# Positive solutions for a class of fractional differential equations at resonance

## Abstract

In this paper, by using the Leggett-Williams norm-type theorem, we consider a m-point boundary value problem for a class of fractional differential equations at resonance. A new result on the existence of solutions for above problem is obtained.

## Introduction

The subject of fractional calculus has gained significant interest and been a valuable tool for both science and engineering (see [13]). In recent years, the fractional boundary value problems (FBVPs for short) have been considered by many authors (see [410] and the references therein). For example, Bai studied a FBVP at non-resonance with $$1<\alpha\leq2$$ (see [10]). FBVPs at resonance were studied by Kosmatov (see [11]) and Jiang (see [12]). But the positive solutions for FBVPs at resonance were studied very few. In [13], Yang and Wang considered the positive solutions of the following FBVP:

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} -D_{0^{+}}^{\alpha}x(t)=f(t,x(t)), \quad t\in[0,1], \\ x(0)=0, \qquad x'(0)=x'(1). \end{array}\displaystyle \right . \end{aligned}

In [14], Chen and Tang studied the positive solution of FBVP as follows:

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} D_{0^{+}}^{\alpha}x(t)=f(t,x(t)),\quad t\in[0,+\infty), \\ x(0)=x'(0)=x''(0)=0, \qquad D_{0^{+}}^{\alpha-1}x(0)=\lim_{t\rightarrow +\infty}D_{0^{+}}^{\alpha-1}x(t). \end{array}\displaystyle \right . \end{aligned}

However, to the best of our knowledge, the fractional differential equations with m-point boundary conditions at resonance have not been considered. Motivated by the papers above, we consider the existence of positive solutions for a m-point FBVP of the form

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} -D_{0^{+}}^{\alpha}x(t)=f(t,x(t)), \quad t\in[0,1], \\ x'(0)=0, \qquad x(1)=\sum_{i=1}^{m-2}\beta_{i} x(\eta_{i}), \end{array}\displaystyle \right . \end{aligned}
(1.1)

where $$D_{0^{+}}^{\alpha}$$ denotes the standard Caputo fractional differential operator of order α, $$1< \alpha\leq2$$, $$\beta_{i} \in\mathbb{R^{+}}$$, $$\sum_{i=1}^{m-2}\beta _{i}=1$$, $$0<\eta_{1}<\eta_{2}<\cdots<\eta_{m-2}<1$$, and $$f:[0,1]\times \mathbb{R} \rightarrow\mathbb{R}$$ is continuous. Obviously, FBVP (1.1) happens to be at resonance under the condition $$\sum_{i=1}^{m-2}\beta_{i}=1$$.

The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions and lemmas. In Section 3, we establish a theorem on the existence of positive solutions for FBVP (1.1) under some restrictions of f, basing on the coincidence degree theory due to [15]. Finally, in Section 4, an example is given to illustrate the main result.

## Preliminaries

For convenience of the reader, we present some definitions, notations, and preliminary statements, which can be found in [2, 16, 17].

Let X and Y be real Banach spaces, $$L : \operatorname{dom} L\subset X\rightarrow Y$$ be a Fredholm operator with index zero, where the index of a Fredholm operator L is defined by

\begin{aligned} \operatorname{Index}L : =\operatorname{dim} \operatorname{Ker}L- \operatorname{dim} \operatorname{Coker}L. \end{aligned}

Suppose $$P: X\rightarrow X$$, $$Q: Y\rightarrow Y$$ be continuous linear projectors such that

\begin{aligned}& \operatorname{Im} P=\operatorname{Ker} L, \qquad \operatorname{Ker} Q= \operatorname{Im} L, \\& X=\operatorname{Ker} L\oplus\operatorname{Ker} P, \qquad Y=\operatorname{Im} L \oplus\operatorname{Im} Q. \end{aligned}

Thus, we see that

\begin{aligned} L|_{\operatorname{dom} L\cap\operatorname{Ker} P}: \operatorname{dom} L\cap\operatorname{Ker} P\rightarrow \operatorname{Im} L \end{aligned}

is invertible. We denote the inverse by $$K_{P}$$. Moreover, by virtue of $$\operatorname{dim} \operatorname{Im}Q=\operatorname{codim} \operatorname{Im}L$$, there exists an isomorphism $$J: \operatorname{Im}Q\rightarrow\operatorname{Ker}L$$. Then we know that the operator equation $$Lx=Nx$$ is equivalent to

$$x=(P+JQN)x+K_{P}(I-Q)Nx,$$

where $$N: X\rightarrow Y$$ be a nonlinear operator.

If Ω is an open bounded subset of X such that $$\operatorname{dom} L\cap\overline{\Omega} \neq\emptyset$$, then the map $$N:X\rightarrow Y$$ will be called L-compact on $$\overline{\Omega}$$ if $$QN:\overline {\Omega}\rightarrow Y$$ is bounded and $$K_{P}(I-Q)N:\overline{\Omega }\rightarrow X$$ is compact.

Let C be a cone in X. Then C induces a partial order in X by

$$x\leq y \quad\mbox{if and only if} \quad y-x\in C.$$

### Lemma 2.1

(see [15])

Let C be a cone in X. Then for every $$u\in C\setminus\{0\}$$ there exists a positive number $$\sigma(u)$$ such that

$$\|x+u\|\geq\sigma(u)\|x\|$$

for all $$x\in C$$.

Let $$\gamma: X\rightarrow C$$ be a retraction, that is, a continuous mapping such that $$\gamma(x)=x$$ for all $$x\in C$$. Set

$$\Psi:=P+JQN+K_{P}(I-Q)N$$

and

$$\Psi_{\gamma}:= \Psi\circ\gamma.$$

### Lemma 2.2

(see [15])

Let C be a cone in X and $$\Omega_{1}$$, $$\Omega_{2}$$ be open bounded subsets of X with $$\overline{\Omega}_{1} \subset\Omega_{2}$$ and $$C\cap (\overline{\Omega}_{2}\setminus\Omega_{1})\neq\emptyset$$. Assume that the following conditions are satisfied:

1. (1)

$$L : \operatorname{dom} L\subset X\rightarrow Y$$ be a Fredholm operator of index zero and $$N:X\rightarrow Y$$ be L-compact on every bounded subset of X,

2. (2)

$$Lx\neq\lambda Nx$$ for every $$(x,\lambda)\in[C\cap\partial \Omega_{2} \cap\operatorname{dom} L]\times(0,1)$$,

3. (3)

γ maps subsets of $$\overline{\Omega}_{2}$$ into bounded subsets of C,

4. (4)

$$\operatorname{deg}([I-(P+JQN)\gamma]|_{\operatorname{Ker} L }, \operatorname{Ker} L \cap \Omega_{2} , 0)\neq0$$,

5. (5)

there exists $$u_{0}\in C\setminus\{0\}$$ such that $$\|x\| \leq \sigma(u_{0})\|\Psi x\|$$ for $$x \in C(u_{0})\cap\partial\Omega_{1}$$, where $$C(u_{0})=\{x\in C: \mu u_{0} \leq x \textit{ for some } \mu>0\}$$ and $$\sigma(u_{0})$$ is such that $$\|x+u_{0}\|\geq\sigma(u_{0})\|x\|$$ for every $$x\in C$$,

6. (6)

$$(P+JQN)\gamma(\partial\Omega_{2}) \subset C$$,

7. (7)

$$\Psi_{\gamma}(\overline{\Omega}_{2}\setminus\Omega_{1})\subset C$$.

Then the equation $$Lx=Nx$$ has at least one solution in $$C\cap(\overline {\Omega}_{2}\setminus\Omega_{1})$$.

### Definition 2.3

(see [17])

The Riemann-Liouville fractional integral operator of order $$\alpha>0$$ of a function x is given by

\begin{aligned} I_{0^{+}}^{\alpha}x(t)=\frac{1}{\Gamma(\alpha)}\int_{0}^{t} (t-s)^{\alpha-1}x(s)\,ds, \end{aligned}

provided that the right side integral is pointwise defined on $$(0,+\infty)$$.

### Definition 2.4

(see [17])

The Caputo fractional derivative of order $$\alpha>0$$ of a continuous function x is given by

\begin{aligned} D_{0^{+}}^{\alpha}x(t)=\frac{1}{\Gamma(n-\alpha)}\int_{0}^{t}(t-s)^{n-\alpha -1} x^{(n)}(s)\,ds, \end{aligned}

where n is the smallest integer greater than or equal to α, provided that the right side integral is pointwise defined on $$(0,+\infty)$$.

### Lemma 2.5

(see [18])

For $$\alpha>0$$, the general solution of the Caputo fractional differential equation

\begin{aligned} D_{0^{+}}^{\alpha}x(t)=0 \end{aligned}

is

\begin{aligned} x(t)=c_{0}+c_{1}t+c_{2}t^{2}+ \cdots+c_{n-1}t^{n-1}, \end{aligned}

where $$c_{i}\in{\mathbb{R}}$$, $$i=0,1,\ldots,n-1$$, here n is the smallest integer greater than or equal to α.

### Lemma 2.6

(see [18])

Suppose that $$D_{0^{+}}^{\alpha}x\in C[0,1]$$, $$\alpha>0$$. Then

\begin{aligned} I_{0^{+}}^{\alpha} D_{0^{+}}^{\alpha}x(t)=x(t)+c_{0}+c_{1}t+c_{2}t^{2}+ \cdots +c_{n-1}t^{n-1}, \end{aligned}

where $$c_{i}\in{\mathbb{R}}$$, $$i=0,1,\ldots,n-1$$, here n is the smallest integer greater than or equal to α.

In this paper, we take $$X=Y=C[0,1]$$ with the norm $$\|x\|_{\infty}=\max_{t\in[0,1]} |x(t)|$$.

Define the operator $$L:\operatorname{dom}L\subset X\rightarrow Y$$ by

\begin{aligned} Lx=-D_{0^{+}}^{\alpha}x, \end{aligned}
(2.1)

where

\begin{aligned} \operatorname{dom} L=\Biggl\{ x\in X: D_{0^{+}}^{\alpha}x\in Y, x'(0)=0, x(1)=\sum_{i=1}^{m-2} \beta_{i} x(\eta_{i})\Biggr\} . \end{aligned}

Let $$N:X\rightarrow Y$$ be the Nemytskii operator

\begin{aligned} Nx(t)=f\bigl(t,x(t)\bigr), \quad \forall t\in[0,1]. \end{aligned}

Then FBVP (1.1) is equivalent to the operator equation

\begin{aligned} Lx=Nx,\quad x\in\operatorname{dom}L. \end{aligned}

## Main result

In this section, a theorem on the existence of positive solutions for FBVP (1.1) will be given.

For simplicity of notation, we set

$$l_{i}(s)= \textstyle\begin{cases} (1-s)^{\alpha-1}-(\eta_{i}-s)^{\alpha-1}, & 0\leq s\leq\eta _{i} \leq1,\\ (1-s)^{\alpha-1}, & 0\leq\eta_{i} \leq s\leq1, \end{cases}$$

and

$$G(t,s)= \textstyle\begin{cases} \frac{1}{\Gamma(\alpha+1)}(1-s)^{\alpha}\\ \quad{}+\frac{\alpha(\Gamma(\alpha+2)-1+(\alpha+1)t^{\alpha})}{\Gamma (\alpha+2)\sum_{i=1}^{m-2}\beta_{i}(1-\eta_{i}^{\alpha})} \sum_{i=1}^{m-2} \beta_{i} l_{i}(s), &0\leq t\leq s \leq1,\\ \frac{1}{\Gamma(\alpha+1)}(1-s)^{\alpha}-\frac{1}{\Gamma(\alpha )}(t-s)^{\alpha-1}\\ \quad{}+\frac{\alpha(\Gamma(\alpha+2)-1+(\alpha+1)t^{\alpha})}{\Gamma (\alpha+2)\sum_{i=1}^{m-2}\beta_{i}(1-\eta_{i}^{\alpha})} \sum_{i=1}^{m-2} \beta_{i} l_{i}(s), & 0\leq s \leq t\leq1. \end{cases}$$

Obviously, $$\max_{0\leq s\leq1}\sum_{i=1}^{m-2}\beta _{i}l_{i}(s)\leq1$$. We denote

$$\kappa:=\min \biggl\{ 1, \frac{\sum_{i=1}^{m-2}\beta_{i}(1-\eta _{i}^{\alpha})}{\alpha}, \frac{1}{\max_{t,s\in[0,1]}G(t,s)} \biggr\} .$$

Thus, one has

\begin{aligned} 1-\frac{\kappa\alpha\sum_{i=1}^{m-2}\beta_{i}l_{i}(s)}{\sum_{i=1}^{m-2}\beta_{i}(1-\eta_{i}^{\alpha})}\geq0,\quad 1-\kappa G(t,s)\geq0. \end{aligned}
(3.1)

### Theorem 3.1

Let $$f:[0,1]\times\mathbb{R}\rightarrow\mathbb{R}$$ be continuous. Suppose that:

(H1):

there exist nonnegative functions $$a,b\in X$$ with $$\frac {\Gamma(\alpha+1)}{2} b_{1}< 1$$ such that

\begin{aligned} \bigl|f(t,u)\bigr|\leq a(t)+b(t)|u|, \quad\forall t\in[0,1], u\in\mathbb{R}, \end{aligned}

where $$b_{1}=\|b\|_{\infty}$$,

(H2):

there exists a constant $$B>0$$ such that

$$uf(t,u)< 0,\quad \forall t\in[0,1], |u|>B,$$
(H3):

$$f(t,u)> -\kappa u$$, for all $$(t,u)\in[0,1]\times[0,\infty)$$,

(H4):

there exist $$r\in(0,+\infty)$$, $$t_{0}\in[0,1]$$, $$M\in(0,1)$$ and continuous function $$h:(0,r]\rightarrow[0,\infty)$$ such that $$f(t,u)\geq h(u)$$ for all $$t\in[0,1]$$, $$u\in(0,r]$$, and $$\frac{h(u)}{u}$$ is non-increasing on $$(0,r]$$ with

$$\frac{h(r)}{r}\int_{0}^{1}G(t_{0},s) \,ds \geq\frac{1-M}{M}.$$
Then FBVP (1.1) has at least one solution in X.

Now, we begin with some lemmas that are useful in what follows.

### Lemma 3.2

Let L be defined by (2.1), then

\begin{aligned}& \operatorname{Ker} L=\bigl\{ x\in X: x(t)=c, \forall t\in[0,1], c \in\mathbb {R}\bigr\} , \end{aligned}
(3.2)
\begin{aligned}& \operatorname{Im} L=\Biggl\{ y\in Y: \sum _{i=1}^{m-2}\beta_{i}\int _{0}^{1}l_{i}(s)y(s)\,ds=0\Biggr\} . \end{aligned}
(3.3)

### Proof

By Lemma 2.5, $$D_{0^{+}}^{\alpha}x(t)=0$$ has solution

\begin{aligned} x(t)=c_{0}+c_{1}t,\quad c_{0},c_{1}\in \mathbb{R}. \end{aligned}

Combining with the boundary conditions of FBVP (1.1), one sees that (3.2) holds.

For $$y\in\operatorname{Im} L$$, there exists $$x\in\operatorname{dom} L$$ such that $$y=Lx\in Y$$. By Lemma 2.6, we have

\begin{aligned} x(t)=-\frac{1}{\Gamma(\alpha)}\int_{0}^{t} {(t-s)^{\alpha-1}} {y(s)}\,ds+c_{0}+c_{1}t,\quad c_{0},c_{1}\in\mathbb{R}. \end{aligned}

Then we get

\begin{aligned} x'(t)=-\frac{1}{\Gamma(\alpha-1)}\int_{0}^{t}{(t-s)^{\alpha-2}} {y(s)}\,ds+c_{1}. \end{aligned}

By the boundary conditions of FBVP (1.1), we see that y satisfies

$$\int_{0}^{1}(1-s)^{\alpha-1}y(s)\,ds=\sum _{i=1}^{m-2}\beta_{i} \int _{0}^{\eta _{i}}(\eta_{i}-s)^{\alpha-1}y(s) \,ds.$$

That is,

\begin{aligned} \sum_{i=1}^{m-2} \beta_{i}\int_{0}^{1}l_{i}(s)y(s) \,ds=0. \end{aligned}
(3.4)

On the other hand, suppose $$y\in Y$$ and satisfies (3.4). Let $$x(t)=-I_{0^{+}}^{\alpha}y(t)+x(0)$$, then $$x\in\operatorname{dom}L$$ and $$D_{0^{+}}^{\alpha}x(t)=-y(t)$$. Thus, $$y\in\operatorname{Im}L$$. Hence (3.3) holds. The proof is complete. □

### Lemma 3.3

Let L be defined by (2.1), then L is a Fredholm operator of index zero, and the linear continuous projector operators $$P:X\rightarrow X$$ and $$Q:Y\rightarrow Y$$ can be defined as

\begin{aligned}& Px(t)=\int_{0}^{1}x(s)\,ds, \quad\forall t\in[0,1],\\& Qy(t)=\frac{\alpha}{\sum_{i=1}^{m-2}\beta_{i}(1-\eta_{i}^{\alpha })}\sum_{i=1}^{m-2} \beta_{i}\int_{0}^{1}l_{i}(s)y(s)\,ds,\quad \forall t\in[0,1]. \end{aligned}

Furthermore, the operator $$K_{P}:\operatorname{Im}L\rightarrow\operatorname{dom}L \cap\operatorname{Ker}P$$ can be written by

\begin{aligned} K_{P}y(t)=\int_{0}^{1} k(t,s)y(s)\,ds,\quad \forall t\in[0,1], \end{aligned}

where

$$k(t,s)= \textstyle\begin{cases} \frac{1}{\Gamma(\alpha+1)}(1-s)^{\alpha}, & 0\leq t\leq s \leq1,\\ \frac{1}{\Gamma(\alpha+1)}(1-s)^{\alpha}-\frac{1}{\Gamma(\alpha )}(t-s)^{\alpha-1}, & 0 \leq s \leq t\leq1. \end{cases}$$
(3.5)

### Proof

Obviously, $$\operatorname{Im} P=\operatorname{Ker} L$$ and $$P^{2}x=Px$$. It follows from $$x=(x-Px)+Px$$ that $$X=\operatorname{Ker} P+\operatorname {Ker} L$$. By a simple calculation, one obtain $$\operatorname{Ker} L\cap\operatorname{Ker} P=\{0\}$$. Thus, we get

\begin{aligned} X=\operatorname{Ker} L\oplus\operatorname{Ker} P. \end{aligned}

For $$y\in Y$$, we have

\begin{aligned} Q^{2}y=Q(Qy)=Qy\cdot\frac{\alpha}{\sum_{i=1}^{m-2}\beta_{i}(1-\eta _{i}^{\alpha})}\sum _{i=1}^{m-2}\beta_{i}\int _{0}^{1}l_{i}(s)\,ds=Qy. \end{aligned}

Let $$y=(y-Qy)+Qy$$, where $$y-Qy\in\operatorname{Ker} Q$$, $$Qy\in\operatorname{Im} Q$$. It follows from $$\operatorname{Ker} Q=\operatorname{Im} L$$ and $$Q^{2}y=Qy$$ that $$\operatorname{Im} Q\cap\operatorname{Im} L=\{ 0 \}$$. Then one has

\begin{aligned} Y=\operatorname{Im} L\oplus\operatorname{Im} Q. \end{aligned}

Thus, we obtain

\begin{aligned} \operatorname{dim} \operatorname{Ker}L=\operatorname{dim} \operatorname{Im}Q =\operatorname{dim} \operatorname{Coker}L=1. \end{aligned}

That is, L is a Fredholm operator of index zero.

Now, we will prove that $$K_{P}$$ is the inverse of $$L|_{\operatorname{dom} L\cap \operatorname{Ker} P}$$. In fact, for $$y\in\operatorname{Im}L$$, we have

\begin{aligned} K_{P}y(t)=-\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}y(s) \,ds+c_{0}, \end{aligned}
(3.6)

where

$$c_{0}=\frac{1}{\Gamma(\alpha+1)}\int_{0}^{1}(1-s)^{\alpha}y(s) \,ds.$$

It is easy to see that $$LK_{P}y=y$$. Moreover, for $$x\in\operatorname{dom} L\cap\operatorname{Ker} P$$, we get $$x'(0)=0$$ and

\begin{aligned} K_{P}Lx(t) =&I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}x(t)-I_{0^{+}}^{\alpha+1} D_{0^{+}}^{\alpha}x(t)|_{t=1} \\ =&x(t)-x(0)-\int_{0}^{1}\bigl(x(s)-x(0)\bigr) \,ds \\ =&x-Px. \end{aligned}
(3.7)

Combining (3.6) with (3.7), we know that $$K_{P}=(L|_{\operatorname {dom}L\cap\operatorname{Ker}P})^{-1}$$. The proof is complete. □

### Lemma 3.4

Assume $$\Omega\subset X$$ is an open bounded subset such that $$\operatorname{dom}L\cap\overline{\Omega}\neq\emptyset$$, then N is L-compact on $$\overline{\Omega}$$.

### Proof

By the continuity of f, we see that $$QN(\overline{\Omega})$$ and $$K_{P}(I-Q)N(\overline{\Omega})$$ are bounded. That is, there exist constants $$A,B>0$$ such that $$|(I-Q)Nx|\leq A$$ and $$|K_{P}(I-Q)Nx|\leq B$$, $$\forall x\in\overline{\Omega}$$, $$t\in[0,1]$$. Thus, one need only prove that $$K_{P}(I-Q)N(\overline{\Omega})\subset X$$ is equicontinuous.

Let $$K_{P,Q}=K_{P}(I-Q)N$$, for $$0\leq t_{1}< t_{2}\leq1$$, $$x\in\overline {\Omega}$$, we get

\begin{aligned} & \bigl\vert (K_{P,Q}x) (t_{2})-(K_{P,Q}x) (t_{1})\bigr\vert \\ &\quad\leq\frac{1}{\Gamma(\alpha)}\biggl\vert \int_{0}^{t_{2}}{(t_{2}-s)^{\alpha -1}} {(I-Q)Nx(s)}\,ds-\int_{0}^{t_{1}}{(t_{1}-s)^{\alpha-1}} {(I-Q)Nx(s)}\,ds \biggr\vert \\ &\quad\leq\frac{A}{\Gamma(\alpha)} \biggl[\int_{0}^{t_{1}}{(t_{2}-s)^{\alpha -1}}-{(t_{1}-s)^{\alpha-1}} \,ds+\int_{t_{1}}^{t_{2}}{(t_{2}-s)^{\alpha-1}} \,ds \biggr]\\ &\quad=\frac{A}{\Gamma( \alpha+1)}\bigl(t_{2}^{\alpha}-t_{1}^{\alpha} \bigr). \end{aligned}

Since $$t^{\alpha}$$ is uniformly continuous on $$[0,1]$$, we see that $$K_{P,Q}N(\overline{\Omega})\subset X$$ is equicontinuous. Thus, we see that $$K_{P,Q}N:\overline{\Omega}\rightarrow X$$ is compact. The proof is completed. □

### Lemma 3.5

Suppose (H1) and (H2) hold, then the set

\begin{aligned} \Omega_{0}=\bigl\{ x\in\operatorname{dom} L: Lx=\lambda Nx, \lambda \in(0,1)\bigr\} \end{aligned}

is bounded.

### Proof

Take $$x\in\Omega_{0}$$, then $$Nx\in\operatorname{Im} L$$. By (3.2), we have

\begin{aligned} \sum_{i=1}^{m-2}\beta_{i}\int _{0}^{1} l_{i}(s)f\bigl(s,x(s)\bigr) \,ds=0. \end{aligned}

Then, by the integral mean value theorem, there exists a constant $$\xi \in(0,1)$$ such that $$f(\xi,x(\xi))=0$$. So, from (H2), we get $$|x(\xi )|\leq B$$. By Lemma 2.6, one has

\begin{aligned}& x(t)= x(0) +\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha -1}D_{0^{+}}^{\alpha}x(s)\,ds,\\& x(\xi)= x(0) +\frac{1}{\Gamma(\alpha)}\int_{0}^{\xi}( \xi-s)^{\alpha -1}D_{0^{+}}^{\alpha}x(s)\,ds. \end{aligned}

Thus, we get

\begin{aligned} x(t)-x(\xi) =&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha -1}D_{0^{+}}^{\alpha}x(s) \,ds\\ &{}-\frac{1}{\Gamma(\alpha)}\int_{0}^{\xi}( \xi-s)^{\alpha -1}D_{0^{+}}^{\alpha}x(s)\,ds, \end{aligned}

which together with (H1) implies that

\begin{aligned} \bigl|x(t)\bigr| \leq& \bigl|x(\xi)\bigr|+\frac{1}{\Gamma(\alpha)}\bigl\| D_{0^{+}}^{\alpha}x\bigr\| _{\infty} \cdot\frac{1}{\alpha}\bigl(t^{\beta}+\xi^{\beta}\bigr)\\ \leq&B+\frac{2}{\Gamma(\alpha+1)}\bigl\| D_{0^{+}}^{\alpha}x\bigr\| _{\infty} \\ \leq&B+\frac{2}{\Gamma(\alpha+1)}\cdot\max_{t\in[0,1]}\bigl|f\bigl(t,x(t)\bigr)\bigr|\\ \leq&B+\frac{2}{\Gamma(\alpha+1)}\bigl(\|a\|_{\infty}+b_{1}\|x \|_{\infty}\bigr), \quad \forall t\in[0,1]. \end{aligned}

That is,

\begin{aligned} \|x\|_{\infty} \leq B+\frac{2}{\Gamma(\alpha+1)}\bigl(\|a\|_{\infty}+b_{1} \|x\| _{\infty}\bigr). \end{aligned}

In view of $$\frac{2}{\Gamma(\alpha+1)}b_{1}<1$$, there exists a constant $$D_{2}>0$$ such that

$$\|x\|_{\infty} \leq D_{2}.$$

Hence, $$\Omega_{0}$$ is bounded. The proof is complete. □

### Proof of Theorem 3.1

Set $$C=\{x\in X: x(t)\geq 0, t\in[0,1]\}$$, $$\Omega_{1}=\{x\in X: r>|x(t)|>M\|x\|_{\infty}, t\in [0,1]\}$$, and $$\Omega_{2}=\{x\in X: \|x\|_{\infty}< R\}$$, where $$R=\max\{ B,D_{2}\}$$. Clearly, $$\Omega_{1}$$, $$\Omega_{2}$$ are open bounded subsets of X and

$$\overline{\Omega}_{1}=\bigl\{ x\in X: r\geq\bigl|x(t)\bigr|\geq M\|x \|_{\infty}, t\in [0,1]\bigr\} \subset\Omega_{2}.$$

From Lemma 3.3, Lemma 3.4, and Lemma 3.5, we see that the conditions (1) and (2) of Lemma 2.2 are satisfied.

Let $$\gamma x(t)=|x(t)|$$ for $$x\in X$$ and $$J=I$$. One can see that γ is a retraction and maps subsets of $$\overline{\Omega}_{2}$$ into bounded subsets of C, which means that the condition (3) of Lemma 2.2 holds.

For $$x\in\operatorname{Ker}L \cap\Omega_{2}$$, we have $$x(t)\equiv c$$. Let

$$H(c,\lambda)=c-\lambda|c|-\frac{\lambda\alpha}{\sum_{i=1}^{m-2}\beta _{i}(1-\eta_{i}^{\alpha})}\sum_{i=1}^{m-2} \beta_{i}\int_{0}^{1}l_{i}(s)f\bigl(s,|c|\bigr) \,ds.$$

From $$H(c,\lambda)=0$$, one has $$c\geq0$$. Moreover, if $$H(R,\lambda )=0$$, we get

$$0\leq R(1-\lambda)=\frac{\lambda\alpha}{\sum_{i=1}^{m-2}\beta_{i}(1-\eta _{i}^{\alpha})}\sum_{i=1}^{m-2} \beta_{i}\int_{0}^{1}l_{i}(s)f(s,R) \,ds,$$

which contradicts (H2). Thus $$H(c,\lambda)\neq0$$ for $$x\in\partial \Omega_{2}$$, $$\lambda\in[0,1]$$. Hence

\begin{aligned} & \operatorname{deg}\bigl(\bigl[I-(P+JQN)\gamma\bigr]|_{\operatorname{Ker}L}, \operatorname{Ker}L\cap\Omega _{2},0\bigr) \\ &\quad=\operatorname{deg}\bigl(H(c,1),\operatorname{Ker}L\cap\Omega_{2},0 \bigr) \\ &\quad=\operatorname{deg}\bigl(H(c,0),\operatorname{Ker}L\cap\Omega_{2},0 \bigr) \\ &\quad=\operatorname{deg}(I,\operatorname{Ker}L\cap\Omega_{2},0) \\ &\quad=1. \end{aligned}

So, the condition (4) of Lemma 2.2 holds.

Let $$x\in\overline{\Omega}_{2}\setminus\Omega_{1}$$, $$t\in[0,1]$$, we have

\begin{aligned} \Psi_{\gamma}x(t) = &\int_{0}^{1}\bigl|x(s)\bigr| \,ds+\frac{\alpha}{\sum_{i=1}^{m-2}\beta_{i}(1-\eta_{i}^{\alpha})}\sum_{i=1}^{m-2} \beta_{i}\int_{0}^{1}l_{i}(s)f \bigl(s,\bigl|x(s)\bigr|\bigr)\,ds \\ &{}+\int_{0}^{1}k(t,s) \Biggl[f\bigl(s,\bigl|x(s)\bigr| \bigr)-\frac{\alpha}{\sum_{i=1}^{m-2}\beta_{i}(1-\eta_{i}^{\alpha})}\sum_{i=1}^{m-2} \beta_{i}\int_{0}^{1}l_{i}( \tau)f\bigl(\tau,\bigl|x(\tau)\bigr|\bigr)\,d\tau \Biggr]\,ds \\ =&\int_{0}^{1}\bigl|x(s)\bigr|\,ds+\int _{0}^{1}G(t,s)f\bigl(s,\bigl|x(s)\bigr|\bigr)\,ds, \end{aligned}

which together with (H3) and (3.1) yields

\begin{aligned} \Psi_{\gamma}x(t) \geq&\int_{0}^{1}\bigl|x(s)\bigr| \,ds- \kappa\int_{0}^{1}G(t,s)\bigl|x(s)\bigr|\,ds =\int_{0}^{1}\bigl(1-\kappa G(t,s)\bigr)\bigl|x(s)\bigr| \,ds\geq0. \end{aligned}

Thus, the condition (7) of Lemma 2.2 holds. In addition, we can prove the condition (6) of Lemma 2.2 holds too by a similar process.

Finally, we will show that the condition (5) of Lemma 2.2 is satisfied. Let $$u_{0}(t)\equiv1$$, $$t\in[0,1]$$, then $$u_{0} \in C\setminus\{ 0\}$$, $$C(u_{0})=\{ x\in C: x(t)>0, t\in[0,1]\}$$ and we can take $$\sigma (u_{0})=1$$. For $$x\in C(u_{0})\cap\partial\Omega_{1}$$, we have $$x(t)>0$$, $$t\in[0,1]$$, $$0<\|x\|_{\infty}\leq r$$, and $$x(t)\geq M\|x\|_{\infty}$$, $$t\in[0,1]$$. So, from (H4), we obtain

\begin{aligned} \Psi x(t_{0}) =&\int_{0}^{1}x(s)\,ds+ \int_{0}^{1}G(t_{0},s)f\bigl(s,x(s) \bigr)\,ds \\ \geq& M\|x\|+\int_{0}^{1}G(t_{0},s)h \bigl(x(s)\bigr)\,ds \\ =& M\|x\|+\int_{0}^{1}G(t_{0},s) \frac{h(x(s))}{x(s)}x(s)\,ds \\ \geq& M\|x\|+\frac{h(r)}{r}\int_{0}^{1}G(t_{0},s)x(s) \,ds \\ \geq& M\|x\|+\frac{h(r)}{r}\int_{0}^{1}G(t_{0},s)M \|x\|\,ds \\ \geq& M\|x\|+(1-M)\|x\| \\ =&\|x\|. \end{aligned}

Then the condition (5) of Lemma 2.2 holds.

Consequently, by Lemma 2.2, the equation $$Lx=Nx$$ has at least one solution $$x^{*}\in C\cap(\overline{\Omega}_{2}\setminus\Omega_{1})$$. Namely, FBVP (1.1) has at least one positive solution in X. The proof is complete. □

## Example

We consider the following FBVP:

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} -D_{0^{+}}^{\frac{3}{2}}x(t)=\sin t-\frac{1}{10}x(t)+10+\sin x(t), \quad t\in[0,1], \\ x'(0)=0,\qquad x(1)=x(\frac{1}{2}). \end{array}\displaystyle \right . \end{aligned}
(4.1)

Thus, we have

\begin{aligned}& l(s)= \textstyle\begin{cases} \sqrt{1-s}-\sqrt{\frac{1}{2}-s}, &0\leq s \leq\frac{1}{2},\\ \sqrt{1-s}, & \frac{1}{2}\leq s\leq1, \end{cases}\displaystyle \\& G(t,s)= \textstyle\begin{cases} \frac{1}{\Gamma(\frac{5}{2})}(1-s)^{\frac{3}{2}} +\frac{3(\Gamma(\frac{7}{2})-1+(\frac{5}{2})t^{\frac {3}{2}})}{2\Gamma(\frac{7}{2})(1-(\frac{1}{2})^{\frac{3}{2}})}l(s), &0\leq t\leq s \leq1,\\ \frac{1}{\Gamma(\frac{5}{2})}(1-s)^{\frac{3}{2}}-\frac{1}{\Gamma(\frac {3}{2})}(t-s)^{\frac{1}{2}} +\frac{3(\Gamma(\frac{7}{2})-1+(\frac{5}{2})t^{\frac {3}{2}})}{2\Gamma(\frac{7}{2})(1-(\frac{1}{2})^{\frac{3}{2}})}l(s), &0\leq s\leq t \leq1. \end{cases}\displaystyle \end{aligned}

Moreover, $$f(t,u)\geq8-\frac{1}{10}u\geq-\frac{1}{4}u$$ for all $$u\geq 0$$, and $$l(s)\leq1$$, $$G(t,s)\leq4$$, $$\kappa=-\frac{1}{4}$$. So, we can find that (H1), (H2), (H3) hold. Next, we take $$t_{0}=0$$, $$h(x)=x$$, and $$M=\frac{2}{3}$$, thus $$G(0,s)= \frac{1}{\Gamma(\frac{5}{2})}(1-s)^{\frac{3}{2}}+\frac{3(\Gamma(\frac {7}{2})-1)}{2\Gamma(\frac{7}{2})(1-(\frac{1}{2})^{\frac{3}{2}})}l(s)$$, $$0\leq s \leq1$$, and $$\int_{0}^{1}G(0,s)\,ds=1$$. Then (H4) is satisfied. According to the above points, by Theorem 3.1, we can conclude that FBVP (4.1) has at least one positive solution.

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## Acknowledgements

The authors would like to thank the referees very much for their helpful comments and suggestions. This work was supported by the National Natural Science Foundation of China (11271364).

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Correspondence to Yanqiang Wu.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Wu, Y., Liu, W. Positive solutions for a class of fractional differential equations at resonance. Adv Differ Equ 2015, 241 (2015). https://doi.org/10.1186/s13662-015-0557-9

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### Keywords

• fractional differential equation
• m-point boundary value problem
• Leggett-Williams norm-type theorem
• resonance