Existence of Positive Solutions for a Class of Higher-Order Caputo Fractional Differential Equation

Abstract

In this paper, the following nonlinear fractional ordinary differential boundary value problem

$$\begin{aligned} \left\{ \begin{array}{ll} D_{0+}^{\alpha } u(t) +\lambda f(t,u(t))+ q(t)=0, &{} 0<t<1, \\ u(0) = u'(1) =u''(0)= \dots = u^{(n-1)}(0)=0, &{} \\ \end{array} \right. \end{aligned}$$

is considered, where $\alpha (n-1 <\alpha \le n)$ is a real number. $\lambda > 0$ is a parameter. $D_{0+}^{\alpha }$ is the standard Caputo differentiation. Some sufficient conditions for the existence of positive solutions to this boundary value problem of nonlinear fractional differential equation are established by nonlinear alternative of Leray–Schauder type and Guo–Krasnoselskii fixed point theorem on cones. As applications, some examples are provided to illustrate our main results.

1 Introduction

In recent years, the fractional differential equations have been of great attention. It is caused both by the intensive development of the theory of fractional calculus itself and by its broader applications such as physics, chemistry, biology, economics, control theory, signal and image processing, biophysics, blood flow phenomena, aerodynamics, fitting of experimental data, etc. There are a large number of papers dealing with the existence or multiplicity of solutions or positive solutions of initial or boundary value problem for some nonlinear fractional differential equations. For details and examples, see [1–6, 8–10, 12, 14–18] and the references therein.

Bai and Qiu [2] discussed the existence of positive solutions to boundary value problem of the following nonlinear fractional differential equation by using Krasnoselskii’s fixed point theorem in a cone and nonlinear alternative of Leray–Schauder

$$\begin{aligned} \left\{ \begin{array}{ll} D_{0^+}^{\alpha }u(t)+f(t,u(t))=0, &{} 0<t<1, \\ u(0)=u'(1)=u''(0)=0, &{} \\ \end{array} \right. \end{aligned}$$

where $\alpha \,(2<\alpha \le 3)$ is a real number. $D_{0^+}^{\alpha }$ is the Caputo’s differentiation. $f:(0,1]\times [0,+\infty )\rightarrow [0,+\infty )$ is singular at $t = 0.$

Zhao [17] investigated the existence of positive solutions for the nonlinear fractional differential boundary value problem

$$\begin{aligned} \left\{ \begin{array}{ll} D_{0^+}^{\alpha }u(t)=\lambda f(u(t)), &{} 0<t<1, \\ u(0)+u'(0)=0, \quad u(1)+u'(1)=0, &{} \\ \end{array} \right. \end{aligned}$$

where $\alpha \,(1<\alpha \le 2)$ is a real number. $D_{0^+}^{\alpha }$ is the standard Caputo fractional derivative. $\lambda >0$ is a parameter. $f:[0,+\infty )\rightarrow (0,+\infty )$ is continuous. By using the properties of the Green’s function and Guo–Krasnoselskii fixed point theorem on cones, the eigenvalue intervals of the nonlinear fractional differential boundary value problem are considered. Some sufficient conditions for the nonexistence and existence of at least one or two positive solutions for this boundary value problem are established.

In this paper, motivated by the above mentioned works, we will study the the existence of at least one or two positive solutions for the following higher-order nonlinear fractional differential boundary value problem (1.1) (abbreviated by BVP (1.1) throughout this paper)

$$\begin{aligned} \left\{ \begin{array}{ll} D_{0+}^{\alpha } u(t) +\lambda f(t,u(t))+ q(t)=0, &{} 0<t<1, \\ u(0)= u'(1) =u''(0)= \dots = u^{(n-1)}(0)=0, &{} \\ \end{array} \right. \end{aligned}$$

(1.1)

where $D_{0^+}^{\alpha }$ is the Caputo fractional derivative of order $\alpha \,(n-1<{\alpha }\le n,\,n\ge 3).$ $f(t,u)\in C([0,1]\times [0,+\infty )\rightarrow [0,+\infty )).$ $\lambda >0$ is a parameter. $q(t)\in C([0,1],[0,+\infty )).$

The rest of this paper is organized as follows. In Sect. 2, we will state some useful definitions, lemmas and the properties of the Green’s function for BVP (1.1). In Sect. 3, some sufficient conditions will be established to guarantee the existence of positive solutions for BVP (1.1). Finally, some examples are also provided to illustrate our main results in Sect. 4.

2 Preliminaries

For the convenience of the reader, we present here the necessary definitions and lemmas.

Definition 2.1

(see [11, 13]) The Riemann–Liouville fractional integral of order $\alpha >0$ of a function $f:(0,+\infty )\rightarrow \mathbb {R}$ is given by

$$\begin{aligned} I_{0+}^{\alpha }f(t)=\frac{1}{\Gamma (\alpha )}\int _0^t(t-s)^{\alpha -1}f(s)ds, \end{aligned}$$

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

Definition 2.2

(see [11, 13]) The Caputo fractional derivative of order $\alpha >0$ of a continuous function $f:(0,+\infty )\rightarrow \mathbb {R}$ is given by

$$\begin{aligned} D_{0+}^{\alpha }f(t)=\frac{1}{\Gamma (n-\alpha )}\int _0^t\frac{f^{(n)}(s)}{(t-s)^{\alpha -n+1}}ds, \end{aligned}$$

where $n-1<{\alpha }\le n,$ provided that the right-hand side is pointwise defined on $(0,+\infty )$.

Lemma 2.3

(see [11]) Assume that $u\in C(0,1)\cap L(0,1)$ with a Caputo fractional derivative of order $\alpha >0$ that belongs to $u\in C^n[0,1],$ then

$$\begin{aligned} I_{0+}^{\alpha }D_{0+}^{\alpha }u(t)=u(t)+C_1+C_2t+\dots +C_nt^{n-1}, \end{aligned}$$

for some $C_{i} \in \mathbb {R}\,(i=1,2,\ldots ,n),$ where n is the smallest integer greater than or equal to $\alpha .$

Lemma 2.4

(see [19]) Let $E$ be a Banach space with $C\subseteq E$ closed and convex. Assume $U$ is a relatively open subset of $C$ with $\theta \in U$ and $T:\overline{U}\rightarrow C$ is a continuous compact map. Then either

• (a) $T$ has a fixed point in $\overline{U},$ or

• (b) there exists $u\in \partial U$ and $\lambda \in (0,1)$ with $u=\lambda Tu.$

Lemma 2.5

(see [7]) Let $E$ be a Banach space, $P\subseteq E$ is a cone, and $\Omega _1$, $\Omega _2$ are two bounded open balls of $E$ centered at the origin with $\theta \in \Omega _1$ and $\overline{\Omega }_1\subset \Omega _2$. Suppose that $A:P\cap (\overline{\Omega }_2\!\setminus \!\Omega _1)\rightarrow P$ is a completely continuous operator such that either

• (i) $\Vert Au\Vert \le \Vert u\Vert$, $u\in P\cap \partial \Omega _1$ and $\Vert Au\Vert \ge \Vert u\Vert$, $u\in P\cap \partial \Omega _2$, or

• (ii) $\Vert Au\Vert \ge \Vert u\Vert$, $u\in P\cap \partial \Omega _1$ and $\Vert Au\Vert \le \Vert u\Vert$, $u\in P\cap \partial \Omega _2$

holds. Then $A$ has at least one fixed point in $P\cap (\overline{\Omega }_2\!\setminus \!\Omega _1)$.

Now we present the Green’s function associated with BVP (1.1).

Lemma 2.6

Given $h\in C[0,1],$ and $n-1 < \alpha \le n,$ the unique solution of

$$\begin{aligned} \left\{ \begin{array}{ll} D_{0^+}^{\alpha }u(t)+h(t)=0, &{} {0<t<1,} \\ u(0) = u'(1) = u''(0)=\dots = u^{(n-1)}(0)=0, &{} \\ \end{array} \right. \end{aligned}$$

(2.1)

is

$$\begin{aligned} u(t)=\int _{0}^{1}G(t,s)h(s)ds, \end{aligned}$$

where

$$\begin{aligned} G(t,s)= \left\{ \begin{array}{ll} \frac{(\alpha -1)t(1-s)^{\alpha -2}-(t-s)^{\alpha -1}}{\Gamma (\alpha )}, &{} 0\le s\le t\le 1, \\ \frac{t(1-s)^{\alpha -2}}{\Gamma (\alpha -1)}, &{} 0\le t\le s\le 1. \\ \end{array} \right. \end{aligned}$$

(2.2)

Proof

By Lemma 2.3 and (2.1), we have

$$\begin{aligned} u(t)&=-I_{0+}^{\alpha }h(t)+C_1+C_{2}t+C_{3}t^2+\dots +C_{n}t^{^{n-1}}\\&=-\frac{1}{\Gamma (\alpha )}\int _0^t(t-s)^{\alpha -1}h(s)ds+C_1+C_{2}t+C_{3}t^2+\dots +C_{n}t^{^{n-1}}. \end{aligned}$$

for some $C_{i}\in \mathbb {R} ,\ \ i=1,2,\ldots ,n.$  From $u(0)=$ $u''(0) =$ $\dots = u^{(n-1)}(0)=0,$ we can obtain $C_{1}=C_{3}=\dots =C_{n}=0.$ Then

$$\begin{aligned} u(t)&=-\frac{1}{\Gamma (\alpha )}\int _0^t(t-s)^{\alpha -1}h(s)ds+C_{2}t. \end{aligned}$$

By $u'(1)=0$, we get

$$\begin{aligned} C_{2}=I_{0+}^{\alpha -1}h(t)=\frac{1}{\Gamma (\alpha -1)}\int _0^1(1-s)^{\alpha -2}h(s)ds. \end{aligned}$$

Therefore, the unique solution of BVP (2.1) is

$$\begin{aligned} u(t)&=-\frac{1}{\Gamma (\alpha )}\int _0^t(t-s)^{\alpha -1}h(s)ds+\frac{1}{\Gamma (\alpha -1)}\int _0^1t(1-s)^{\alpha -2}h(s)ds\\&=\int _0^t\left[ \frac{t(1-s)^{\alpha -2}}{\Gamma (\alpha -1)}-\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\right] h(s)ds+\int _t^1\frac{t(1-s)^{\alpha -2}}{\Gamma (\alpha -1)}h(s)ds\\&=\int _0^1G(t,s)h(s)ds, \end{aligned}$$

where G(t,s) is defined by (2.2). The proof is complete. $\square$

Lemma 2.7

the function $G(t,s)$ is defined by (2.2) have the following properties.

• (i) $G(t,s) \ge 0$ is continuous for all t, s $\in [0,1]$, $G(t,s)>0$ for all t, s $\in (0,1)$;

• (ii) $G(t,s) \le G(1,s)$ for each t, s $\in [0,1],$ and $G(t,s)\ge tG(1,s), \forall \,t, s\in [0,1]$.

Proof

(i) Obviously, $G(t,s)$ is continuous. When $0\le s\le t\le 1,$ in the light of $n-1<\alpha \le n,$ $n\ge 3,$ we can obtain

$$\begin{aligned} (\alpha -1)t(1-s)^{\alpha -2}\ge t(1-s)^{\alpha -2}\ge t(t-s)^{\alpha -2}\ge (t-s)^{\alpha -1}. \end{aligned}$$

Therefore, $G(t,s)\ge 0.$ When $0\le t\le s\le 1,$ it is easy to see that $G(t,s)\ge 0.$ Thus, we get $G(t,s)\ge 0$ for all $t, s \in [0,1],$ and $G(t,s)>0$ for all t, s $\in (0,1).$

(ii) On the one hand, since

$$\begin{aligned} \frac{\partial G(t,s)}{\partial t}= \left\{ \begin{array}{l@{\quad }l} \frac{(\alpha -1)[(1-s)^{\alpha -2}-(t-s)^{\alpha -2}]}{\Gamma (\alpha )}\ge 0, &{} 0\le s\le t\le 1, \\ \frac{(1-s)^{\alpha -2}}{\Gamma (\alpha -1)}\ge 0, &{} 0\le t\le s\le 1, \\ \end{array} \right. \end{aligned}$$

the function $G(t,s)$ is nondecreasing for $t\in [0,1],$ we have $G(t,s)\le G(1,s)$. On the other hand, if $t\ge s,$ we have

$$\begin{aligned} \frac{G(t,s)}{G(1,s)}&=\frac{(\alpha -1)t(1-s)^{\alpha -2}-(t-s)^{\alpha -1}}{(\alpha -1)(1-s)^{\alpha -2}-(1-s)^{\alpha -1}}\\&\ge \frac{(\alpha -1)t(1-s)^{\alpha -2}-(t-s)(1-s)^{\alpha -2}}{(\alpha -1)(1-s)^{\alpha -2}-(1-s)(1-s)^{\alpha -2}}\\&=\frac{(\alpha -2)t+s}{\alpha -2+s}\ge \frac{(\alpha -2)t+ts}{\alpha -2+s}=t. \end{aligned}$$

If $t\le s,$ we have

$$\begin{aligned} \frac{G(t,s)}{G(1,s)}&=\frac{(\alpha -1)t(1-s)^{\alpha -2}}{(\alpha -1)(1-s)^{\alpha -2}-(1-s)^{\alpha -1}}\\&\ge \frac{(\alpha -1)t(1-s)^{\alpha -2}-(t-s)^{\alpha -1}}{(\alpha -1)(1-s)^{\alpha -2}-(1-s)^{\alpha -1}}\\&\ge t. \end{aligned}$$

Therefore, we have $G(t,s)\ge tG(1,s), \forall \,t, s\in [0,1].$ The proof is complete. $\square$

From Lemma 2.7, we can obtain the following useful inequality.

Corollary 2.8

Let the function $G(t,s)$ be defined by (2.2), then

$$\begin{aligned} \min _{a\le t\le b} G(t,s)\ge tG(1,s)\ge aG(1,s), \forall \,s\in [0,1], [a,b]\subset (0,1). \end{aligned}$$

(2.3)

3 Existence of Positive Solutions

In this section, we will discuss the existence of positive solutions for BVP (1.1).

Let $E= \{u(t):u(t)\in C[0,1]\}$ denote a real Banach space with the norm $\Vert \cdot \Vert$ defined by $\Vert u\Vert = \max _{0\le t\le 1}\big |u(t)\big |$. Define the cone $P\subset E$ by

$$\begin{aligned} P&=\{ u\in E:u(t)\ge 0 \}. \end{aligned}$$

Let

$$\begin{aligned} K&= \big \{ u\in P:u\ge 0, \min _{t\in [a,b]}u(t)\ge a\Vert u\Vert ,[a,b]\subset (0,1) \big \}, \end{aligned}$$

(3.1)

$$\begin{aligned} K_{r}&=\{u\in K:\Vert u\Vert < r\}, \,\, \partial K_{r}=\{u\in K:\Vert u\Vert \ =r \}. \end{aligned}$$

(3.2)

From Lemma 2.6, we can obtain the following lemma.

Lemma 3.1

Suppose that $f(t,u)$ is continuous, then $u\in E$ is a solution of boundary value problem (1.1) if and only if $u\in E$ is a solution of the integral equation

$$\begin{aligned} u(t)&=\int _0^1G(t,s)\left[ \lambda f(s,u(s))+q(s)\right] ds. \end{aligned}$$

Define $T:E\rightarrow E$ be the operator defined as

$$\begin{aligned} (Tu)(t)&=\int _0^1G(t,s)\left[ \lambda f(s,u(s))+q(s)\right] ds. \end{aligned}$$

(3.3)

Then by Lemma 3.1, the fixed point of operator $T$ defined by (3.3) coincides with the solution of system (1.1).

Lemma 3.2

Let $f(t,u)$ be continuous on $(0,1)\times [0,+\infty )\rightarrow [0,+\infty ),$ then $T:P \rightarrow P$ and $T:K \rightarrow K$ defined by (3.3) are completely continuous.

Proof

i) Let $u\in P,$ in view of nonnegativity and continuity of functions $G(t,s),$ $q(t)$, $f(t,u(t))$ and $\lambda >0,$ we conclude that $T:P \rightarrow P$ is continuous.

Let $\Omega \subset P$ be bounded, that is, there exists a positive constant $h>0$ such that $\Vert u\Vert \le h$ for all $u \in \Omega .$ Let

$$\begin{aligned} M&=\max \{\big |\lambda f(s,u(s))+q(s)\big |+1:0\le s \le 1,0\le u\le h\}. \end{aligned}$$

Then we have

$$\begin{aligned} \Vert Tu\Vert&=\max _{0\le t\le 1}(Tu)(t)\le \int _0^1G(1,s)\left[ \lambda f(s,u(s))+q(s)\right] ds\\&\le M\int _0^1G(1,s)ds. \end{aligned}$$

Hence, $T(\Omega )$ is uniformly bounded.

Since $G(t,s)$ is continuous on $[0,1]\times [0,1]$, it is uniformly continuous on $[0,1]\times [0,1]$. Thus, for fixed $s\in [0,1]$ and for any $\varepsilon >0$, there exists a constant $\delta > 0$, such that any $t_1, t_2 \in [0,1]$ and $|t_1-t_2| < \delta$,

$$\begin{aligned} |G(t_1,s)-G(t_2,s)|&< \frac{\varepsilon }{M}. \end{aligned}$$

Then

$$\begin{aligned} |Tu(t_2)-Tu(t_1)|&\le M\int _0^1\left| G(t_1,s)-G(t_2,s)\right| ds<\varepsilon . \end{aligned}$$

That is to say, $T(P)$ is equicontinuous. By the means of the Arzela-Ascoli theorem, we have $T:P \rightarrow P$ is completely continuous.

ii) For any $u \in K,$ Lemma 2.7 implies that

$$\begin{aligned} (Tu)(t)&\ge t\int _0^1G(1,s)\left[ \lambda f(s,u(s))+q(s)\right] ds, \ \ t\in [0,1]. \end{aligned}$$

On the other hand

$$\begin{aligned} \Vert Tu\Vert&=\max _{0\le t\le 1}(Tu)(t)\le \int _0^1G(1,s)\left[ \lambda f(s,u(s))+q(s)\right] ds. \end{aligned}$$

Then $(Tu)(t)\ge t\Vert Tu\Vert ,$ which implies $T(K)\subset K$.

According to the Ascoli-Arzela theorem, we can easily get that $T:K \rightarrow K$ is completely continuous operator. The proof is complete. $\square$

Theorem 3.3

Assume that $f(t,u)$ is continuous on $(0,1)\times [0,+\infty )\rightarrow [0,+\infty ),$ and for all $t\in [0,1],$ there exist a function $m(t)>0$ such that

$$\begin{aligned} \big | f(t,u_{1})-f(t,u_{2}) \big |\le&m(t)\big |u_{1}-u_{2}\big |, \end{aligned}$$

for all $t\in (0,1),u_{1},u_{2}\in [0,\infty ).$ Then BVP (1.1) has a unique positive solution if

$$\begin{aligned} 0<\lambda <\frac{1}{\int _0^1G(1,s)m(s)ds}. \end{aligned}$$

(3.4)

Proof

For all $u\in P,$ by the nonnegativeness of $G(t,s),$ $\lambda ,$ $q(t)$ and $f(t,u),$ we have $(Tu)(t)\ge 0.$ Hence, $T(P)\subset P$. Let $\rho =\lambda \int _0^1G(1,s)m(s)ds$, From Lemma 2.7, we obtain

$$\begin{aligned} \Vert Tu_{2}-Tu_{1}\Vert&=\max _{t\in [0,1]}\big |Tu_{2}-Tu_{1}\big |\\&=\max _{t\in [0,1]}\left| \int _0^1G(t,s) [\lambda f(s,u_{1}(s))+q(s)-\lambda f(s,u_{2}(s))-q(s)]ds\right| \\&\le \int _0^1 \left| G(1,s)\lambda [f(s,u_{1}(s))-f(s,u_{2}(s))]\right| ds\\&\le \lambda \int _0^1G(1,s)m(s)ds\Vert u_{1}-u_{2}\Vert \\&=\rho \Vert u_{1}-u_{2}\Vert . \end{aligned}$$

By Lemma 3.2, $T$ is completely continuous. In the light of (3.4), we have $\rho \in (0,1).$ Therefore, in view of Banach fixed point theorem, the operator $T$ has a unique fixed point in $P,$ which is the unique positive solution of BVP (1.1). This completes the proof. $\square$

Theorem 3.4

Assume that $q(t)$ is continuous on $(0,1)\rightarrow [0,+\infty ).$ $f(t,u)$ is continuous on $(0,1)\times [0,+\infty )\rightarrow [0,+\infty ).$ Further suppose that there exist $c_1(t)>0,c_2(t)>0,$ for all $t\in [0,1],$ such that the following conditions $(H_1)$ and $(H_2)$ hold.

• $(H_{1})$ $\big |f(t,u(t))\big |\le c_{1}(t)+c_{2}(t)\big |u(t)\big |.$

• $(H_{2})$ $0<\lambda < 1/C_1, C_2<\infty , C_3 <\infty$, where $C_{1}=\int _0^1G(1,s)c_{2}(s)ds,C_{2}=\int _0^1G(1,s)c_{1}(s)ds, \quad C_{3}=\int _0^1G(1,s)q(s)ds.$

Then BVP (1.1) has at least one positive solution $u$ belonging to

$$\begin{aligned} Q&= \left\{ u\in P | \ \Vert u\Vert < \frac{\lambda C_{2}+C_{3}}{1-\lambda C_{1}}\right\} . \end{aligned}$$

Proof

Let $Q= \left\{ u\in P :\Vert u\Vert < r \right\}$ with $r={(\lambda C_{2}+C_{3})}/{(1-\lambda C_{1})}.$ Define the operator $T:Q\rightarrow P$ as (3.3). Let $u\in Q,$ that is, $\Vert u\Vert < r.$ From Lemma 2.7, we obtain

$$\begin{aligned} \Vert Tu\Vert&=\max _{t\in [0,1]}\big |Tu(t)\big | =\max _{t\in [0,1]}\left| \int _0^1G(t,s)[\lambda f(s,u(s))+q(s)]ds\right| \\&\le \int _0^1G(1,s)\lambda ( c_{1}(s)+c_{2}(s)\big |u(s)\big |)ds+\int _0^1G(1,s)q(s)ds\\&\le \lambda \int _0^1G(1,s)c_{1}(s)ds+\lambda \int _0^1G(1,s)c_{2}(s)ds\Vert u\Vert +\int _0^1G(1,s)q(s)ds\\&=\lambda C_{2}+\lambda C_{1}\Vert u\Vert +C_{3}< r. \end{aligned}$$

Thus, $Tu\in \overline{Q}.$ From Lemma 3.2, we have $T:Q\rightarrow \overline{Q}$ is completely continuous.

Consider the eigenvalue problem

$$\begin{aligned} u&=\lambda ^* Tu, \quad \lambda ^* \in (0,1). \end{aligned}$$

(3.5)

Under the assumption that $u$ is a solution of (3.5) for a $\lambda ^*\in (0,1),$ one obtains

$$\begin{aligned} \Vert u\Vert&=\Vert \lambda ^* Tu\Vert =\lambda ^* \max _{t\in [0,1]}\left| \int _0^1G(t,s)[\lambda f(s,u(s))+q(s)]ds\right| \\&\le \int _0^1G(1,s)\lambda ( c_{1}(s)+c_{2}(s)\big |u(s)\big |)ds+\int _0^1G(1,s)q(s)ds\\&\le \lambda \int _0^1G(1,s)c_{1}(s)ds+\lambda \int _0^1G(1,s)c_{2}(s)ds\Vert u\Vert +\int _0^1G(1,s)q(s)ds\\&=\lambda C_{2}+\lambda C_{1}\Vert u\Vert +C_{3}< r. \end{aligned}$$

So $\Vert u\Vert \ne r$, which is contradiction with $u\in \partial Q$. According to Lemma 2.4, $T$ has a fixed point $u\in \overline{Q}.$ Therefore, BVP (1.1) has at least one positive solution. This completes the proof. $\square$

In the rest of the paper, for the convenience of presentation, we introduce some notations as follows.

$$\begin{aligned} F_\delta&=\limsup \limits _{u\rightarrow \delta } \max \limits _{t\in [0,1]}\frac{f(t,u)}{u},\ \ \ f_\delta =\liminf \limits _{u\rightarrow \delta } \min \limits _{t\in [0,1]} \frac{f(t,u)}{u}, \end{aligned}$$

where $\delta$ denotes 0 or $+\infty$, and let

$$\begin{aligned} A&= \int _0^1G(1,s)ds, \quad B = a^2\int _a^bG(1,s)ds, \quad C=\int _0^1G(1,s)q(s)ds. \end{aligned}$$

Theorem 3.5

Assume that $f_0B>F_\infty A$ holds, then for each

$$\begin{aligned} \lambda \in ((f_0B)^{-1},(F_\infty A)^{-1}), \end{aligned}$$

(3.6)

BVP (1.1) has at least one positive solution. $($Particularly, $(f_0B)^{-1}=0$ if $f_0=+\infty$ and $(F_\infty A)^{-1}=+\infty$ if $F_\infty =0\,).$

Proof

Let operator $T$ defined by (3.3) is completely continuous in the corresponding cone. Let $\lambda$ satisfy (3.6) and $\varepsilon >0$ be such that

$$\begin{aligned} ((f_0-\varepsilon )B)^{-1}\le \lambda \le ((F_\infty +\varepsilon )A)^{-1}. \end{aligned}$$

(3.7)

From the definition of $f_0,$ we see that there exists $r_1 > 0$ such that $f(t,u)\ge (f_0 - \varepsilon )u,$ for all $t\in [0,1], u\in [0,r_1].$ Then, for $t\in [0,1], u\in \partial K_{r_1},$ we get from (2.3), (3.1) and (3.7) that

$$\begin{aligned} \Vert Tu\Vert&=\max _{0\le t \le 1}(Tu)(t) =\max _{0\le t \le 1}\int _0^1G(t,s)\left[ \lambda f(s,u(s))+q(s)\right] ds\\&\ge a\lambda \int _a^bG(1,s)f(s,u(s))ds \ge a\lambda \int _a^bG(1,s)(f_0 - \varepsilon )u(s)ds\\&\ge a^2\lambda (f_0 - \varepsilon )\Vert u\Vert \int _a^bG(1,s)ds \ge ((f_0-\varepsilon )B)^{-1}(f_0 - \varepsilon )\Vert u\Vert B = \Vert u\Vert . \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert Tu\Vert \ge \Vert u\Vert , \quad u\in \partial K_{r_1}. \end{aligned}$$

(3.8)

On the other hand, from the definition of $F_\infty ,$ we see that there exists $\overline{R}_1 > 0$ such that $f(t,u)\le (F_\infty + \varepsilon _1)u$ for all $t\in [0,1], u\in (\overline{R}_1,+\infty )$, where $0<\varepsilon _1 <\varepsilon .$ In term of (3.7), we have $\lambda \le ((F_\infty +\varepsilon )A)^{-1}<((F_\infty +\varepsilon _1)A)^{-1}.$ Thus, $\lambda A(F_\infty + \varepsilon _1) < 1.$ Set $M = \max _{t\in [0,1], u\in [0,\overline{R}_1]}f(t,u)$, Then, $f(t,u)\le M + (F_\infty + \varepsilon _1)u$.

Choose $R_1 >\max \{r_1, \overline{R}_1, (\lambda MA+C)(1-\lambda A(F_\infty + \varepsilon _1))^{-1}\}$. Then for $t\in [0,1], u\in \partial K_{R_1}$, from Lemma 2.7 and $\Vert u\Vert = \max _{0\le t\le 1}\big |u(t)\big | = R_1,$ we also get

$$\begin{aligned} \Vert Tu\Vert&=\max _{0\le t \le 1}(Tu)(t) =\max _{0\le t \le 1}\int _0^1G(t,s)\left[ \lambda f(s,u(s))+q(s)\right] ds\\&\le \lambda \int _0^1G(1,s)[M+(F_\infty + \varepsilon _1)]u(s)ds + \int _0^1G(1,s)q(s)ds\\&\le \lambda M\int _0^1G(1,s)ds + \lambda (F_\infty + \varepsilon _1)\Vert u\Vert \int _0^1G(1,s)ds + \int _0^1G(1,s)q(s)ds\\&= \lambda MA+\lambda (F_\infty + \varepsilon _1)R_1A + C\\&\le [1-\lambda A(F_\infty + \varepsilon _1)]R_1+\lambda (F_\infty + \varepsilon _1)R_1A\\&=R_1=\Vert u\Vert . \end{aligned}$$

So, we have

$$\begin{aligned} \Vert Tu\Vert \le \Vert u\Vert , \quad u\in \partial K_{R_1}. \end{aligned}$$

(3.9)

Applying Lemma 2.5 to (3.8) and (3.9), yields that $T$ has a fixed point $\overline{u}\in P\cap (\overline{K}_{R_1}\!\setminus \! K_{r_1})$. Thus it follows that BVP (1.1) has a positive solution $\overline{u}.$ We complete the proof of Theorem 3.5. $\square$

Similarly, we have the following result.

Theorem 3.6

Assume that $f_\infty B>F_0 A$ holds, then for each

$$\begin{aligned} \lambda \in ((f_\infty B)^{-1},(F_0 A)^{-1}), \end{aligned}$$

(3.10)

BVP (1.1) has at least one positive solution. $($Particularly, $(f_\infty B)^{-1}=0$ if $f_\infty =+\infty$ and $(F_0 A)^{-1}=+\infty$ if $F_0=0).$

Theorem 3.7

Suppose there exist $r_2>\max \{r_1,C\}$, where $r_1>0$, such that

$$\begin{aligned} \max _{0\le u\le r_2, 0\le t \le 1}f(t,u(t))\le \frac{r_2-C}{\lambda A},\quad \min _{ar_1\le u\le r_1, a\le t \le b}f(t,u(t))\ge \frac{r_1}{\lambda B}. \end{aligned}$$

Then BVP (1.1) has at least a positive solutions $u\in P$ with $r_1 \le \Vert u\Vert \le r_2.$

Proof

On the one hand, choose $r_1$ with $r_1>0$, then for $u\in P\cap \partial K_{r_1},$ $t\in [a,b],$ we have

$$\begin{aligned} \Vert Tu\Vert&=\max _{0\le t \le 1}(Tu)(t) =\max _{0\le t \le 1}\int _0^1G(t,s)\left[ \lambda f(s,u(s))+q(s)\right] ds\\&\ge a\lambda \int _a^bG(1,s)f(s,u(s))ds \ge a^2\lambda \int _a^bG(1,s)f(s,u(s))ds\\&\ge a^2\lambda \int _a^bG(1,s)\min _{ar_1\le u\le r_1}f(s,u(s))ds \ge \lambda B\frac{r_1}{\lambda B}=r_1= \Vert u\Vert . \end{aligned}$$

On the other hand, choose $r_2$ with $r_2>\max \{r_1,C\},$ then for $u\in P\cap \partial K_{r_2},$ $t\in [0,1]$, we have

$$\begin{aligned} \Vert Tu\Vert&=\max _{0\le t \le 1}(Tu)(t) \le \int _0^1G(1,s)\left[ \lambda f(s,u(s))+q(s)\right] ds\\&\le \lambda \int _0^1G(1,s)\max _{0\le u\le r_2}f(s,u(s))ds+\int _0^1G(1,s)q(s)ds\\&\le \lambda A\frac{r_2-C}{\lambda A}+C=r_2= \Vert u\Vert . \end{aligned}$$

Thus, applying Lemma 2.5, BVP (1.1) has a positive solution $u\in P$ with $r_1 \le \Vert u\Vert \le r_2.$ The proof is complete. $\square$

For the reminder of the paper, we will need the following condition.

• $(H_3)$ $\sup \limits _{r>0}\min \limits _{u\in [ar,r], t\in [a,b]}f(t,u) > 0.$

Denote

$$\begin{aligned} \lambda _1=\frac{1}{A}\sup \limits _{r>0}\frac{r}{\max \limits _{0\le u\le r, 0\le t\le 1}f(t,u)},\end{aligned}$$

(3.11)

$$\begin{aligned} \lambda _2=\frac{1}{B}\inf \limits _{r>0}\frac{r}{\min \limits _{ar\le u\le r, a\le t\le b}f(t,u)}. \end{aligned}$$

(3.12)

In view of the continuity of $f(t,u)$ and $(H_3),$ we have $0<\lambda _1,$ $\lambda _2\le +\infty .$

Theorem 3.8

Assume $(H_3)$ holds. If $f_0=+\infty ,$ $f_\infty =+\infty$, then BVP (1.1) has at least two positive solutions for each $\lambda \in (0,\lambda _1)$.

Proof

Define

$$\begin{aligned} h(r)=\frac{r}{A\max \limits _{0\le u\le r, 0\le t\le 1}f(t,u)}. \end{aligned}$$

By the continuity of $f(t,u),$ $f_0=+\infty$ and $f_\infty =+\infty$, we have that $h:$ $(0,+\infty )$ $\rightarrow$ $(0,+\infty )$ is continuous and

$$\begin{aligned} \lim _{r\rightarrow 0}h(r)=\lim _{r\rightarrow +\infty }h(r)=0. \end{aligned}$$

By (3.11), there exists $r_0\in (0,+\infty )$, such that

$$\begin{aligned} h(r_0)=\sup _{r>0}h(r)=\lambda _1, \end{aligned}$$

then for $\lambda \in (0,\lambda _1)$, there exist constants $c_1,$ $c_2(0<$ $c_1<$ $r_0<$ $c_2<+\infty )$ with

$$\begin{aligned} h(c_1)=h(c_2)=\lambda . \end{aligned}$$

Thus,

$$\begin{aligned} f(t,u)\le \frac{c_1}{\lambda A}=\frac{(c_1+C)-C}{\lambda A},\quad u\in [0,c_1],\,\, t\in [0,1],\end{aligned}$$

(3.13)

$$\begin{aligned} f(t,u)\le \frac{c_2}{\lambda A}=\frac{(c_2+C)-C}{\lambda A},\quad u\in [0,c_2],\,\, t\in [0,1]. \end{aligned}$$

(3.14)

On the other hand, applying the conditions $f_0=+\infty ,$ $f_\infty =+\infty$, there exist constants $d_1, d_2(0<d_1<c_1<r_0<c_2<c_2+C<d_2<+\infty )$ with

$$\begin{aligned} \frac{f(t,u)}{u}\ge \frac{1}{a\lambda A},\quad u\in (0,d_1]\cup [ad_2,+\infty ),\,\, t\in [a,b]. \end{aligned}$$

Then we derive

$$\begin{aligned} \min _{ad_1\le u\le d_1, a\le t \le b}f(t,u)\ge \frac{d_1}{\lambda A} \end{aligned}$$

(3.15)

and

$$\begin{aligned} \min _{ad_2\le u\le d_2, a\le t \le b}f(t,u)\ge \frac{d_2}{\lambda A}. \end{aligned}$$

(3.16)

By Theorem combing with either (3.13) and (3.15), or (3.14) and (3.16) respectively, we derive (1.1) be at least two positive solutions $u_1$ and $u_2$ with $d_1\le \Vert u_1\Vert \le c_1+C, c_2+C\le \Vert u_2\Vert \le d_2.$ This completes the proof. $\square$

Corollary 3.9

Assume $(H_3)$ holds. If $f_0=+\infty , f_\infty =+\infty$, then for each $\lambda \in (0,\lambda _1)$, BVP (1.1) has at least one positive solution.

Theorem 3.10

Assume $(H_3)$ holds. If $f_0=0, f_\infty =0,$ then BVP (1.1) has at least two positive solutions for each $\lambda \in (\lambda _2,+\infty ).$

Proof

Define

$$\begin{aligned} g(r)=\frac{r}{B\min \limits _{ar\le u\le r, a\le t\le b}f(t,u)}. \end{aligned}$$

By the continuity of $f(t,u), f_0=0$ and $f_\infty =0$, we have that $g:(0,+\infty )\rightarrow (0,+\infty )$ is continuous and

$$\begin{aligned} \lim _{r\rightarrow 0}g(r)=\lim _{r\rightarrow +\infty }g(r)=+\infty . \end{aligned}$$

By (3.12), there exists $r_0\in (0,+\infty )$ such that

$$\begin{aligned} g(r_0)=\inf _{r>0}g(r)=\lambda _2. \end{aligned}$$

Then for $\lambda \in (\lambda _2,+\infty ),$ there exist constants $d_1, d_2(0<d_1<r_0<d_2<+\infty )$ with

$$\begin{aligned} g(d_1)=g(d_2)=\lambda . \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} f(t,u)\ge \frac{d_1}{\lambda B},\quad u\in [ad_1,d_1], \,\,t\in [a,b] \end{aligned}$$

(3.17)

and

$$\begin{aligned} f(t,u)\le \frac{d_2}{\lambda B},\quad u\in [ad_2,d_2], \,\,t\in [a,b]. \end{aligned}$$

(3.18)

On the other hand, applying the conditions $f_0=0,$ $f_\infty =0$, there exist constants $c_1, c_2(0<c_1<c_1+C<d_1<d_1<r_0<d_2<c_2<c_2+C<+\infty )$ with

$$\begin{aligned} \frac{f(t,u)}{u}\le \frac{1}{\lambda A},\quad u\in [0,c_1]\cup [c_2,+\infty ), t\in [0,1] \end{aligned}$$

Then

$$\begin{aligned} \max _{0\le u\le c_1, 0\le t \le 1}f(t,u)\le \frac{c_1}{\lambda A}=\frac{(c_1+C)-C}{\lambda A}. \end{aligned}$$

(3.19)

Let

$$\begin{aligned} M=\max _{0\le u\le c_2, 0\le t \le 1}f(t,u), \quad c_2\ge \lambda AM. \end{aligned}$$

It is easily seen that

$$\begin{aligned} \max _{0\le u\le c_2, 0\le t \le 1}f(t,u)\le \frac{c_2}{\lambda A}=\frac{(c_2+C)-C}{\lambda A}. \end{aligned}$$

(3.20)

By Theorem 3.7 combining with either (3.17) and (3.19), or (3.18) and (3.20) respectively, we derive BVP (1.1) be at least two positive solutions $u_1$ and $u_2$ with $c_1+C\le \Vert u_1\Vert \le d_1, d_2\le \Vert u_2\Vert \le c_2+C.$ The proof is complete. $\square$

Corollary 3.11

Assume $(H_3)$ holds. If $f_0=0, f_\infty =0,$ then for each $\lambda \in (\lambda _2,+\infty ),$ BVP (1.1) has at least one positive solution.

Corollary 3.12

Assume $(H_3)$ holds. If $f_0=+\infty , f_\infty =l,$ or $f_\infty =+\infty , f_0=l,$ then for any $\lambda \in \big (0,(lA)^{-1} \big ),$ BVP (1.1) has at least one positive solution.

Corollary 3.13

Assume $(H_3)$ holds. If $f_0=0, f_\infty =l,$ or $f_\infty =0, f_0=l,$ then for any $\lambda \in \big ((alB)^{-1},+\infty \big ),$ BVP (1.1) has at least one positive solution.

4 Some Examples

Example 4.1

Consider the following nonlinear fractional differential equations:

$$\begin{aligned} \left\{ \begin{array}{ll} -D_{0^+}^{\frac{5}{2}}u(t)=\frac{\lambda }{1+t^2}\left[ u(t)\right] ^a+\frac{1}{2^t}, &{} 0<t<1, \\ u(0)=u'(1)=u''(0)=0, &{} \\ \end{array} \right. \end{aligned}$$

(4.1)

where $f(t,u)=u^a/(1+t^2),$ $\lambda >0,$ $q(t)=\frac{1}{2^t}>0.$

• (i) As $0<a<1$. By simple computation, we have $f_0=+\infty , F_\infty =0.$ Thus it follows that BVP (4.1) has at least a positive solution for $\lambda >0$ by Theorem 3.5.

• (ii) As $1<a<\infty .$ By simple computation, we have $F_0=0, f_\infty =+\infty .$ Thus it follows that BVP (4.1) has at least a positive solution for $\lambda >0$ by Theorem 3.6.

Example 4.2

Consider the system of nonlinear fractional differential equations:

$$\begin{aligned} \left\{ \begin{array}{ll} -D_{0^+}^{\frac{5}{2}}u(t)=\lambda \frac{u(t)\ln u(t)}{1+t^2}+\frac{1}{2^t}, &{} 0<t<1, \\ u(0)=u'(1)=u''(0)=0, &{} \\ \end{array} \right. \end{aligned}$$

(4.2)

where $f(t,u)=\left| \frac{u\ln u}{1+t^2}\right| ,$ $q(t)=\frac{1}{2^t},\lambda >0.$ Take $[a,b]=[\frac{1}{4},\frac{3}{4}]\subset [0,1],$ $r=1.$ It is easy to see that $(H_3)$ holds. By simple computation, we have

$$\begin{aligned} f_0&=\liminf \limits _{u\rightarrow 0} \min \limits _{t\in [0,1]}\left| \frac{u\ln u}{(1+t^2)u}\right| =\liminf \limits _{u\rightarrow 0}\frac{|\ln u|}{2}=+\infty ,\\ f_\infty&=\liminf \limits _{u\rightarrow \infty } \min \limits _{t\in [0,1]}\left| \frac{u\ln u}{(1+t^2)u}\right| =\liminf \limits _{u\rightarrow \infty }\frac{|\ln u|}{2}=+\infty . \end{aligned}$$

For $0<u\le 1, f(t,u)=\frac{-u\ln u}{1+t^2},$ we can obtain $f(t,u)$ arrives at maximum at $u=1/e, t=0.$ By (3.11), we have

$$\begin{aligned} \lambda _1=\sup \frac{1}{A\max \limits _{0\le u\le 1, 0\le t\le 1}\left| \frac{u(t)\ln u(t)}{1+t^2}\right| }=\frac{e}{A}=6.0225359. \end{aligned}$$

Thus it follows that BVP (4.2) has at least two positive solutions for $\lambda \in (0,6.0225359)$ by Theorem 3.8.