partial \varOmega _{2}$,

then the operatorThas at least one fixed point in$P\cap (\overline{\varOmega _{2}}\setminus \varOmega _{1})$.

Lemma 2.5

([29])

Suppose$\sigma \in (0,\frac{1}{2})$is a fixed number, for each$z\in P$and$\theta \in [\sigma ,1-\sigma ]$ (Pis defined in Lemma 3.1), there exists a constant$\lambda \in (0,1)$that satisfies

$$ \Vert z_{\theta } \Vert _{[-\tau ,0]}\geq \lambda \Vert z \Vert _{[0,1]}, \quad\quad \Vert z \Vert _{[0,1]}= \sup _{t\in [0,1]} \bigl\vert z(t) \bigr\vert . $$

3 Main results

Next, the problem of positive solutions for Eq. (1) are studied. For convenience, some notations and hypotheses are presented as follows:

$$\begin{aligned}& g_{\infty }=\lim_{z\in C_{\tau }^{+}, \Vert z \Vert _{[-\tau ,0]}\rightarrow + \infty } \frac{g(t,z)}{ \Vert z \Vert _{[-\tau ,0]}}, \quad\quad g_{0}=\lim_{z\in C_{ \tau }^{+}, \Vert z \Vert _{[-\tau ,0]}\rightarrow 0^{+}} \frac{g(t,z)}{ \Vert z \Vert _{[-\tau ,0]}}, \\& A=\frac{\lambda }{2} \int _{\sigma }^{1-\sigma }G(1,\theta )\,d\theta , \lambda \in (0,1), \sigma \in \biggl(0,\frac{1}{2}\biggr), \qquad B=M \int _{0}^{1}G(1, \theta )\,d\theta ; \end{aligned}$$

$(C_{1})$:

$\phi (t)\geq 0$ on $[-\tau ,0]$;

$(C_{2})$:

$g(t,z)\geq 0$ for $t\in [0,1]$ and $z\in C_{\tau }^{+}$;

$(C_{3})$:

$g_{0}=g_{\infty }=+\infty$;

$(C_{4})$:

$g_{0}=g_{\infty }=0$;

$(C_{5})$:

if there exists a constant $m\geq \Vert \phi \Vert _{[-\tau ,0]}>0$, then

$$ g(t,z)\leq \frac{m}{B}, \quad \Vert z \Vert _{[-\tau ,0]}\in [0,m], t\in [0,1]. $$

$(C_{6})$:

if there exists a constant $n\geq \Vert \phi \Vert _{[-\tau ,0]}>0$, then

$$ g(t,z)\geq \frac{\lambda n}{A}, \quad \Vert z \Vert _{[-\tau ,0]}\in [ \lambda n,n], t\in [\sigma ,1-\sigma ]. $$

On $C[-\tau ,1]$ define an operator T

$$\begin{aligned} Tz(t)= \textstyle\begin{cases} \int _{0}^{1}G(t,\theta )g(\theta ,z_{\theta })\,d\theta ,&t\in [0,1], \\ \phi (t),&t\in [-\tau ,0]. \end{cases}\displaystyle \end{aligned}$$

Lemma 3.1

If$(C_{1})$, $(C_{2})$hold andPis a cone in Banach space$X=C[-\tau ,1]$with norm$\Vert z \Vert _{[-\tau ,1]}=\max_{t\in [-\tau ,1]} \vert z(t) \vert$as follows:

$$ P=\bigl\{ z\in X\vert z\geq 0, z \textit{ is concave down on } [0,1]\bigr\} , $$

then the following conclusions are true.

• (i) $T(P)\subseteq P$;

• (ii) $T:P\rightarrow P$is completely continuous.

Proof

It is easy to check that (i) holds and T is continuous. So we only prove that (ii) is true. Assume that H be a bounded subset in P, which is to say there exists $l>0$ such that $\Vert z \Vert \leq l$ for all $z\in H$. Let

$$ N=\sup_{t\in [0,1],z\in [0,l] } \bigl\vert g(t,z_{t}) \bigr\vert +1. $$

Then, for $z\in H$, we have

$$ \bigl\vert Tz(t) \bigr\vert = \biggl\vert \int _{0}^{1}G(t,\theta )g(\theta ,z_{\theta }) \,d\theta \biggr\vert \leq \frac{2N}{(2-k)\varGamma (\beta )}. $$

That is, $T(H)$ is uniformly bounded.

Let $z\in H$ and $t_{1}< t_{2}$, $t_{1}, t_{2}\in [-\tau ,1]$. If $0\leq t_{1}< t_{2}\leq 1$, then

$$\begin{aligned}& \bigl\vert (Tz)^{\prime }(t) \bigr\vert \\& \quad = \biggl\vert - \int ^{t}_{0}\frac{(t-\theta )^{\beta -2}}{\varGamma (\beta -1)}g( \theta ,z_{\theta })\,d\theta + \int _{0}^{1} \frac{2(1-\theta )^{\beta -1}(\beta -k+k \theta )}{(2-k)\varGamma (\beta +1)}g( \theta ,z_{\theta })\,d\theta \biggr\vert \\& \quad \leq \int ^{t}_{0}\frac{(t-\theta )^{\beta -2}}{\varGamma (\beta -1)} \bigl\vert g( \theta ,z_{\theta }) \bigr\vert \,d\theta + \int _{0}^{1} \frac{2(1-\theta )^{\beta -1}(\beta -k+k \theta )}{(2-k)\varGamma (\beta +1)} \bigl\vert g( \theta ,z_{\theta }) \bigr\vert \,d\theta \\& \quad \leq N\biggl[\frac{t^{\beta -1}}{\varGamma (\beta )}+ \frac{2(\beta +1-k)}{(2-k)\varGamma (\beta +2)}\biggr] \\& \quad \leq N \frac{\beta (\beta +1)(2-k)+2(\beta +1-k)}{(2-k)\varGamma (\beta +2)}]:=N_{0}. \end{aligned}$$

Hence

$$ \bigl\vert Tz(t_{2})-Tz(t_{1}) \bigr\vert \leq \int _{t_{1}}^{t_{2}} \bigl\vert (Tz)^{\prime }( \theta ) \bigr\vert \,d \theta \leq N_{0}(t_{2}-t_{1}). $$

If $-\tau \leq t_{1}< t_{2}\leq 0$, then

$$ \bigl\vert Tz(t_{2})-Tz(t_{1}) \bigr\vert = \bigl\vert \phi (t_{2})-\phi (t_{1}) \bigr\vert . $$

If $-\tau \leq t_{1}<0<t_{2}\leq 1$, then

$$\begin{aligned} \bigl\vert Tz(t_{2})-Tz(t_{1}) \bigr\vert =& \bigl\vert Tz(t_{2})-Tz(0) \bigr\vert + \bigl\vert Tz(0)-Tz(t_{1}) \bigr\vert \\ \leq & \int ^{1}_{0} \bigl\vert G(t_{2},\theta )-G(0,\theta ) \bigr\vert \bigl\vert g(\theta ,z_{ \theta }) \bigr\vert \,d\theta + \bigl\vert \phi (0)-\phi (t_{1}) \bigr\vert \\ \leq & \frac{2N(\beta +1-k)}{(2-k)\varGamma (\beta +2)}t_{2}+ \bigl\vert \phi (0)- \phi (t_{1}) \bigr\vert \\ < & \frac{2N(\beta +1-k)}{(2-k)\varGamma (\beta +2)} \vert t_{2}-t_{1} \vert + \bigl\vert \phi (0)- \phi (t_{1}) \bigr\vert . \end{aligned}$$

Therefore, $T(H)$ is equicontinuous. On the basis of the Ascoli–Arzelà theorem we conclude that $T(H)$ is relatively compact. The conclusion has been proved. □

Theorem 3.1

If$(C_{1})$, $(C_{2})$, $(C_{3})$and$(C_{5})$are satisfied, then Eq. (1) has at least two positive solutions$z_{1}$and$z_{2}$with

$$ 0\leq \Vert z_{1} \Vert _{[-\tau ,1]}< m< \Vert z_{2} \Vert _{[-\tau ,1]}. $$

Proof

Suppose that $(C_{5})$ holds. Let $\varOmega _{m}=\{z\in P: \Vert z \Vert _{[-\tau ,1]}< m\}$, for any $z\in P\cap \partial \varOmega _{m}$, we have

$$\begin{aligned} (Tz) (t) =& \textstyle\begin{cases} \int _{0}^{1}G(t,\theta )g(\theta ,z_{\theta })\,d\theta , &0\leq t \leq 1, \\ \phi (t), &-\tau \leq t\leq 0, \end{cases}\displaystyle \\ \leq & \textstyle\begin{cases} \frac{m}{B}M\int _{0}^{1}G(1,\theta )\,d\theta , &0\leq t\leq 1, \\ \Vert \phi \Vert _{[-\tau ,0]}, &-\tau \leq t\leq 0, \end{cases}\displaystyle \\ \leq & \textstyle\begin{cases} m, &0\leq t\leq 1, \\ \Vert \phi \Vert _{[-\tau ,0]}, &-\tau \leq t\leq 0, \end{cases}\displaystyle \\ \leq& \Vert z \Vert _{[-\tau ,1]}, \end{aligned}$$

which yields

$$ \Vert Tz \Vert _{[-\tau ,1]}\leq \Vert z \Vert _{[-\tau ,1]}, \quad \text{for } z\in P\cap \partial \varOmega _{m}. $$

Suppose that $(C_{3})$ holds. Since $g_{0}=\infty$, we may choose $\Vert \phi \Vert _{[-\tau ,0]}< m_{1}< m$, such that $g(t,z)\geq K \Vert z \Vert _{[-\tau ,0]}$, for $0\leq \Vert z \Vert _{[-\tau ,0]}\leq m_{1}$, where $K>0$ satisfies $KA\geq 1$.

Let $\varOmega _{m_{1}}=\{z\in P: \Vert z \Vert _{[-\tau ,1]}< m_{1}\}$, for any $z\in P\cap \partial \varOmega _{m_{1}}$, we have

$$\begin{aligned} (Tz) \biggl(\frac{1}{2}\biggr) \geq & \int _{\sigma }^{1-\sigma }G\biggl(\frac{1}{2}, \theta \biggr)g(\theta ,z_{\theta })\,d\theta \geq K \int _{\sigma }^{1- \sigma }G\biggl(\frac{1}{2},\theta \biggr) \Vert z_{\theta } \Vert _{[-\tau ,0]}\,d\theta \\ \geq & \frac{K\lambda }{2} \int _{\sigma }^{1-\sigma }G(1,\theta ) \Vert z \Vert _{[0,1]}\,d\theta =\frac{K\lambda }{2} \int _{\sigma }^{1-\sigma }G(1, \theta ) \Vert z \Vert _{[-\tau ,1]}\,d\theta \\ \geq & \Vert z \Vert _{[-\tau ,1]}, \end{aligned}$$

which yields

$$ \Vert Tz \Vert _{[-\tau ,1]}\geq \Vert z \Vert _{[-\tau ,1]}, \quad \text{for } z\in P\cap \partial \varOmega _{m_{1}}. $$

Next, since $g_{\infty }=\infty$, we may choose $m_{2}>m> \Vert \phi \Vert _{[-\tau ,0]}$, such that $g(t,z)\geq L \Vert z \Vert _{[-\tau ,0]}$, for $\Vert z \Vert _{[-\tau ,0]}\geq \lambda m_{2}$, where $L>0$ satisfies $LA\geq 1$.

Let $\varOmega _{m_{2}}=\{z\in P: \Vert z \Vert _{[-\tau ,1]}< m_{2}\}$, for any $z\in P\cap \partial \varOmega _{m_{2}}$, we have

$$\begin{aligned} (Tz) \biggl(\frac{1}{2}\biggr) \geq & \int _{\sigma }^{1-\sigma }G\biggl(\frac{1}{2}, \theta \biggr)g(\theta ,z_{\theta })\,d\theta \geq L \int _{\sigma }^{1- \sigma }G\biggl(\frac{1}{2},\theta \biggr) \Vert z_{\theta } \Vert _{[-\tau ,0]}\,d\theta \\ \geq & \frac{L\lambda }{2} \int _{\sigma }^{1-\sigma }G(1,\theta ) \Vert z \Vert _{[0,1]}\,d\theta =\frac{L\lambda }{2} \int _{\sigma }^{1-\sigma }G(1, \theta ) \Vert z \Vert _{[-\tau ,1]}\,d\theta \\ \geq & \Vert z \Vert _{[-\tau ,1]}, \end{aligned}$$

which yields

$$ \Vert Tz \Vert _{[-\tau ,1]}\geq \Vert z \Vert _{[-\tau ,1]}, \quad \text{for } z\in P\cap \partial \varOmega _{m_{2}}. $$

Therefore, the conclusion has been proved by (i) and (ii) of Lemma 2.4. □

Theorem 3.2

If$(C_{1})$, $(C_{2})$, $(C_{4})$and$(C_{6})$are satisfied, then Eq. (1) has at least two positive solutions$z_{1}$and$z_{2}$with

$$ 0\leq \Vert z_{1} \Vert _{[-\tau ,1]}< n< \Vert z_{2} \Vert _{[-\tau ,1]}. $$

Proof

Suppose that $(C_{6})$ holds. Letting $\varOmega _{n}=\{z\in P: \Vert z \Vert _{[-\tau ,1]}< n\}$, for any $z\in P\cap \partial \varOmega _{n}$, we have

$$\begin{aligned} (Tz) \biggl(\frac{1}{2}\biggr) \geq & \int _{\sigma }^{1-\sigma }G\biggl(\frac{1}{2}, \theta \biggr)g(\theta ,z_{\theta })\,d\theta \geq \frac{1}{2} \int _{\sigma }^{1- \sigma }G(1,\theta )g(\theta ,z_{\theta })\,d\theta \\ \geq & \frac{n\lambda }{2A} \int _{\sigma }^{1-\sigma }G(1,\theta )\,d \theta =n= \Vert z \Vert _{[-\tau ,1]}, \end{aligned}$$

which yields

$$ \Vert Tz \Vert _{[-\tau ,1]}\geq \Vert z \Vert _{[-\tau ,1]}, \quad \text{for } z\in P\cap \partial \varOmega _{n}. $$

Suppose that $(C_{4})$ holds. Since $g_{0}=0$, we may choose $\Vert \phi \Vert _{[-\tau ,0]}< n_{1}<n$, such that $g(t,z)\leq C \Vert z \Vert _{[-\tau ,0]}$, for $0\leq \Vert z \Vert _{[-r,0]}\leq n_{1}$, where $C>0$ satisfies $CB\leq 1$.

Let $\varOmega _{n_{1}}=\{z\in P: \Vert z \Vert _{[-\tau ,1]}< n_{1}\}$, for any $z\in P\cap \partial \varOmega _{m_{2}}$, we have

$$\begin{aligned} (Tz) (t) =& \textstyle\begin{cases} \int _{0}^{1}G(t,\theta )g(\theta ,z_{\theta })\,d\theta , &0\leq t \leq 1, \\ \phi (t), &-\tau \leq t\leq 0, \end{cases}\displaystyle \\ \leq & \textstyle\begin{cases} CM\int _{0}^{1}G(1,\theta ) \Vert z_{\theta } \Vert _{[-\tau ,0]}\,d\theta , &0 \leq t\leq 1, \\ \Vert \phi \Vert _{[-r,0]}, &-\tau \leq t\leq 0, \end{cases}\displaystyle \\ \leq & \textstyle\begin{cases} CB \Vert z \Vert _{[-\tau ,1]}, &0\leq t\leq 1, \\ \Vert \phi \Vert _{[-r,0]}, &-\tau \leq t\leq 0, \end{cases}\displaystyle \\ \leq &\Vert z \Vert _{[-\tau ,1]}, \end{aligned}$$

which yields

$$ \Vert Tz \Vert _{[-\tau ,1]}\leq \Vert z \Vert _{[-\tau ,1]}, \quad \text{for } z\in P\cap \partial \varOmega _{n_{1}}. $$

In addition, since $g_{\infty }=0$, there exists $R>n$, such that $g(t,z)\leq D \Vert z \Vert _{[-\tau ,0]}$, for $\Vert z \Vert _{[-\tau ,0]}>R$, where $D>0$ satisfies $(D+1)B\leq 1$.

Choose a constant $n_{2}>0$, such that $n_{2}>\max \{n, \Vert \phi \Vert _{[-\tau ,0]},\max \{g(\theta ,z_{\theta })\vert 0 \leq \Vert z_{\theta } \Vert _{[-\tau ,0]}\leq R\}B\}$. Let $\varOmega _{n_{2}}=\{z\in P: \Vert z \Vert _{[-\tau ,1]}< n_{2}\}$, for any $z\in P\cap \partial \varOmega _{n_{2}}$, we have

$$\begin{aligned}& (Tz) (t) \\& \quad\leq \textstyle\begin{cases} \int _{ \Vert z_{\theta } \Vert _{[-\tau ,0]}>R}MG(1,\theta )g(\theta ,z_{ \theta })\,d\theta \\ \quad{} +\int _{0\leq \Vert z_{\theta } \Vert _{[-\tau ,0]}\leq R}MG(1,\theta )g( \theta ,z_{\theta })\,d\theta , &0\leq t\leq 1, \\ \phi (t), &-\tau \leq t\leq 0, \end{cases}\displaystyle \\& \quad\leq \textstyle\begin{cases} \{Dn_{2}+\max \{g(\theta ,z_{\theta })\vert 0\leq \Vert z_{\theta } \Vert _{[-\tau ,0]} \leq R\}\}B, &0\leq t\leq 1, \\ \Vert \phi \Vert _{[-\tau ,0]}, &-\tau \leq t\leq 0, \end{cases}\displaystyle \\& \quad \leq \textstyle\begin{cases} n_{2}, &0\leq t\leq 1, \\ \Vert \phi \Vert _{[-\tau ,0]}, &-\tau \leq t\leq 0, \end{cases}\displaystyle \\& \quad \leq n_{2}= \Vert z \Vert _{[-\tau ,1]}, \end{aligned}$$

which yields

$$ \Vert Tz \Vert _{[-\tau ,1]}\leq \Vert z \Vert _{[-\tau ,1]}, \quad \text{for } z\in P\cap \partial \varOmega _{n_{2}}. $$

Therefore, the conclusion has been proved by (i) and (ii) of Lemma 2.4. □

From the ideas in the proofs of Theorem 3.1 and Theorem 3.2, we have Theorem 3.3 and Theorem 3.4.

Theorem 3.3

If$(C_{1})$, $(C_{2})$are satisfied and the conditions$g_{0}=\infty$, $g_{\infty }=0$hold, then Eq. (1) has at least one positive solution.

Theorem 3.4

If$(C_{1})$, $(C_{2})$are satisfied and the conditions$g_{0}=0$, $g_{\infty }=\infty$hold, then Eq. (1) has at least one positive solution.

4 Some examples

Example 4.1

We consider the following equation:

$$ \textstyle\begin{cases} {}^{c}D^{\frac{10}{3}}z(t)=-z^{\frac{1}{3}}(t-\frac{1}{2}), \quad t\in [0,1], \\ z(t)=t^{6}, \quad t\in [-\frac{1}{2},0], \\ z(0)=z^{\prime \prime }(0)=z^{\prime \prime \prime }(0)=0, \quad \quad z(1)=\frac{1}{2}\int _{0}^{1}z(\theta )\,d\theta , \end{cases} $$

(6)

where $\alpha =\frac{10}{3}$, $k=\frac{1}{2}$, $\tau =\frac{1}{2}$, $g(t,z)=z^{ \frac{1}{3}}(-\frac{1}{2})$; since

$$ \frac{g(t,z)}{ \Vert z \Vert _{[-\frac{1}{2},0]}} = \frac{z^{\frac{1}{3}}(-\frac{1}{2})}{ \Vert z \Vert _{[-\frac{1}{2},0]}}\leq \frac{ \Vert z \Vert ^{\frac{1}{3}}_{[-\frac{1}{2},0]}}{ \Vert z \Vert _{[-\frac{1}{2},0]}} = \Vert z \Vert ^{-\frac{2}{3}}_{[-\frac{1}{2},0]}\rightarrow 0, \quad \text{as } \Vert z \Vert _{[- \frac{1}{2},0]}\rightarrow +\infty $$

we have $g_{\infty }=0$. In addition, there exists a constant $c>0$ with $z(t)\geq c \Vert z \Vert _{[-r,0]}$,

$$ \frac{g(t,z)}{ \Vert z \Vert _{[-\frac{1}{2},0]}}= \frac{z^{\frac{1}{3}}(-\frac{1}{2})}{ \Vert z \Vert _{[-\frac{1}{2},0]}}\geq c \frac{ \Vert z \Vert ^{\frac{1}{3}}_{[-\frac{1}{2},0]}}{ \Vert z \Vert _{[-\frac{1}{2},0]}}=c \Vert z \Vert ^{-\frac{2}{3}}_{[-\frac{1}{2},0]}\rightarrow +\infty , \quad \text{as } \Vert z \Vert _{[- \frac{1}{2},0]}\rightarrow 0. $$

Thus, $g_{0}=+\infty$, Eq. (6) has at least one positive solution by Theorem 3.3.

Example 4.2

We consider the following equation:

$$ \textstyle\begin{cases} {}^{c}D^{\frac{7}{2}}z(t)=-(z^{\frac{1}{2}}(t-\frac{1}{3})+z^{2}(t- \frac{1}{3})), \quad t\in [0,1], \\ z(t)=t^{4}, \quad t\in [-\frac{1}{3},0], \\ z(0)=z^{\prime \prime }(0)=z^{\prime \prime \prime }(0)=0, \quad \quad z(1)=\int _{0}^{1}z(\theta )\,d\theta , \end{cases} $$

(7)

where $\alpha =\frac{7}{2}$, $k=1$, $\tau =\frac{1}{3}$, $g(t,z)=z^{\frac{1}{2}}(- \frac{1}{3})+z^{2}(-\frac{1}{3})$, and there exists a constant $c>0$ with $z(t)\geq c \Vert z \Vert _{[-r,0]}$; since

$$\begin{aligned}& \frac{g(t,z)}{ \Vert z \Vert _{[-\frac{1}{3},0]}}= \frac{z^{\frac{1}{2}}(-\frac{1}{3})+z^{2}(-\frac{1}{3})}{ \Vert z \Vert _{[-\frac{1}{3},0]}} \geq c \frac{ \Vert z \Vert ^{\frac{1}{2}}_{[-\frac{1}{3},0]}+ \Vert z \Vert ^{2}_{[-\frac{1}{3},0]}}{ \Vert z \Vert _{[-\frac{1}{3},0]}} \rightarrow + \infty , \quad \text{as } \Vert z \Vert _{[-\frac{1}{3},0]}\rightarrow + \infty , \\& \frac{g(t,z)}{ \Vert z \Vert _{[-\frac{1}{3},0]}}= \frac{z^{\frac{1}{2}}(-\frac{1}{3})+z^{2}(-\frac{1}{3})}{ \Vert z \Vert _{[-\frac{1}{3},0]}} \geq c \frac{ \Vert z \Vert ^{\frac{1}{2}}_{[-\frac{1}{3},0]}+ \Vert z \Vert ^{2}_{[-\frac{1}{3},0]}}{ \Vert z \Vert _{[-\frac{1}{3},0]}} \rightarrow + \infty , \quad \text{as } \Vert z \Vert _{[-\frac{1}{3},0]}\rightarrow 0, \end{aligned}$$

we have $g_{0}=+\infty$, $g_{\infty }=+\infty$. Thus the condition $(C_{3})$ holds. Furthermore, $M=\frac{2\alpha }{k(\alpha -2)}=\frac{14}{3}$, $\int _{0}^{1}G(1,s)\,ds= \int _{0}^{1} \frac{(1-s)^{\alpha -1}(k[2(s-1)+\alpha ])}{(2-k)\varGamma (\alpha +1)}\,ds= \frac{5}{9\varGamma (\frac{9}{2})}$, $B=M\int _{0}^{1}G(1,s)\,ds= \frac{224}{567\sqrt{\pi }}$. Taking $m=2$, then when $0\leq \Vert z \Vert _{[-\frac{1}{3},0]}\leq 2$, we have $g(t,z)\leq 6\leq \frac{m}{B}=\frac{567\sqrt{\pi }}{112}$, which implies the condition $(C_{5})$ holds. Hence by Theorem 3.1, Eq. (7) has at least two positive solutions $z_{1}$ and $z_{2}$ with

$$ 0< \Vert z \Vert _{[-\frac{1}{3},1]}< 2< \Vert z \Vert _{[-\frac{1}{3},1]}. $$

5 Conclusion

In this paper, on the basis of the Guo–Krasnoselskii theorem, the sufficient conditions ensure that the existence and multiplicity of positive solutions are obtained. This research method can be extended to many fractional boundary value problems. It is worth noting that the equation involves time delay and an integral boundary value condition, to be compared to much previous work, which has never been considered. In addition, our work is inspiring for future research as regards triple positive solutions of fractional boundary value problems with delay.