In this paper, we study the existence and multiplicity of positive solutions for the fractional boundary value problem

$ \begin{equation}\label{problem} \begin{cases} \textbf{D}_{0+}^\alpha u(t)+f(t, u(t))=0, \ 0 < t < 1, \\ u^{(i)}(0)=u(1)=0, \ 0\le i\le n-2, \end{cases} \end{equation} $

(1.1)

where $n \in \mathbb{N}$ and $n\ge 3$, $\alpha\in (n-1, n]$ is a real number, $\textbf{D}_{0+}^\alpha$ is the standard Riemann-Liouville fractional derivative of order $\alpha$ and $f\in C([0,1]\times [0, +\infty), [0, +\infty))$.

In view of fractional differential equation's modeling capabilities in engineering, science, economy, and other fields, the last few decades has resulted in a rapid development of the theory of fractional differential equation, see the recent books [1-5]. This may explain the reason that the last few decades have witnessed an overgrowing interest in the research of such problems, with many papers in this direction published. Recently, there are some papers dealing with the existence of solutions (or positive solutions) of nonlinear fractional differential equation by the use of techniques of nonlinear analysis (fixed point theorems, Leray-Schauder theory, upper and lower solution method, etc.), see for example [6-11] and the references therein.

In [6], Zhao et al. considered the existence on multiple positive solutions for the nonlinear fractional differential equation boundary value problem

$ \begin{equation} \begin{cases} \textbf{D}_{0+}^\alpha u(t)+f(t, u(t))=0, \ 0 < t < 1, \\ u(0)=u'(0)=u'(1)=0, \end{cases} \end{equation} $

(1.2)

where $2 < \alpha \le 3$ is a real number, $\textbf{D}_{0+}^\alpha$ is the Riemann-Liouville fractional derivative. By the properties of the Green function, the lower and upper solution method and fixed point theorem, some new existence criteria for singular and nonsingular fractional differential equation boundary value problem are established.

Meanwhile, we also note that conjugate boundary value problem of integer order differential equation has been extensively studied, see [12-15] and the references therein. However, to the best of our knowledge, there is very little known about the existence of positive solutions for fractional conjugate boundary value problem (1.1). To our knowledge, only [9-11] were devoted to this direction. In [9], Yuan considered the semipositone conjugate fractional boundary value problem (1.1) with a parameter $\lambda$ and $f: [0,1]\times [0, +\infty)\to (-\infty, +\infty)$ is a sign-changing continuous function. He first given the properties of Green's function of (1.1), and then derived an interval of $\lambda$ such that any $\lambda$ lying in this interval, the semipositone boundary value problem (1.1) has multiple positive solutions. In [10, 11] He adopted the same method in [9] to discuss the existence of multiple positive solutions for $(n-1, 1)$-type semipositone conjugate and integral boundary value problems for coupled systems of nonlinear fractional differential equations, respectively.

In this paper, we utilize Krasnoselskii-Zabreiko fixed point theorem to establish our main results based on a priori estimates achieved by developing some inequalities associated with fractional Green's function. It is well known that a cone plays a very important role involving the existence of solutions (positive solutions) for differential equations. It is difficult to structure a cone for speciality of Green's function for fractional equation. In this work, we first study the properties of the Green's function and obtain an inequality about it, and then structure a cone associated with the inequality. Based on this, we obtained some easily verifiable sufficient criteria to ensure the existence and multiplicity of positive solutions for (1.1). Thus our results improve and extend the corresponding ones in [6-11].

As is known to all, the Riemann-Liouville fractional derivative $\textbf{D}^\alpha_{0+}$ is defined by

$ \textbf{D}^\alpha_{0+}y(t):=\frac{1}{\Gamma (n-\alpha)}\left(\frac{\rm{d}}{{\rm{d}} t}\right)^n\int_0^t\frac{y(s){\rm{d}} s}{(t-s)^{\alpha-n+1}}, $

where $\Gamma$ is the gamma function and $n=[\alpha]+1$. For more details of fractional calculus, we refer the reader to the recent books such as [1-5].

Lemma 2.1(see [9, Lemma 3.1])  Let $f$ be determined by (1.1). Then the problem (1.1) is equivalent to

$ u(t)=\int_0^1 G(t, s)f(s, u(s)){\rm{d}} s, $

where

$ \begin{equation}\label{G}G(t, s):=\frac{1}{\Gamma(\alpha)}\begin{cases} t^{\alpha-1}(1-s)^{\alpha-1}-(t-s)^{\alpha-1},& 0\le s\le t\le 1, \\ t^{\alpha-1}(1-s)^{\alpha-1},& 0\le t\le s\le 1. \end{cases}\end{equation} $

(2.1)

Here $G(t, s)$ is called the Green's function for the boundary value problem (1.1).

Lemma 2.2(see [9, Lemma 3.2])   $G(t, s)\in C([0,1]\times [0,1], [0, +\infty))$ has the following properties:

(R1) $G(t, s)=G(1-s, 1-t)$, for $(t, s)\in [0,1]\times [0,1]$,

(R2) $\Gamma{(\alpha)}k(t)q(s)\le G(t, s)\le (\alpha-1)q(s)$, for $(t, s)\in [0,1]\times [0,1]$,

(R3) $\Gamma{(\alpha)}k(t)q(s)\le G(t, s)\le (\alpha-1)k(t)$, for $(t, s)\in [0,1]\times [0,1]$, where

$ \begin{equation}\label{kq}k(t)=\frac{t^{\alpha-1}(1-t)}{\Gamma{(\alpha)}}, \quad q(s)=\frac{s(1-s)^{\alpha-1}}{\Gamma{(\alpha)}}.\end{equation} $

(2.2)

Lemma 2.3  Let $\mathcal {K}_1:= \frac{\alpha\Gamma{(\alpha+1)}}{\Gamma{(2\alpha+2)}}\ \text{and} \ \mathcal {K}_2:= \frac{\alpha-1}{\Gamma{(\alpha+2)}}$. Then the following inequality holds:

$ \begin{equation}\label{gin}\mathcal {K}_1q(s)\le \int_0^1 G(t, s)q(t){\rm{d}} t\le \mathcal {K}_2 q(s), \forall s\in [0,1].\end{equation} $

(2.3)

Let

$\begin{eqnarray*} E:=C[0,1], \ \ \|u\|:=\max\limits_{t\in [0,1]}|u(t)|, \ \ P:=\{u\in E:u(t)\geq 0, \forall t \in [0,1]\}.\end{eqnarray*} $

Then $(E, \|\cdot\|)$ becomes a real Banach space and $P$ is a cone on $E$. We denote $B_\rho :=\{u \in E: \|u\| <\rho\}$ for $\rho> 0$ in the sequel. Now, note that $u$ solves (1.1) if and only if $u$ is a fixed point of the operator

$ (Au)(t):=\int_0^1 G(t, s)f(s, u(s)){\rm{d}} s. $

Clearly, $A:P\to P$ is a completely continuous operator. Define the completely continuous linear operators $L:E\to E$ by

$ (Lu)(t):=\int_0^1 G(t, s)u(s){\rm{d}} s. $

Then $L$ is also a positive operator, i.e. $L(P) \subset P$. Let $\omega=(\alpha-1)^{-1}\mathcal {K}_1 $ and a cone on $E$ as follows

$ P_{0}=\left\{u\in P:\int_0^1 u(t)q (t){\rm{d}} t\geq \omega\|u\|\right\}. $

Next, we shall prove that

Lemma 2.4   $L(P)\subset P_0$.

Proof  By (R2) of Lemma 2.2, we have

$ (Lu)(t)=\int_0^1 G(t, s)u(s){\rm{d}} s\le (\alpha-1)\int_0^1 q(s)u(s){\rm{d}} s. $

On the other hand, from (2.3), we find

$ \begin{align} & \int_{0}^{1}{(Lu)}(t)q(t)\rm{d}\mathit{t}=\int_{0}^{1}{\left( \int_{0}^{1}{\mathit{G}}(\mathit{t}, \mathit{s})\mathit{u}(\mathit{s})\rm{d}\mathit{s} \right)}\mathit{q}(\mathit{t})\rm{d}\mathit{t} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ge {{\mathcal{K}}_{1}}\int_{0}^{1}{q}(s)u(s)\rm{d}\mathit{s}\ge \frac{{{\mathcal{K}}_{1}}}{\alpha -1}\|\mathit{Lu}\| \\ \end{align} $

and thus $\int_0^1 (L u)(t)q(t) {\rm{d}} t\ge \omega\|Lu\|$, that is, $L(P)\subset P_0$. This completes the proof.

From now on, let $r(L)$ denote the spectral radii of $L$. By Gelfand's theorem, we could easily have the following result.

Lemma 2.5   $0 < r(L)\le \mathcal {K}_2$.

Lemma 2.6(see [16])  Let $E$ be a real Banach space and $W$ a cone of $E$. Suppose that $A:(\overline{B}_R\backslash B_r)\cap W\to W$ is a completely continuous operator with $0 < r < R$. If either

(1) $Au\nleqslant u$ for each $\partial B_r\cap W$ and $Au\ngeqslant u$ for each $\partial B_R\cap W$ or

(2) $Au\ngeqslant u$ for each $\partial B_r\cap W$ and $Au\nleqslant u$ for each $\partial B_R\cap W$.

Then $A$ has at least one fixed point on $(\overline{B}_R\backslash B_r)\cap W$.

For convenience, we now list our hypotheses on $f$. Let $\lambda_1:=1/\mathcal {K}_1>0, \ \lambda_2:=1/\mathcal {K}_2>0$.

(H1) $\liminf\limits_{u\to \infty}\frac{f(t, u)}{u}> \lambda_1$ uniformly with respect to $t\in [0,1]$.

(H2) $\limsup\limits_{u\to 0^+}\frac{f(t, u)}{u} < \lambda_2$ uniformly with respect to $t\in [0,1]$.

(H3) $\liminf\limits_{u\to 0^+}\frac{f(t, u)}{u}> \lambda_1$ uniformly with respect to $t\in [0,1]$.

(H4) $ \limsup\limits_{u\to \infty}\frac{f(t, u)}{u} < \lambda_2$ uniformly with respect to $t\in [0,1]$.

(H5) There exists a number $\rho> 0$ such that the inequality $f(t, u)\le \zeta\rho$ holds whenever $u\in [0, \rho]$, $\zeta\in(0, (\alpha-1)^{-1}\Gamma(\alpha+2))$, and $t\in [0,1]$.

Theorem 3.1  Suppose that (H1) and (H2) are satisfied, then (1.1) has at least one positive solution.

Proof  By (H1), there exist $\varepsilon > 0$ and $b > 0$ such that $f(t, u)\geq (\lambda_1 + \varepsilon)u - b$ for all $u\geq 0$ and $t\in [0,1]$. This implies

$ \begin{equation}\label{Ineq-2}(Au)(t)\geq (\lambda_1+\varepsilon)\int_0^1G(t, s) u(s){\rm{d}} s- b\int_0^1G(t, s) {\rm{d}} s\end{equation} $

(3.1)

for all $u\in P$. Let

$ \mathbf{M}_1:=\{u\in P: u\ge Au\}. $

We shall prove that $\mathbf M_1$ is bounded in $P$. Indeed, $u\in \mathbf{M}_1$, along with (3.1), leads to

$ u(t)\geq (\lambda_1+\varepsilon)\int_0^1G(t, s) u(s){\rm{d}} s- b\int_0^1G(t, s) {\rm{d}} s. $

Multiply by $q(t)$ on both sides of the above and integrate over $[0,1]$ and use (2.3) to obtain

$ \int_0^1 u(t) q(t){\rm{d}} t \geq (\lambda_1 + \varepsilon)\lambda_1^{-1}\int_0^1 u(t)q(t){\rm{d}} t-b\lambda_2^{-1}\int_0^1 q(t){\rm{d}} t $

and thus $\int_0^1 u(t)q(t){\rm{d}} t\leq \frac{\varepsilon^{-1}b(\alpha-1)\Gamma{(2\alpha+2)}}{\alpha\Gamma{(\alpha+1)}\Gamma^2{(\alpha+2)}}$ for all $u\in\mathbf M_1$. Note that we have $\mathbf M_1\subset P_0$ by Lemma 2.4. This together with the preceding inequality implies

$ \|u\|\leq \frac{\varepsilon^{-1}b(\alpha-1)^2\Gamma^2{(2\alpha+2)}}{\alpha^2\Gamma^2{(\alpha+1)}\Gamma^2{(\alpha+2)}} $

for all $u\in\mathbf M_1$, which establishes the boundedness of $\mathbf M_1$, as required. Taking $R> \sup \mathbf M_1$, we obtain

$ \begin{equation}\label{1}u\ngeqslant Au, \ \forall u\in \partial B_R\cap P.\end{equation} $

(3.2)

On the other hand, by (H2), there exist $r\in (0, R)$ and $\varepsilon\in (0, \lambda_2)$ such that $f(t, u) \leq (\lambda_2-\varepsilon)u$ for all $u\in [0, r]$ and $t\in [0,1]$. This implies

$ \begin{equation}\label{Ineq}(Au)(t)\leq (\lambda_2-\varepsilon)\int_0^1G(t, s) u(s){\rm{d}} s \end{equation} $

(3.3)

for all $u\in\overline{B}_r\cap P$. Let

$ \mathbf{M}_2:=\{u\in \overline{B}_r\cap P: u\le Au \}. $

Now, we claim $\mathbf{M}_2=\{0\}$. Indeed, if there exist $u_0 \in \partial B_r \cap P$, then this together with (3.3) leads to

$ u_0(t)\leq (\lambda_2-\varepsilon)\int_0^1G(t, s) u_0(s){\rm{d}} s. $

Multiply by $q(t)$ on both sides of the preceding inequality and integrate over $[0,1]$ and use (2.3) to obtain

$ \int_0^1 u_0(t)q(t){\rm{d}} t \leq (\lambda_2-\varepsilon)\lambda_2^{-1} \int_0^1 u_0(t)q(t){\rm{d}} t $

and thus $\int_0^1 u_0(t)q(t){\rm{d}} t=0$, whence $u_0(t)\equiv 0$, contradicting $u_0 \in \partial B_r \cap P$. Therefore,

$ \begin{equation}\label{2}u \nleqslant Au, \forall u\in \partial B_r\cap P.\end{equation} $

(3.4)

Now Lemma 2.6 indicates that the operator $A$ has at least one fixed point on $(B_R\setminus \overline{B}_r)\cap P$. Therefore (1.1) has at least one positive solution, which completes the proof.

Theorem 3.2  If (H3) and (H4) are satisfied, then (1.1) has at least one positive solution.

Proof  By (H3), there exist $r>0$ and $\varepsilon>0$ such that

$ \begin{equation}\label{sublinearly-1} f(t, u)\geq (\lambda_1+\varepsilon) u, \text{ for all }u\in [0, r]\text{ and }t\in [0,1].\end{equation} $

(3.5)

This implies

$ \begin{equation}\label{Ineq-1} (Au)(t)\geq(\lambda_1+\varepsilon)\int_0^1G(t, s) u(s){\rm{d}} s \end{equation} $

(3.6)

for all $u\in \overline B_{r}\cap P$. Let $\mathbf{M}_3:= \{u\in \overline{B}_r\cap P: u\ge Au\}.$ We claim that $\mathbf{M}_3=\{0\}$. Indeed, if the claim is false, then there exists $u_1\in \partial B_{r}\cap P$ such that $u_1\geq Au_1$. Combining with (3.6), we obtain

$ u_1(t)\geq (\lambda_1+\varepsilon)\int_0^1G(t, s) u_1(s){\rm{d}} s. $

Multiply by $q(t)$ on both sides of the above and integrate over [0,1] and use (2.3) to obtain

$ \int_0^1 u_1(t)q(t){\rm{d}} t \ge (\lambda_1+\varepsilon)\lambda_1^{-1}\int_0^1 u_1(t)q(t){\rm{d}} t $

and thus $\int_0^1 u_1(t)q(t){\rm{d}} t= 0$, whence $u_1(t)\equiv 0$, contradicting $u_1 \in \partial B_r \cap P$. Consequently,

$ \begin{equation}\label{3}u\ngeqslant Au, \forall u\in \partial B_r\cap P.\end{equation} $

(3.7)

In addition, by (H4), there exist $\varepsilon\in (0, \lambda_2)$ and $m>0$ such that

$ \begin{equation}\label{sublinearly-2}f(t, u)\leq (\lambda_2-\varepsilon) u+m, \text{ for all }u\geq 0\text{ and }t\in [0,1].\end{equation} $

(3.8)

Let $ \mathbf{M}_4:=\{u\in P: \ u\le Au\}. $ We shall prove that $\mathbf{M}_4$ is bounded in $P$. Indeed, if $u\in \mathbf{M}_4$, then we have

$ \begin{equation} u(t)\le (Au)(t) = \int_0^1G(t, s) f(s, u(s)){\rm{d}} s\\ \leq \int_0^1G(t, s) ((\lambda_2-\varepsilon) u(s)+m){\rm{d}} s=(\lambda_2-\varepsilon)(Lu)(t)+u_0(t), \end{equation} $

(3.9)

where $u_0\in P\setminus\{0\}$ being defined by $u_0(t)=m\int_0^1G(t, s) {\rm{d}} s$. Notice $r((\lambda_2-\varepsilon)L)<1$ by Lemma 2.5. This implies the inverse operator of $I-(\lambda_2-\varepsilon) L$ exists and equals

$ (I-(\lambda_2-\varepsilon) L)^{-1}=I+(\lambda_2-\varepsilon)L+(\lambda_2-\varepsilon)^2L^2+\cdots+(\lambda_2-\varepsilon)^nL^n+\cdots, $

from which we obtain $(I-(\lambda_2-\varepsilon)L)^{-1}(P)\subset P$. Applying this to (3.9) gives $u\leq (I-(\lambda_2-\varepsilon)L)^{-1}u_0$ for all $u\in \mathbf M_4$. This proves the boundedness of $\mathbf{M}_4$, as required. Choosing $R>\sup \{\| u \|: u\in\mathbf M_4\}$ and $R>\rho$, we have

$ \begin{equation}\label{4}u\nleqslant Au, \ \forall u\in\partial B_R\cap P.\end{equation} $

(3.10)

Now Lemma 2.6 implies that $A$ has at least one fixed point on $(B_R\setminus \overline{B}_r)\cap P$. Therefore (1.1) has at least one positive solution, which completes the proof.

Theorem 3.3  Suppose that (H1), (H3) and (H5) hold, then (1.1) has at least two positive solutions.

Proof  By (H5), we have

$\begin{eqnarray*} \|Au\|&=&\max\limits_{t\in [0,1]} \int_0^1 G(t, s)f(s, u(s)){\rm{d}} s\\ &\le& (\alpha-1)\zeta\rho\int_0^1 q(s){\rm{d}} s = \frac{(\alpha-1)\zeta\rho}{\Gamma{(\alpha+2)}} < \rho=\|u\|\end{eqnarray*} $

for all $u\in \partial B_\rho \cap P$, from which we obtain

$ \begin{equation}\label{5}u\nleqslant Au, \forall u\in \partial B_\rho\cap P.\end{equation} $

(3.11)

On the other hand, by (H1) and (H3), we may take $R>\rho$ and $r\in (0, \rho)$ so that (3.2) and (3.7) hold (see the proofs of Theorems 3.1 and 3.2). Combining (3.2), (3.7) and (3.11), we conclude, together with Lemma 2.6, $A$ has at least two fixed points, one on $(B_R\backslash \overline{B}_{\rho})\cap P$ and the other on $(B_\rho\backslash \overline{B}_{r})\cap P$. Hence (1.1) has at least two positive solutions in $ P \backslash \{0\}$. This completes the proof.

In this section, we offer some interesting examples to illustrate our main results.

Example 4.1  Let $f(t, u)=u^\alpha, t\in[0,1], u\in\mathbb R^+$, where $\alpha\in (0, 1)\cup (1, \infty)$. If $\alpha\in (1, \infty)$, then (H1) and (H2) are satisfied. If $\alpha\in (0, 1)$, then (H3) and (H4) are satisfied. By Theorems 3.1 or 3.2, (1.1) has at least one positive solution.

Example 4.2  Let

$ f(t, u)=\left\{ \begin{align} & \frac{{{\lambda }_{2}}}{2}{{u}^{\alpha }}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\le u\le 1, \\ & 2{{\lambda }_{1}}{{u}^{\alpha }}-2{{\lambda }_{1}}+\frac{{{\lambda }_{2}}}{2}, \ \ \ \ \ \ \ \ u\ge 1, \\ \end{align} \right. $

where $\alpha\ge 1$. Now (H1) and (H2) are satisfied. By Theorem 3.1, (1.1) has at least one positive solution.

Example 4.3  Let

$ f(t, u)=\left\{ \begin{align} & 2{{\lambda }_{1}}{{u}^{\beta }}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\le u\le 1, \\ & \frac{{{\lambda }_{2}}}{2}{{u}^{\beta }}+2{{\lambda }_{1}}-\frac{{{\lambda }_{2}}}{2}, \ \ \ \ \ \ u\ge 1, \\ \end{align} \right. $

where $0<\beta\le 1$. Now (H3) and (H4) are satisfied. By Theorem 3.2, (1.1) has at least one positive solution.

Example 4.4  Let $f(t, u)=\lambda (u^a+u^b)$, where $0<a<1<b$, $0<\lambda< (\alpha-1)^{-1}\Gamma(\alpha+2)$. (H1), (H3) and (H5) are satisfied with $\rho=1$. By Theorem 3.3, (1.1) has at least two positive solutions.