Positive solutions for a class of two-term fractional differential equations with multipoint boundary value conditions

Abstract

In this article, we study the existence of positive solutions to a class of two-term fractional nonlocal boundary value problems. The existence and multiplicity of positive solutions are established by means of fixed point index theory. The nonlinearity $f(t,x)$ permits a singularity at $t = 0,1$ and $x=0$.

1 Introduction

In this paper, we consider the following two-term fractional differential equation boundary value problems:

$$ \textstyle\begin{cases} -D^{\alpha }_{0+}u(t)+bu(t)=f(t,u(t)), \quad 0< t< 1, \\ u(0)=0,\qquad u(1)=\sum_{i=1}^{m-2}\eta _{i} u(\xi _{i}), \end{cases} $$

(1.1)

where $D^{\alpha }_{0+}$ is the standard Riemann–Liouville derivative, $1 < \alpha \leq 2$, $b> 0$, $\eta _{i}>0$, $0 < \xi _{1}< \cdots < \xi _{m-2} < 1$, $\sum^{m-2}_{i=1}\eta _{i}\xi _{i}^{\alpha -1}\leq 1$, $f:(0,1)\times (0,+\infty )\rightarrow [0,+\infty )$ is continuous and may be singular at $t = 0,1$ and $x=0$.

Fractional differential equation boundary value problems (FBVPs) have attracted a great deal of attention during the past decades. The literature on boundary value problems of fractional differential equations is now much enriched (see [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19]). Multi-term fractional differential equations appear in the mathematical models of many real world problems. For example, multi-term fractional differential equations have been used to model various types of visco-elastic damping ([20,21,22]). In [20], the authors introduced the Bagley–Torvik equation:

$$ a_{1}D^{2}x(t) + a_{2}D^{\frac{3}{2}}x(t) +a_{3}x(t)=f(t), $$

to describe model for the motion of thin plate in Newtonian fluid. In [22], the authors investigated the endolymph equation:

$$ D^{2}x(t) + a_{1}Dx(t) + a_{2}D^{\frac{1}{2}}x(t) +a_{3}x(t)=-g(t), $$

which can be used to describe model for the response of the semicircular canals to the angular acceleration. The existing literature on multi-term fractional differential equations equipped with initial conditions is quite wide. However, the boundary value problems of multi-term fractional differential equations needs to be investigated. For some recent developments on Caputo type multi-term FBVPs, we mention the papers [23, 24] and the references cited therein.

In [9], by using the technique of [25], we rewrite the original resonant problems as an equivalent non-resonant two-term FBVPs. Combining with the properties of the Green’s function we derived, the existence and uniqueness results of positive solutions are obtained by using of the fixed point index theory and iterative technique. It is well known that the suitable cone plays an important role in seeking positive solutions, which is usually depended on the positive properties of the Green function. However, there are much more difficulties in dealing with the Green functions of fractional-order boundary value problems than ordinary-order problems, especially for the case that $1< \alpha <2$. In [26], we established some new positive properties of the Green function for a class of two-term fractional differential equation with Dirichlet-type boundary value conditions. It should be noted that the properties of the Green function derived in [9] is not suitable for some methods of nonlinear analysis to be used. By employing height functions of the nonlinear term on special bounded sets together with Leggett–Williams and Krasnosel’skii fixed point theorems, Zhang and Zhong [27] established the existence of triple positive solutions for a class of fractional differential equations with integral conditions.

Motivated by the above work, in this paper we aim to establish the existence of positive solutions to the FBVP (1.1). Our work presented in this paper has the following features. Firstly, we consider few cases of Riemann–Liouville type two-term FBVPs which has been studied before. Secondly, some new properties of the Green function for the case that $1< \alpha <2$ have been discovered to deal with the difficulties related to the Green function for this case. Thirdly, the FBVP (1.1) possesses a singularity, that is, $f(t,x)$ may be singular at $t = 0,1$ and $x =0$.

2 Basic definitions and preliminaries

For the convenience of the reader, we present some preliminaries and lemmas.

Definition 2.1

([28])

The fractional integral of order $\alpha > 0$ of a function $u:(0,+\infty )\rightarrow R$ is given by

$$ I^{\alpha }_{0+}u(t)=\frac{1}{\varGamma (\alpha )} \int ^{t}_{0}(t-s)^{ \alpha -1}u(s)\,ds $$

provided that the right-hand side is point-wise defined on $(0,+ \infty )$.

Definition 2.2

([28])

The Riemann–Liouville fractional derivative of order $\alpha > 0$ of a function $u:(0,+\infty )\rightarrow R$ is given by

$$ D^{\alpha }_{0+}u(t)=\frac{1}{\varGamma (n-\alpha )} \biggl( \frac{d}{dt} \biggr) ^{n} \int ^{t}_{0}(t-s)^{n-\alpha -1}u(s)\,ds, $$

where $n=[\alpha ]+1$, $[\alpha ]$ denotes the integer part of number α, provided that the right-hand side is point-wise defined on $(0,+\infty )$.

In [9], we have proved that the function

$$ g(t)=:\frac{\alpha -2}{\varGamma (\alpha -1)}+\sum^{+\infty }_{k=1} \frac{t ^{k}}{\varGamma ((k+1)\alpha -2)}, $$

has a unique positive root $b^{\ast }$. Throughout this paper, we assume that the following conditions hold:

$(A_{1})$:

$b\in (0, b^{\ast }]$ is a constant.

$(A_{2})$:

$f:(0,1)\times (0,+\infty )\rightarrow [0,+\infty )$ is continuous. In addition, for any $R\geq r>0$, there exists $\varPsi _{r,R} \in L^{1}[0,1]\cap C(0,1)$ such that

$$ f(t,x)\leq \varPsi _{r,R}(t),\quad \forall t \in (0,1), x\in \bigl[ rt ^{\alpha -1},R\bigr]. $$

Lemma 2.1

The unique solution of the problem

$$ \textstyle\begin{cases} -D^{\alpha }_{0+}u(t)+bu(t)=y(t) , \quad 0< t< 1,\\ u(0)=0, \qquad u(1)=\sum_{i=1}^{m-2}\eta _{i} u(\xi _{i}), \end{cases} $$

can be expressed by

$$ u(t)= \int _{0}^{1}K(t,s)y(s)\,ds, $$

where

$$\begin{aligned}& K(t,s)=K_{1}(t,s)+t^{\alpha -1}E_{\alpha ,\alpha } \bigl(bt^{\alpha }\bigr)q(s), \\& q(s)=\frac{\sum_{i=1}^{m-2}\eta _{i} K_{1}(\xi _{i},s)}{ E_{\alpha , \alpha }(b)-\sum_{i=1}^{m-2}\eta _{i}\xi _{i}^{\alpha -1}E_{\alpha , \alpha }(b\xi _{i}^{\alpha }) }, \\& K_{1}(t,s)= \textstyle\begin{cases} \frac{(t-ts)^{\alpha -1}E_{\alpha ,\alpha }(bt^{\alpha })E_{\alpha ,\alpha }(b(1-s)^{\alpha })}{E_{\alpha ,\alpha }(b)}, & 0\leq t \leq s\leq 1 ,\\ \frac{(t-ts)^{\alpha -1}E_{\alpha ,\alpha }(bt^{ \alpha })E_{\alpha ,\alpha }(b(1-s)^{\alpha })}{E_{\alpha ,\alpha }(b)} \\ \quad {}-\frac{(t-s)^{\alpha -1}E_{\alpha ,\alpha }(b(t-s)^{\alpha })E _{\alpha ,\alpha }(b)}{E_{\alpha ,\alpha }(b)},& 0\leq s\leq t \leq 1, \end{cases}\displaystyle \end{aligned}$$

here

$$ E_{\alpha ,\alpha }(x)=\sum^{+\infty }_{k=0} \frac{x^{k}}{\varGamma ((k+1) \alpha )} $$

is the Mittag-Leffler function.

Proof

Noticing that $\eta _{i}>0$, $0 < \xi _{i} < 1$ and $\sum_{i=1}^{m-2} \eta _{i} \xi _{i}^{\alpha -1}\leq 1$, we have

$$ \begin{aligned}[b] E_{\alpha ,\alpha }(b)-\sum _{i=1}^{m-2}\eta _{i}\xi _{i}^{\alpha -1}E _{\alpha ,\alpha }\bigl(b\xi _{i}^{\alpha }\bigr) &=\sum^{+\infty }_{k=0} \frac{b ^{k} [1-\sum_{i=1}^{m-2}\eta _{i} \xi _{i}^{(k+1)\alpha -1} ]}{ \varGamma ((k+1)\alpha )} \\ &\geq \sum^{+\infty }_{k=0} \frac{b^{k}\sum_{i=1}^{m-2}\eta _{i} \xi _{i}^{\alpha -1}(1- \xi _{i}^{k\alpha })}{ \varGamma ((k+1)\alpha )}>0. \end{aligned} $$

The proof is similar to Lemma 2.1 in [9], we omit it here. □

Lemma 2.2

The function $K(t,s)$ satisfies the following properties:

• (1) $K(t,s) > 0$, $\forall t,s\in (0,1)$;

• (2) $h_{2}(s)t^{\alpha -1}\leq K(t,s) \leq h_{1}(s)t^{\alpha -1}$, $\forall t,s\in [0,1]$, where

  $$ h_{1}(s)=(1-s)^{\alpha -1}E_{\alpha ,\alpha } \bigl(b(1-s)^{\alpha }\bigr)+E_{ \alpha ,\alpha }(b)q(s),\qquad h_{2}(s)=\frac{q(s)}{\varGamma (\alpha )}. $$

Proof

The proof is similar to Lemma 2.2 in [9], we omit it here. □

Lemma 2.3

([26])

The function $K_{1}(t,s)$ has the following properties:

• (1) $K_{1}(t,s) > 0$, $\forall t,s\in (0,1)$;

• (2) $K_{1}(t,s)=K_{1}(1-s,1-t)$, $\forall t,s\in [0,1]$;

• (3) $K_{1}(t,s) \leq E_{\alpha ,\alpha }(b) s (1-s)^{\alpha -1}t ^{\alpha -2}$, $\forall s\in [0,1]$, $t\in (0,1]$;

• (4) $K_{1}(t,s) \geq M_{1} s (1-s)^{\alpha -1}(1-t)t^{\alpha -1}$, $\forall t,s\in [0,1]$, where

  $$ M_{1}=\min \biggl\{ \frac{1}{[\varGamma (\alpha )]^{2}E_{\alpha ,\alpha }(b)}, (\alpha -1)^{2}E_{\alpha ,\alpha }(b) \biggr\} . $$

Lemma 2.4

The function $K(t,s)$ has the following properties:

• (1) $K(t,s) \leq t^{\alpha -2}E_{\alpha ,\alpha }(b) [s (1-s)^{ \alpha -1}+q(s)]$, $\forall s\in [0,1]$, $t\in (0,1]$;

• (2) $K(t,s) \geq M t^{\alpha -1}[s (1-s)^{\alpha -1}+q(s)]$, $\forall t,s\in [0,1]$, where

  $$ M=\min \biggl\{ \frac{1}{2\varGamma (\alpha )}, M_{1}, \frac{M _{1}}{2\varGamma (\alpha )}\times \frac{\sum_{i=1}^{m-2}\eta _{i}\xi _{i} ^{\alpha -1}-\sum_{i=1}^{m-2}\eta _{i} \xi _{i}^{\alpha }}{ E_{\alpha , \alpha }(b)-\sum_{i=1}^{m-2}\eta _{i}\xi _{i}^{\alpha -1}E_{\alpha , \alpha }(b\xi _{i}^{\alpha })} \biggr\} . $$

Proof

From (3) of Lemma 2.3, we have

$$ \begin{aligned}[b] K(t,s) &=K_{1}(t,s)+t^{\alpha -1}E_{\alpha ,\alpha } \bigl(bt^{\alpha }\bigr)q(s) \\ &\leq t^{\alpha -2}E_{\alpha ,\alpha }(b) s (1-s)^{\alpha -1}+t ^{\alpha -1}E_{\alpha ,\alpha }(b)q(s) \\ & \leq t^{\alpha -2}E_{\alpha ,\alpha }(b) \bigl[s (1-s)^{\alpha -1}+q(s) \bigr]. \end{aligned} $$

Therefore (1) holds.

On the other hand, it follows from (4) of Lemma 2.3 that

$$ \begin{aligned}[b] q(s) &=\frac{\sum_{i=1}^{m-2}\eta _{i} K_{1}(\xi _{i},s)}{ E_{\alpha , \alpha }(b)-\sum_{i=1}^{m-2}\eta _{i}\xi _{i}^{\alpha -1}E_{\alpha , \alpha }(b\xi _{i}^{\alpha })} \\ & \geq \frac{\sum_{i=1}^{m-2}\eta _{i} (1- \xi _{i})\xi _{i}^{\alpha -1}}{ E_{\alpha ,\alpha }(b)-\sum_{i=1}^{m-2} \eta _{i}\xi _{i}^{\alpha -1}E_{\alpha ,\alpha }(b\xi _{i}^{\alpha })}M _{1} s (1-s)^{\alpha -1} \\ & =\frac{\sum_{i=1}^{m-2}\eta _{i}\xi _{i} ^{\alpha -1}-\sum_{i=1}^{m-2}\eta _{i} \xi _{i}^{\alpha }}{ E_{\alpha , \alpha }(b)-\sum_{i=1}^{m-2}\eta _{i}\xi _{i}^{\alpha -1}E_{\alpha , \alpha }(b\xi _{i}^{\alpha })}M_{1} s (1-s)^{\alpha -1}. \end{aligned} $$

(2.1)

Denote

$$ M_{2}=\min \biggl\{ \frac{M_{1}}{2\varGamma (\alpha )}\times \frac{\sum_{i=1}^{m-2}\eta _{i}\xi _{i}^{\alpha -1}-\sum_{i=1}^{m-2}\eta _{i} \xi _{i}^{\alpha }}{ E_{\alpha ,\alpha }(b)-\sum_{i=1}^{m-2}\eta _{i}\xi _{i}^{\alpha -1}E_{\alpha ,\alpha }(b\xi _{i}^{\alpha })}, M_{1} \biggr\} . $$

(2.2)

From (2.1) and (2.2), one has

$$ \frac{t^{\alpha -1}}{2\varGamma (\alpha )}q(s)-M_{2} s (1-s)^{\alpha -1}t ^{\alpha } \geq \frac{t^{\alpha -1}}{2\varGamma (\alpha )}q(s)-M_{2} s (1-s)^{ \alpha -1}t^{\alpha -1}\geq 0. $$

Then

$$ \begin{aligned}[b] K(t,s) &=K_{1}(t,s)+t^{\alpha -1}E_{\alpha ,\alpha } \bigl(bt^{\alpha }\bigr)q(s) \\ &\geq M_{1} s (1-s)^{\alpha -1}(1-t)t^{\alpha -1}+t^{\alpha -1}E _{\alpha ,\alpha }\bigl(bt^{\alpha }\bigr)q(s) \\ &\geq M_{1} s (1-s)^{\alpha -1}(1-t)t ^{\alpha -1}+t^{\alpha -1} \frac{1}{\varGamma (\alpha )}q(s) \\ &\geq M _{2} s (1-s)^{\alpha -1}(1-t)t^{\alpha -1}+ \frac{t^{\alpha -1}}{ \varGamma (\alpha )}q(s) \\ &= M_{2} s (1-s)^{\alpha -1}t^{\alpha -1}+ \frac{t ^{\alpha -1}}{2\varGamma (\alpha )}q(s)+\frac{t^{\alpha -1}}{2\varGamma ( \alpha )}q(s)-M_{2} s (1-s)^{\alpha -1}t^{\alpha } \\ &\geq M_{2} s (1-s)^{ \alpha -1}t^{\alpha -1}+ \frac{t^{\alpha -1}}{2\varGamma (\alpha )}q(s) \\ & \geq \min \biggl\{ \frac{1}{2\varGamma (\alpha )}, M_{2} \biggr\} t ^{\alpha -1}\bigl[s (1-s)^{\alpha -1}+q(s)\bigr] \\ &=M t^{\alpha -1}\bigl[s (1-s)^{ \alpha -1}+q(s)\bigr]. \end{aligned} $$

So (2) holds. □

By Lemma 2.2 and Lemma 2.4, we have the following lemma.

Lemma 2.5

The function $K^{\ast }(t,s)=:t^{2-\alpha }K(t,s)$ satisfies:

• (1) $K^{\ast }(t,s) > 0$, $\forall t,s\in (0,1)$;

• (2) $K^{\ast }(t,s)\leq h_{1}(s)t$, $\forall t,s\in [0,1]$;

• (3) $K^{\ast }(t,s)\leq E_{\alpha ,\alpha }(b) [s (1-s)^{ \alpha -1}+q(s)]$, $\forall s\in [0,1]$, $t\in (0,1]$;

• (4) $K^{\ast }(t,s)\geq Mt [s (1-s)^{\alpha -1}+q(s) ]$, $\forall t,s\in [0,1]$.

Let $E=C[0,1]$ be endowed with the maximum norm $\| u\| = \max_{0\leq t\leq 1}| u(t)|$, θ is the zero element of E, $B_{r}=\{u\in E : \| u\|< r\}$. Define a cone P by

$$ P= \biggl\{ u\in E : \exists l_{u}>0, \text{ such that } l _{u}t\geq u(t)\geq \frac{M \Vert u \Vert }{E_{\alpha ,\alpha }(b)}t , t \in [0,1] \biggr\} . $$

Define the height functions as follows:

$$\begin{aligned}& \psi (t,r)=\min \biggl\{ f\bigl(t,t^{\alpha -2}x\bigr):\frac{M r}{E_{\alpha , \alpha }(b)}t \leq x\leq r \biggr\} ; \\ & \varPsi (t,r)=\max \biggl\{ f\bigl(t,t^{\alpha -2}x\bigr):\frac{M r}{E_{\alpha , \alpha }(b)}t \leq x\leq r \biggr\} . \end{aligned}$$

For convenience, we list here some assumptions to be used later:

$(H_{1})$:

there exist $r_{1}>0$ and a nonnegative function $b_{1} \in L^{1}[0,1]$ with $\int _{0}^{1}b_{1}(s)\,ds>0$, such that

$$ f\bigl(t,t^{\alpha -2}x\bigr) \geq b_{1}(t)x , \quad \forall (t,x)\in (0,1) \times (0,r_{1}]; $$

$(H_{2})$:

there exist $r_{2}>0$ and a nonnegative function $b_{2} \in L^{1}[0,1]$ with $\int _{0}^{1}b_{2}(s)\,ds>0$, such that

$$ f\bigl(t,t^{\alpha -2}x\bigr) \leq b_{2}(t)x , \quad \forall (t,x)\in (0,1) \times [r_{2},+\infty ); $$

$(H_{3})$:

there exists $r_{3}>0$ such that

$$ \int _{0}^{1}\bigl[s (1-s)^{\alpha -1}+q(s) \bigr]\psi (t,r_{3})>\frac{r_{3}}{M}; $$

$(H_{4})$:

there exist $r_{4}>0$ such that

$$ \int _{0}^{1}\bigl[s (1-s)^{\alpha -1}+q(s) \bigr]\varPsi (t,r_{4})< \frac{r_{4}}{E _{\alpha ,\alpha }(b)}. $$

Define operators A, $L_{1}$ and $L_{2}$ as follows:

$$\begin{aligned}& A(u) (t)= \int _{0}^{1}K^{\ast }(t,s)f \bigl(s,s^{\alpha -2}u(s)\bigr)\,ds, \\& L_{i}u(t)= \int _{0}^{1}K^{\ast }(t,s)b_{i} (s)u(s)\,ds, \quad i=1,2. \end{aligned}$$

Lemma 2.6

For any $r>0$, $A: P\setminus B_{r} \rightarrow P$ is completely continuous.

Proof

The proof is similar to Lemma 2.3 in [13], we omit it here. □

By the extension theorem of a completely continuous operator (see Theorem 2.7 of [29]), for any $r>0$, there exists an extension operator $\widetilde{A}:P \rightarrow P$, which is still completely continuous. Without loss of the generality, we still write it as A. By virtue of the Krein–Rutmann theorem and Lemma 2.5, we have the following lemma.

Lemma 2.7

$L_{i}:P\rightarrow P$ ($i=1,2$) are completely continuous linear operator. Moreover, the spectral radius $r(L_{i})> 0$ and $L_{i}$ has a positive eigenfunction $\varphi _{i}$ corresponding to its first eigenvalue $(r(L_{i}))^{-1}$, that is, $L_{i}\varphi _{i}=r(L _{i})\varphi _{i}$.

Lemma 2.8

([29])

Let P be a cone in a Banach space E, and Ω be a bounded open set in E. Suppose that $A: \overline{ \varOmega }\cap P\rightarrow P$ is a completely continuous operator. If there exists $u_{0}\in P$ with $u_{0}\neq \theta$ such that

$$ u-Au\neq \lambda u_{0},\quad \forall \lambda \geq 0,x\in \partial \varOmega \cap P, $$

then $i(A,\varOmega \cap P,P)=0$.

Lemma 2.9

([29])

Let P be a cone in a Banach space E, and Ω be a bounded open set in E. Suppose that $A: \overline{ \varOmega }\cap P\rightarrow P$ is a completely continuous operator. If

$$ Au\neq \lambda u,\quad \forall \lambda \geq 1,u\in \partial \varOmega \cap P, $$

then $i(A,\varOmega \cap P,P)=1$.

Lemma 2.10

([29])

Let P be a cone in a Banach space E, and Ω be a bounded open set in E. Suppose that $A: \overline{ \varOmega }\cap P\rightarrow P$ is a completely continuous operator. If

$$\begin{aligned}& \inf_{u\in \partial \varOmega \cap P} \Vert Au \Vert >0, \\& Au\neq \lambda u,\quad \forall \lambda \in (0, 1],u\in \partial \varOmega \cap P, \end{aligned}$$

then $i(A,\varOmega \cap P,P)=0$.

3 Main results

Theorem 3.1

Assume that there exist $r_{2}>r_{1}>0$ such that $(H_{1})$ and $(H_{2})$ hold, and

$$ 0< r(L_{2})< 1 \leq r(L_{1}). $$

Then FBVP (1.1) has at least one positive solution.

Proof

For any $u\in \partial B_{r_{1}}\cap P$, it follows from $(H_{1})$ that

$$ Au(t)= \int _{0}^{1}K^{\ast }(t,s)f \bigl(s,s^{\alpha -2}u(s)\bigr)\,ds \geq \int _{0}^{1}K^{\ast }(t,s)b_{1}(s)u(s)\,ds=L_{1}u(t). $$

Suppose that A has no fixed points on $\partial B_{r_{1}}\cap P$ (otherwise, the proof is finished). In the following, we will show that

$$ u-Au \neq \mu \varphi _{1},\quad \forall u \in \partial B_{r_{1}}\cap P, \mu > 0, $$

(3.1)

in which $\varphi _{1}$ is the positive eigenfunction of $L_{1}$ satisfying $L_{1}\varphi _{1}=r(L_{1})\varphi _{1}$. If otherwise, there exist $\mu _{0}> 0$ and $u_{1}\in \partial B_{r_{1}}\cap P$ such that

$$ u_{1}-Au_{1}= \mu _{0} \varphi _{1}. $$

Therefore,

$$ u_{1}=Au_{1}+\mu _{0} \varphi _{1}\geq \mu _{0} \varphi _{1}. $$

Set

$$ \mu ^{\ast }=\sup \{\mu :u_{1}\geq \mu \varphi _{1}\}. $$

It is clear that $\mu ^{\ast }\geq \mu _{0}$, and $u_{1}\geq \mu ^{ \ast }\varphi _{1}$. Since $L_{1}$ is nondecreasing linear operator, one has

$$ L_{1}u_{1}\geq \mu ^{\ast }L_{1} \varphi _{1}=\mu ^{\ast }r(L_{1})\varphi _{1}\geq \mu ^{\ast }\varphi _{1}. $$

Then

$$ u_{1}=Au_{1}+\mu _{0}\varphi _{1}\geq L_{1}u_{1}+\mu _{0} \varphi _{1} \geq \bigl(\mu ^{\ast }+\mu _{0} \bigr)\varphi _{1}, $$

which contradicts the definition of $\mu ^{\ast }$. So (3.1) holds. It follows from Lemma 2.8 that

$$ i(A,B_{r_{1}}\cap P,P)=0. $$

(3.2)

Denote

$$ W= \{u\in P\backslash B_{r_{1}} \mid u=\mu Au,0 \leq \mu \leq 1 \}. $$

Next, we will prove that W is bounded.

For any $u \in W$, one has

$$ f\bigl(t,t^{\alpha -2}u(t)\bigr)\leq b_{2}(t)u(t)+f \bigl(t,t^{\alpha -2}\tilde{u}(t)\bigr), $$

in which $\tilde{u}(t)=\min \{u(t),r_{2}\}$. It is easy to see that

$$ \frac{M r_{1}}{E_{\alpha ,\alpha }(b)}t\leq \tilde{u}(t)\leq u(t) \leq l_{u}t. $$

Therefore,

$$ l_{u}\geq l_{u} t^{\alpha -1}\geq t^{\alpha -2}\tilde{u}(t)\geq \frac{M r_{1}}{E_{\alpha ,\alpha }(b)}t^{\alpha -1}=:r_{0}t^{\alpha -1}. $$

Then

$$ u(t)=\mu Au(t)\leq Au(t) \leq L_{2}u(t)+A\tilde{u}(t) \leq L_{2}u(t)+M _{3}, $$

here

$$ M_{3}= \int _{0}^{1}h_{1}(s)\varPsi _{r_{0},l_{u}} (s)\,ds. $$

Then

$$ (I-L_{2})u(t)\leq M_{3},t\in [0,1]. $$

(3.3)

By $r(L_{2})< 1$, the inverse operator of $(I-L_{2})$ can be expressed by

$$ (I-L_{2})^{-1}=I+L_{2}+L_{2}^{2}+ \cdots +L_{2}^{n}+\cdots . $$

Therefore, (3.3) yields $u(t)\leq (I-L_{2})^{-1}M_{3}\leq M_{3}\| (I-L _{2})^{-1} \|$, $t\in [0,1]$, so W is bounded.

Choose $R> \max \{r_{2},M_{3}\| (I-L_{2})^{-1} \|\}$. From Lemma 2.9, one has

$$ i(A,B_{R}\cap P,P)=1. $$

(3.4)

It follows from (3.2) and (3.4) that

$$ i\bigl(A,(B_{R}\backslash \bar{B}_{r_{1}})\cap P,P \bigr)=i(A,B_{R}\cap P,P)-i(A,B _{r_{1}}\cap P,P)=1. $$

Then A has a fixed point $u^{\ast }\in (B_{R}\backslash \bar{B}_{r _{1}})\cap P$, that is,

$$ u^{\ast }(t)=Au^{\ast }(t)= \int _{0}^{1}K^{\ast }(t,s)f \bigl(s,s^{\alpha -2}u ^{\ast }(s)\bigr)\,ds. $$

It is easy to check that $t^{\alpha -2}u^{\ast }(t)$ is a positive solution of FBVP (1.1). □

Theorem 3.2

Assume that there exist $r_{4}>r_{3}>0$ such that $(H_{3})$ and $(H_{4})$ hold. Then FBVP (1.1) has at least one positive solution.

Proof

For any $u \in \partial B_{r_{3}}\cap P$, it follows from $(H_{3})$ and Lemma 2.5 that

$$ \begin{aligned}[b] Au &\geq Mt \int _{0}^{1}\bigl[s (1-s)^{\alpha -1}+q(s) \bigr]f\bigl(s,s^{\alpha -2}u(s)\bigr)\,ds \\ &\geq Mt \int _{0}^{1}\bigl[s (1-s)^{\alpha -1}+q(s) \bigr]\psi (t,r_{3})\,ds>r _{3}t. \end{aligned} $$

(3.5)

Therefore,

$$ \inf_{u\in \partial B_{r_{3}}\cap P} \Vert Au \Vert \geq r_{3}>0. $$

$\forall \lambda \in (0, 1]$, $u\in \partial B_{r_{3}}\cap P$, we have $\lambda u\leq u\leq r_{3}$. This with (3.5) implies

$$ Au\neq \lambda u,\quad \forall \lambda \in (0, 1],u\in \partial B_{r_{3}} \cap P. $$

It follows from Lemma 2.10 that

$$ i(A,B_{r_{3}}\cap P,P)=0. $$

(3.6)

Next, we prove that

$$ Au\neq \mu u,\quad \forall u \in \partial B_{r_{4}}\cap P,\mu \geq 1. $$

If otherwise, there exists $u_{1}\in \partial B_{r_{4}}\cap P$, $\mu _{0} \geq 1$ such that $Au_{1}= \mu _{0} u_{1}$. From $(H_{4})$ and Lemma 2.5, we have

$$ \begin{aligned}[b] u_{1}&\leq \mu _{0} u_{1}=Au_{1}\leq E_{\alpha ,\alpha }(b) \int _{0} ^{1}\bigl[s (1-s)^{\alpha -1}+q(s) \bigr]f\bigl(s,s^{\alpha -2}u_{1}(s)\bigr)\,ds \\ &\leq E _{\alpha ,\alpha }(b) \int _{0}^{1}\bigl[s (1-s)^{\alpha -1}+q(s) \bigr]\psi (t,r _{4})\,ds< r_{4}, \end{aligned} $$

which contradicts $\|u_{1}\|=r_{4}$. Then, by Lemma 2.9, we have

$$ i(A,B_{r_{4}}\cap P,P)=1. $$

(3.7)

Equations (3.6) and (3.7) yield

$$ i\bigl(A,(B_{r_{4}}\backslash \bar{B}_{r_{3}})\cap P,P \bigr)=i(A,B_{r_{4}} \cap P,P)-i(A,B_{r_{3}}\cap P,P)=1. $$

Then A has a fixed point $u^{\ast }\in (B_{R}\backslash \bar{B}_{r _{1}})\cap P$. Clearly, $t^{\alpha -2}u^{\ast }(t)$ is a positive solution of FBVP (1.1). □

Theorem 3.3

Assume that there exist $r_{3}>r_{4}>r_{1}>0$ such that $(H_{1})$, $(H_{3})$ and $(H_{4})$ hold with $r(L_{1})\geq 1$. Then FBVP (1.1) has at least two positive solutions.

Proof

Suppose that $\exists r'_{1}\in (0,r_{1})$ such that A has no fixed points on $\partial B_{r'_{1}}\cap P$ (otherwise, the proof is finished). By the proof of Theorem 3.1 and Theorem 3.2, we have

$$ i(A,B_{r'_{1}}\cap P,P)=0,\qquad i(A,B_{r_{4}}\cap P,P)=1,\qquad i(A,B _{r_{3}}\cap Q,Q)=0. $$

Therefore,

$$\begin{aligned}& i\bigl(A,(B_{r_{4}}\backslash \bar{B}_{r'_{1}})\cap P,P \bigr)=i(A,B_{r_{4}} \cap P,P)-i(A,B_{r'_{1}}\cap P,P)=1, \\& i\bigl(A,(B_{r_{3}}\backslash \bar{B}_{r_{4}})\cap P,P \bigr)=i(A,B_{r_{3}} \cap P,P)-i(A,B_{r_{4}}\cap P,P)=-1. \end{aligned}$$

Then FBVP (1.1) has at least two positive solutions. □

Theorem 3.4

Assume that there exist $r_{2}>r_{3}>r_{4}>r_{1}>0$ such that $(H_{1})$–$(H_{4})$ hold with $r(L_{1})\geq 1 >r(L_{2})>0$. Then FBVP (1.1) has at least three positive solutions.

Proof

By Theorem 3.3 and (3.4), we get

$$\begin{aligned}& i(A,B_{r'_{1}}\cap P,P)=0,\qquad i(A,B_{r_{4}}\cap P,P)=1, \\& i(A,B_{r_{3}}\cap Q,Q)=0,\qquad i(A,B_{R}\cap P,P)=1. \end{aligned}$$

Therefore,

$$\begin{aligned}& i\bigl(A,(B_{r_{4}}\backslash \bar{B}_{r'_{1}})\cap P,P \bigr)=i(A,B_{r_{4}} \cap P,P)-i(A,B_{r'_{1}}\cap P,P)=1, \\& i\bigl(A,(B_{r_{3}}\backslash \bar{B}_{r_{4}})\cap P,P \bigr)=i(A,B_{r_{3}} \cap P,P)-i(A,B_{r_{4}}\cap P,P)=-1, \\& i\bigl(A,(B_{R}\backslash \bar{B}_{r_{3}})\cap P,P \bigr)=i(A,B_{R}\cap P,P)-i(A,B _{r_{3}}\cap P,P)=1. \end{aligned}$$

Then FBVP (1.1) has at least three positive solutions. □

4 Example

Example 4.1

Consider the following problem:

$$ \textstyle\begin{cases} -D^{\frac{3}{2}}_{0+}u(t)+\frac{1}{5}u(t)=f(t,u(t)) , \quad 0< t< 1,\\ u(0)=0,\qquad u(1)=3 u(\frac{1}{9}), \end{cases} $$

(4.1)

where

$$ f(t,x)=\frac{(t-\frac{1}{2})^{2}}{\sqrt{t(1-t)}} \bigl[x^{ \frac{1}{2}}+x^{-\frac{1}{2}} \bigr]. $$

For any $t\in [0,+\infty )$, noticing the function $\varGamma (\cdot )$ is strictly increasing on $[2,+\infty )$, we have

$$ \begin{aligned}[b] g(t) &=-\frac{1}{2\sqrt{\pi }}+t+\sum ^{+\infty }_{k=2}\frac{t^{k}}{ \varGamma (\frac{3}{2}k-\frac{1}{2})} \\ & \leq -\frac{1}{2\sqrt{\pi }}+t+ \sum^{+\infty }_{k=2} \frac{t^{k}}{\varGamma (k)} =-\frac{1}{2\sqrt{ \pi }}+t \Biggl[1+\sum ^{+\infty }_{k=1}\frac{t^{k}}{k!} \Biggr] \\ & =-\frac{1}{2\sqrt{ \pi }}+te^{t}. \end{aligned} $$

Therefore $g(\frac{1}{5})\leq -\frac{1}{2\sqrt{\pi }}+\frac{1}{5}e ^{\frac{1}{5}}\thickapprox -0.282+ 0.243= -0.039<0$, which implies $\frac{1}{5}< b^{\ast }$.

For any $R\geq r>0$, $\forall t \in (0,1)$, $x\in [ r\sqrt{t},R]$, we have

$$ f(t,x)\leq \frac{(t-\frac{1}{2})^{2}}{\sqrt{t(1-t)}} \bigl[\sqrt{R}+r ^{-\frac{1}{2}}t^{-\frac{1}{4}} \bigr]. $$

It is clear that

$$ f\bigl(t,t^{-\frac{1}{2}}x\bigr)=\frac{(t-\frac{1}{2})^{2}}{\sqrt{t(1-t)}} \bigl[t^{-\frac{1}{4}}x^{\frac{1}{2}}+t^{\frac{1}{4}}x^{-\frac{1}{2}} \bigr], \quad \forall (t,x)\in (0,1)\times (0,+\infty ). $$

Therefore, we have

$$\begin{aligned}& f\bigl(t,t^{-\frac{1}{2}}x\bigr)\geq (1-t)^{-\frac{1}{2}}\biggl(t- \frac{1}{2}\biggr)^{2}t ^{-\frac{1}{4}}x^{-\frac{1}{2}}, \quad \forall (t,x)\in (0,1)\times (0,1]; \\& f\bigl(t,t^{-\frac{1}{2}}x\bigr)\leq \frac{(t-\frac{1}{2})^{2}}{\sqrt{t(1-t)}} \bigl[t^{-\frac{1}{4}}+t ^{\frac{1}{4}} \bigr]x^{\frac{1}{2}},\quad \forall (t,x)\in (0,1) \times [1,+\infty ). \end{aligned}$$

Denote

$$ b_{3}(t)=(1-t)^{-\frac{1}{2}} \biggl(t-\frac{1}{2} \biggr)^{2}t^{- \frac{1}{4}},\qquad b_{4}(t)= \frac{(t-\frac{1}{2})^{2}}{ \sqrt{t(1-t)}} \bigl(t^{-\frac{1}{4}}+t^{\frac{1}{4}} \bigr). $$

Let

$$\begin{aligned}& L_{3}u(t)= \int _{0}^{1}K^{\ast }(t,s)b_{3} (s)u(s)\,ds, \\& L_{4}u(t)= \int _{0}^{1}K^{\ast }(t,s)b_{4} (s)u(s)\,ds. \end{aligned}$$

It follows from Lemma 2.7 that $r(L_{3}),r(L_{4})>0$.

Set

$$ r_{1}=\min \bigl\{ \bigl[r(L_{3})\bigr]^{\frac{2}{3}},1 \bigr\} ,\qquad r_{2}=1+\bigl[r(L_{4})\bigr]^{2}, $$

and

$$ b_{1}(t)=r_{1}^{-\frac{3}{2}}b_{3}(t), \qquad b_{2}(t)=r_{2}^{- \frac{1}{2}}b_{4}(t). $$

Then

$$\begin{aligned}& f\bigl(t,t^{-\frac{1}{2}}x\bigr)\geq b_{1}(t)x,\quad \forall (t,x)\in (0,1) \times \bigl(0,r_{1}]; \\& f\bigl(t,t^{-\frac{1}{2}}x\bigr)\leq b_{2}(t)x,\quad \forall (t,x) \in (0,1) \times [r_{2},+\infty \bigr). \end{aligned}$$

It is easy to see that

$$\begin{aligned}& r(L_{1})=r_{1}^{-\frac{3}{2}}r(L_{3}) \geq 1, \\& 0< r(L_{2})= r_{2}^{-\frac{1}{2}}r(L_{4})< 1. \end{aligned}$$

So the assumptions of Theorem 3.1 are satisfied. Thus Theorem 3.1 ensures that FBVP (4.1) has at least one positive solution.

5 Conclusions

In this article, we consider a class of Riemann–Liouville type two-term fractional nonlocal boundary value problems for the case that $1< \alpha <2$. Some new properties of the Green function have been discovered to construct an exact cone. By using fixed point index theory on the exact cone, the existence and multiplicity of positive solutions are established. The nonlinearity $f(t,x)$ permits a singularity at $t = 0,1$ and $x=0$.