A Study on Positive Solutions to Nonlinear Fractional Differential Equations

Abstract

The purpose of this article concerns a type of nonlinear boundary value problem of fractional differential equations with the Caputo derivative. Using the fixed point index theory of condensing mapping in cones, the existence results of positive solutions for such a problem are obtained.

1. Introduction

Fractional differential equations have recently aroused great interest among people. This is caused by the deep development of fractional calculus theory itself and the application of this construction in various fields of science and engineering, such as control, porous media, electromagnetic, and other fields (see [1,2,3] and their references).

Ordinary differential equations or partial differential equations have received widespread attention due to their importance in engineering, physics, material mechanics, and chemotaxis mechanisms (for example, see [4,5] and their references).

Due to the fact that fractional order models are more accurate in differential equations than integer order models, and have a profound physical background and rich theoretical foundation, they have attracted widespread attention. In recent years, many scholars have studied the properties of using Banach operators to solve partial fractional differential equation boundary value problems, such as Guo–Krasnoselskii fixed point theorem, monotone iterative technique, and so on [6,7,8].

In recent years, positive solutions for nonlinear differential equations for integer order equations, including two-point and three-point boundary value problems, have been widely applied; see [8,9]. In these works, nonlinearity includes derivative terms.

On the other hand, many scholars have studied positive solutions for various fractional differential equations with various boundary conditions. Many scholars are concerned about these issues, and readers can refer to [1,3,10,11,12,13,14,15,16,17]. In these works, nonlinearity does not include derivative terms.

In 2014, Cabada and HAMDi [12] considered the existence results of positive solutions for fractional differential equations of the form

$$\left\{\begin{array}{c}{D}_{0^{+}}^{\gamma }x\left(t\right)+f(t,x\left(t\right))=0,0<t<1,\hfill \\ x\left(0\right)=x^{\prime }\left(0\right)=0,x\left(1\right)=\eta {\int }_0^{1}x\left(t\right)dt,\hfill \end{array}\right.$$

where $2<\gamma \le 3,0<\eta <\gamma$, $D_{0^{+}}^{\gamma }$ is Riemann–Liouville fractional derivative. Some existence results of positive solutions were obtained.

In 2017, Cabada et al. [18] considered the existence of a positive solutions for the nonlinear fractional differential equation

$$\left\{\begin{array}{c}{}^{c}D_{0^{+}}^{\gamma }x\left(t\right)+f(t,x\left(t\right))=0,0<t<1,\hfill \\ x\left(0\right)=x^{\prime }\left(0\right)=0,x\left(1\right)=\eta {\int }_0^{1}x\left(s\right)ds,\hfill \end{array}\right.$$

where $0\le \eta <1$, ${}^{c}D_{0^{+}}^{\gamma }$ is the Caputo fractional derivative of order $\gamma$, and f is a continuous function.

In 2014, Zhang et al. [19] studied the existence of positive solutions for nonlinear fractional differential equations

$$\left\{\begin{array}{c}{}^{c}D_{0^{+}}^{\gamma }x\left(t\right)+g\left(t\right)f(t,x\left(t\right))=0,0<t<1,\hfill \\ x\left(0\right)=x^{\prime }\left(0\right)=x^{″}\left(0\right)=0,x\left(1\right)=\beta {\int }_0^{\alpha }u\left(s\right)ds,\hfill \end{array}\right.$$

where $3<\gamma \le 4$, $0<\beta \le 1,$${}^{c}D_{0^{+}}^{\gamma }$ is the Caputo fractional derivative. They obtain some existence results of positive solution by means of $u_0$-bounded function and the fixed point index theory.

In 2018, Chen et al. [13] investigated positive solutions for fractional differential equations

$$\left\{\begin{array}{c}{D}_{0^{+}}^{\gamma }x\left(t\right)+f(t,x\left(t\right))=0,0<t<1,\hfill \\ x\left(0\right)=x^{\prime }\left(0\right)=0,x\left(1\right)=\eta {\int }_0^{1}x\left(s\right)ds,\hfill \end{array}\right.$$

where both $2<\gamma \le 3$ and $0<\eta <\gamma$, $D_{0^{+}}^{\gamma }$ are Riemann–Liouville fractional derivative.

In 2022, Gou [20] studied positive solutions for nonlinear fractional differential equations

$$\left\{\begin{array}{c}{D}_{0^{+}}^{\gamma }x\left(t\right)+f(t,x\left(t\right),x^{\prime }\left(t\right))=0,t\in (0,1),n-1<\gamma \le n,\hfill \\ x^{\left(i\right)}\left(0\right)=0,i=0,1,2,\dots ,n-2,\hfill \\ {\left(D_{0^{+}}^{\eta }x\left(t\right)\right)}_{t=1}=0,2\le \eta \le n-2,\hfill \end{array}\right.$$

where $n>4$. We investigated the existence of positive solution by means of the theory of fixed point index on a special cone in $C^1([0,1],E)$.

As an obvious fact, there are few positive solutions that result in fractional differential equation conditions, and we also refer to [16]. In this work, the authors are concerned with positive solutions for equation

$$\left\{\begin{array}{c}{}^{c}D_{0^{+}}^{\gamma }x\left(t\right)+f(t,x\left(t\right))=0,0<t<1,\hfill \\ x\left(0\right)=x^{″}\left(0\right)=0,x\left(1\right)=\eta {\int }_0^{1}x\left(s\right)ds,\hfill \end{array}\right.$$

where $2<\gamma <3,0<\eta <2$, ${}^{c}D_{0^{+}}^{\gamma }$ is the Caputo fractional derivative of order $\gamma$ and $f(t,x):[0,1]\times [0,\infty )\to [0,\infty )$ is a continuous function, by employing to Guo–Krasnoselskii fixed point theorem.

Since only positive solutions are useful for many applications, inspired and motivated by those previous works, we deal with the existence of positive solutions for nonlinear fractional differential equations in an ordered Banach space, E,

$$\left\{\begin{array}{c}{}^{c}D_{0^{+}}^{\gamma }x\left(t\right)+f(t,x\left(t\right))=0,0<t<1,\hfill \\ x\left(0\right)=x^{″}\left(0\right)=0,x\left(1\right)=\eta {\int }_0^{1}x\left(s\right)ds,\hfill \end{array}\right.$$

(1)

where $2<\gamma <3,0<\eta <2$, ${}^{c}D_{0^{+}}^{\gamma }$ is the Caputo fractional derivative of order $\gamma$ and $f(t,x):[0,1]\times P\to P$ is a continuous function, P is a positive cone in Banach space E. By means of the theory of fixed point index on a special cone in $C\left(\right[0,1],E)$, the existence results of positive solution of (1) are obtained.

In previous these works, the positivity of the corresponding Green’s function $G(t,s)$ plays an important role. The positivity guarantees that the concerned problem can be converted to a fixed point problem of a cone mapping in $C\left(J\right)$, where $J=[0,1]$. Hence, these authors can apply the fixed point theorems of cone mapping to obtain the existence of positive solutions. But in this paper, our argument methods are the fixed point index theory of condensing mapping. Because of the properties of condensing mapping, the nonlinear term f requires the addition of compactness conditions, i.e., the nonlinear term f satisfies the measure of noncompactness condition.

We now outline the main contributions, significance, and novelty of this article as follows:

(1)

Our purpose in this paper is to give some existence results for positive solution to the problem (1).

(2)

The sublinear growth of the nonlinearity f are described by inequality conditions instead of the usual upper and lower limits conditions.

(3)

Our discussion is based on the fixed point index theory in cones. Some of our ideas are inspired by the give ones in the recent paper [13].

(4)

We use the inequality conditions to describe the growth of the nonlinearity f. Our inequality conditions are weaker than the usual conditions described by the upper and lower limits. Moreover, our conditions are easy to verify for some nonlinear terms.

2. Preliminaries

Let E be ordered Banach space with partial order “≤” and positive cone P.

Denote by $C(J,E)$ all continuous functions and the norm ${\parallel u\parallel }_C={max}_{t\in J}\parallel x\left(t\right)\parallel$, where $J=[0,1]$. Then, $C(J,E)$ is an ordered Banach space induced by the convex cone $C(J,P)=\{x\in C(J,E\left)\right|x\left(t\right)\in P,t\in J\}$, and $C(J,P)$ is also a normal cone in $C(J,E)$. For more details on the cone and partial order, see [21,22,23].

Definition 1 

([22]). A cone P in E is called normal if, for $m,n\in E$,

$$\theta \le m\le n\Rightarrow \parallel m\parallel \le N\parallel n\parallel ,$$

where $N>0$ is normality constant.

Definition 2 

([1]). The Riemann–Liouville fractional integral of order γ of a function x is defined as

$$I_{0+}^{\gamma }x\left(t\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^t{(t-s)}^{\gamma -1}x\left(s\right)ds,t>0,\gamma >0,$$

provided that such integral exists, where $\mathsf{\Gamma }\left(\alpha \right)$ is the gamma function.

Definition 3 

([1]). The Riemann–Liouville fractional derivative of order γ for a function x is defined by

$$D_{0^{+}}^{\gamma }x\left(t\right)={\displaystyle \frac{1}{\mathsf{\Gamma }(n-\gamma )}}{\displaystyle \frac{d^n}{d{t}^n}}{\int }_0^t{\displaystyle \frac{x\left(s\right)}{{(t-s)}^{\gamma +1-n}}}ds,t>0,n-1<\gamma \le n.$$

Definition 4 

([1]). The Caputo fractional derivative of order γ for a function x can be written as

$${}^{c}D_{0^{+}}^{\gamma }x\left(t\right)=D_{0^{+}}^{\gamma }\left[x\left(t\right)-\sum _{k=0}^{n-1}{\displaystyle \frac{t^k}{k!}}{x}^{\left(k\right)}\left(0\right)\right],t>0,n-1<\gamma \le n,$$

where $n=\left[\gamma \right]+1$ and $\left[\gamma \right]$ denotes the integer part of γ.

If x is an abstract function with values in E, then the integrals which appear in Definitions 2–4 are taken in Bochner’s sense. A measurable function $x:J\to E$ is Bochner integrable if $\parallel x\parallel$ is Lebesgue integrable.

Remark 1. 

In the case where $x\left(t\right)\in C^n([0,1],E)$, then

$${}^{c}D_{0^{+}}^{\gamma }x\left(t\right)={\displaystyle \frac{1}{\mathsf{\Gamma }(n-\gamma )}}{\int }_0^t{(t-s)}^{n-\gamma -1}{x}^{\left(n\right)}\left(s\right)ds=I_{0+}^{n-\gamma }{f}^n\left(t\right),t>0,n-1<\gamma \le n.$$

In the following, we obtain the exact expression of the Green’s function associated with the following fractional differential equation:

$$\left\{\begin{array}{c}{}^{c}D_{0^{+}}^{\gamma }x\left(t\right)+m\left(t\right)=0,t\in J,\hfill \\ x\left(0\right)=x^{″}\left(0\right)=0,x\left(1\right)=\eta {\int }_0^{1}x\left(s\right)ds.\hfill \end{array}\right.$$

(2)

The Lemma is the following. In the proof of this lemma, we use [24] (Theorem 2.1). But, in opposition to [24], values of the function x belong to a Banach space under our consideration, and we provide a proof here.

Lemma 1. 

If $2\le \gamma \le 3$, $\eta \ne 2$. Then, for $m\in C(J,E)$, Problem (2) has a unique solution x, given by the expression

$$\begin{array}{c}\hfill x\left(t\right)={\int }_0^1\tilde{G}(t,s)m\left(s\right)ds:=\tilde{T}m\left(t\right),\end{array}$$

(3)

where the Green’s function is given by

$$\begin{array}{c}\hfill \tilde{G}(t,s)=\left\{\begin{array}{c}{\displaystyle \frac{2t{(1-s)}^{\gamma -1}(\gamma -\eta +\eta s)-(2-\eta )\gamma {(t-s)}^{\gamma -1}}{(2-\eta )\mathsf{\Gamma }(\gamma +1)}},0\le s\le t\le 1,\hfill \\ {\displaystyle \frac{2t{(1-s)}^{\gamma -1}(\gamma -\eta +\eta s)}{(2-\eta )\mathsf{\Gamma }(\gamma +1)}},0\le t\le s\le 1.\hfill \end{array}\right.\end{array}$$

(4)

Proof. 

In view of [24] (Theorem 2.1), we obtain that $x\left(t\right)$ given by (3) is the solution to (2).

$x_1\left(t\right),x_2\left(t\right)\in C(J,E)$ is solution to (2). For any $\phi \in E^{*}$, ($E^{*}$ is a dual space of E), set $r\left(t\right)=\phi (x_2\left(t\right)-x_1\left(t\right))$, then $r\left(t\right)$ is a solution of the following equation:

$$\left\{\begin{array}{c}{}^{c}D_{0^{+}}^{\gamma }r\left(t\right)=0,t\in J,\hfill \\ r\left(0\right)=r^{″}\left(0\right)=0,\hfill \\ r\left(1\right)=\eta {\int }_0^{1}r\left(s\right)ds.\hfill \end{array}\right.$$

After a simple calculation, we obtain $r\left(t\right)\equiv 0$. By the arbitrary of $\phi \in E^{*}$, we obtain $x_2\left(t\right)-x_1\left(t\right)\equiv 0$, i.e., $x_1\left(t\right)\equiv x_2\left(t\right)$ on J. Then, $x\left(t\right)$ is a unique solution to (2). □

Lemma 2 

([24]). Let $\tilde{G}(t,s)$ be the Green’s function related to Problem (3), which is given by Expression (4). Then, for all $\alpha \in (2,3)$, the following properties are fulfilled:

(i) 

$\tilde{G}(0,s)=\tilde{G}(t,1)=0$ for all $t,s\in [0,1]$ and $\eta \ne 2$.

(ii) 

$\tilde{G}(1,s)=0$ for all $s\in [0,1]$ if and only if $\eta =0$.

(iii) 

$\tilde{G}(1,s)>0$ for all $s\in (0,1)$ if and only if $\eta \in (0,2)$.

(iv) 

$\tilde{G}(t,0)>0$ for all $t\in (0,1)$ if and only if $\eta \in [0,2)$.

(v) 

$\tilde{G}(t,s)>0$ for all $t,s\in (0,1)$ if and only if $\eta \in [0,2)$.

(vi) 

$\tilde{G}(t,s)\le {\displaystyle \frac{2}{(2-\eta )\mathsf{\Gamma }\left(\gamma \right)}}$ for all $t,s\in [0,1]$ and $\eta \in [0,2)$.

(vii) 

$t\tilde{G}(1,s)\le \tilde{G}(t,s)\le {\displaystyle \frac{2\gamma }{\eta (\gamma -2)}}\tilde{G}(1,s)$ for all $t,s\in (0,1),2<\gamma <3$ and $0<\eta <2$.

Lemma 3. 

The operator $\tilde{T}:C(J,E)\to C(J,E)$ defined by (3) satisfies

$$\parallel \tilde{T}{\parallel }_C<{\displaystyle \frac{2(\gamma -\eta )}{(2-\gamma )\mathsf{\Gamma }(\gamma +2)}}+{\displaystyle \frac{2\eta }{(2-\eta )\mathsf{\Gamma }(\gamma +3)}}=M_1.$$

Proof. 

For $m\in C(J,E)$, $t\in J$, we obtain

$$\begin{array}{cc}\hfill \parallel \left(\tilde{T}m\right)\left(t\right)\parallel & \le \parallel {\int }_0^1\tilde{G}(t,s)m\left(s\right)ds\parallel \hfill \\ \hfill & \le ({\int }_0^t{\displaystyle \frac{2t{(1-s)}^{\gamma -1}(\gamma -\eta +\eta s)}{(2-\eta )\mathsf{\Gamma }(\gamma +1)}}ds-{\int }_0^t{\displaystyle \frac{(2-\eta )\gamma {(t-s)}^{\gamma -1}}{(2-\eta )\mathsf{\Gamma }(\gamma +1)}}ds\hfill \\ \hfill & +{\int }_t^1{\displaystyle \frac{2t{(1-s)}^{\gamma -1}(\gamma -\eta +\eta s)}{(2-\eta )\mathsf{\Gamma }(\gamma +1)}}ds){\parallel m\parallel }_C\hfill \\ \hfill & =\left({\displaystyle \frac{2t(\gamma -\eta )}{(2-\gamma )\mathsf{\Gamma }(\gamma +2)}}+{\displaystyle \frac{2\eta t}{(2-\eta )\mathsf{\Gamma }(\gamma +3)}}-{\displaystyle \frac{t^{\gamma }}{\mathsf{\Gamma }(\gamma +1)}}\right){\parallel m\parallel }_C\hfill \\ \hfill & \le \left({\displaystyle \frac{2(\gamma -\eta )}{(2-\gamma )\mathsf{\Gamma }(\gamma +2)}}+{\displaystyle \frac{2\eta }{(2-\eta )\mathsf{\Gamma }(\gamma +3)}}\right){\parallel m\parallel }_C.\hfill \end{array}$$

Thus, we have

$$\parallel \tilde{T}{\parallel }_C<{\displaystyle \frac{2(\gamma -\eta )}{(2-\gamma )\mathsf{\Gamma }(\gamma +2)}}+{\displaystyle \frac{2\eta }{(2-\eta )\mathsf{\Gamma }(\gamma +3)}}:\triangleq M_1.$$

This completes the proof of Lemma 3. □

For any $M\subset C(J,E)$ and $t\in J$, let $M\left(t\right)=\left\{x\right(t\left)\right|x\in M\}$, then $M\left(t\right)\subset E$. Denote $\alpha (·)$ by the Kuratowski measure of noncompactness. For more details, see [22,23,25,26].

Lemma 4 

([27]). Let E be a metric space, and $M\subset E$ be bounded. Then, there exists a countable set $M_0\subset M$, such that $\alpha \left(M\right)\le 2\alpha \left(M_0\right)$.

Lemma 5 

([23]). Let E be a Banach space, and let $M\subset C(J,E)$ be equicontinuous and bounded, then $\alpha \left(M\right(t\left)\right)$ is continuous on J, and

$$\alpha \left(M\right)=\underset{t\in J}{max}\alpha \left(M\left(t\right)\right).$$

Lemma 6 

([28]). Let $M={\left\{x_n\right\}}_{n=1}^{\infty }\subset C(J,E)$ be a bounded and countable set, and suppose that there exists a function $m\in L^1(J,{\mathbb{R}}^{+})$ such that, for every $n\in N$,

$$\parallel x_n\left(t\right)\parallel \le m\left(t\right),a.e.t\in J.$$

Then, $\alpha \left(M\right(t\left)\right)$ is Lebesgue integrable on J, and

$$\alpha \left(\left\{{\int }_J{x}_n\left(t\right)dt:n\in \mathbb{N}\right\}\right)\le 2{\int }_J\alpha \left(M\left(t\right)\right)dt.$$

Define the operator $\tilde{Q}:C(J,E)\to C(J,E)$ by

$$\begin{array}{c}\hfill \left(\tilde{Q}x\right)\left(t\right)={\int }_0^1\tilde{G}(t,s)f(s,x\left(s\right))ds.\end{array}$$

(5)

Obviously, $\tilde{Q}:C(J,E)\to C(J,E)$, from Lemma 1, and the fixed points of $\tilde{Q}$ coincide with the solution of (1).

We say that $\tilde{Q}$ is condensing if

$$\alpha \left(\tilde{Q}\left(M\right)\right)<\alpha \left(M\right),\forall M\subset \mathsf{\Omega }\mathrm{bounded}\mathrm{with}\alpha \left(M\right)>0.$$

Lemma 7. 

For any $R>0$, let $\tilde{B}(\theta ,R)\subset E$ be a closed set with $\theta$ as the center and R as the radius, the function $f:J\times \tilde{B}(\theta ,R)\to E$ is continuous and bounded, if the following the condition holds:

(F0) 

There exists $L>0$ satisfying $0<4L<{\displaystyle \frac{1}{M_1}}$, such that, for $M\subset \tilde{B}(\theta ,R)$,

$$\alpha \left(f\right(t,M\left)\right)\le L\left(\alpha \right(M\left)\right),t\in J$$

Then, $\tilde{Q}$ is a condensing mapping, where $M_1$ is given by in Lemma 3.

Proof. 

Step 1: We prove that $\tilde{Q}$ maps bounded sets into bounded sets in $C(J,E)$. By (F0), for any $R>0$, there is $H>0$ such that for $x\in \tilde{B}(\theta ,R)$, ${\parallel f(·,x(·))\parallel }_C\le H$, we obtain

$$\begin{array}{cc}\hfill \parallel \tilde{Q}x\left(t\right)\parallel & =\parallel {\int }_0^1\tilde{G}(t,s)f(s,x\left(s\right))ds\parallel \le H{\int }_0^1\tilde{G}(t,s)ds\le {\displaystyle \frac{2H}{(2-\eta )\mathsf{\Gamma }\left(\gamma \right)}}.\hfill \end{array}$$

Let $\tilde{M}={\displaystyle \frac{2H}{(2-\rho )\mathsf{\Gamma }\left(\alpha \right)}}$; we obtain

$$\parallel \tilde{Q}x\left(t\right)\parallel \le \tilde{M}.$$

Hence, we have

$$\parallel \tilde{Q}{x\parallel }_C\le \tilde{M}.$$

Step 2: We show that $\tilde{Q}$ is equicontinuous in $C(J,E)$. Since $\tilde{G}(t,s)$ is continuous, then $\tilde{G}(t,s)$ are uniformly continuous. Let $t_2,t_1\in [0,1]$ with $t_1<t_2$. For $ϵ>0$, $\exists \delta >0$, $|t_1-t_2|<\delta$, we have

$$|\tilde{G}(t_2,s)-\tilde{G}(t_1,s)|<{\displaystyle \frac{ϵ}{H}}.$$

For $x\in \tilde{B}(\theta ,R)$, we obtain

$$\begin{array}{cc}\hfill \parallel \tilde{Q}x\left(t_2\right)-\tilde{Q}x\left(t_1\right)\parallel & =\parallel {\int }_0^1(\tilde{G}(t_2,s)-\tilde{G}(t_1,s))f(s,x\left(s\right))ds\parallel \hfill \\ \hfill & \le H{\int }_0^1|\tilde{G}(t_2,s)-\tilde{G}(t_1,s)|ds<ϵ.\hfill \end{array}$$

Hence,

$$\parallel \tilde{Q}x\left(t_2\right)-\tilde{Q}x\left(t_1\right){\parallel }_C\le ϵ.$$

Step 3: We prove that $\tilde{Q}$ is a condensing mapping. For set $M\subset C(J,E)$, by Lemma 4, $\exists M_0=\left\{x_n\right\}\subset M$,

$$\alpha \left(\tilde{Q}\left(M\right)\right)\le 2\alpha \left(\tilde{Q}\left(M_0\right)\right).$$

By Lemma 5, we have

$$\alpha \left(\tilde{Q}\left(M_0\right)\right)=\underset{t\in J}{max}\alpha \left(\tilde{Q}\left(M_0\right)\left(t\right)\right).$$

For $t\in J$, we obtain

$$\begin{array}{cc}\hfill \alpha \left(\tilde{Q}\left(M_0\right)\left(t\right)\right)& =\alpha \left(\left\{{\int }_0^1\tilde{G}(t,s)f(s,x_n\left(s\right))ds|n\in \mathbb{N}\right\}\right)\hfill \\ \hfill & \le 2{\int }_0^1\alpha \left(\{\tilde{G}(t,s)f(s,x_n\left(s\right))|n\in \mathbb{N}\}\right)ds\hfill \\ \hfill & \le 2{\int }_0^1\tilde{G}(t,s)\alpha \left(f(s,M_0\left(s\right))\right)ds.\hfill \end{array}$$

By condition (F0), we obtain

$$\alpha \left(f(s,M_0\left(s\right))\right)\le L\alpha \left(M_0\left(s\right)\right)\le L\alpha \left(M_0\right))\le L\alpha \left(M\right).$$

Thus,

$$\alpha \left(\tilde{Q}\left(M_0\right)\right)\le 2L\alpha \left(M\right){\int }_0^1\tilde{G}(t,s)ds.$$

Moreover,

$$\begin{array}{cc}\hfill \alpha \left(\tilde{Q}\left(M_0\right)\left(t\right)\right)& =\underset{t\in J}{max}\alpha \left(\tilde{Q}\left(M_0\right)\left(t\right)\right)\hfill \\ \hfill & \le 2L\alpha \left(M\right)\underset{t\in J}{max}{\int }_0^1\tilde{G}(t,s)ds\hfill \\ \hfill & <2L{M}_1\alpha \left(M\right).\hfill \end{array}$$

Therefore, from this, we have

$$\begin{array}{c}\hfill \alpha \left(\tilde{Q}\left(M\right)\right)\le 2\alpha \left(M_0\right)<4L{M}_1\alpha \left(M\right)<\alpha \left(M\right).\end{array}$$

Thus, $\tilde{Q}$ is a condensing mapping.

Let cone

$$P=\left\{x\in C(J,E)|x\left(t\right)\ge {\displaystyle \frac{t\eta (\gamma -2)}{2\gamma }}x\left(\tau \right),\forall t,\tau \in J\right\}\subset C(J,E).$$

□

Lemma 8. 

If $f(J\times P)\subset P$, then $\tilde{Q}\left(P\right)\subset P$.

Proof. 

For $x\in P$, $t,\tau \in J$, by Lemma 2, we obtain

$$\begin{array}{cc}\hfill \left(\tilde{Q}x\right)\left(t\right)& ={\int }_0^1\tilde{G}(t,s)f(s,x\left(s\right))ds\hfill \\ \hfill & \ge t{\int }_0^1\tilde{G}(1,s)f(s,x\left(s\right))ds\hfill \\ \hfill & \ge {\displaystyle \frac{t\eta (\gamma -2)}{2\gamma }}x\left(\tau \right){\int }_0^1\tilde{G}(\tau ,s)f(s,x\left(s\right))ds\hfill \\ \hfill & ={\displaystyle \frac{t\eta (\gamma -2)}{2\gamma }}x\left(\tau \right).\hfill \end{array}$$

Hence, $\tilde{Q}\left(P\right)\subset P$. This completes the proof of Lemma 8. □

By the definitions of $\tilde{Q}$, the positive solution of Problem (1) is equivalent to the nonzero fixed point of Q. We will find the nonzero fixed point of A by using the fixed point index theory of condensing mapping in cones.

To find the nonzero fixed point of $\tilde{Q}$ defined by (5), we recall some concepts and conclusions on the fixed point index of condensing mapping in cones, which will be used in the argument later. Let E be a Banach space, and let P be a closed convex cone in E. Assume that it is a bounded open subset of E with boundary $\partial \mathsf{\Omega }$ and $P\cap \overline{\mathsf{\Omega }}\ne \varnothing$. Let $\tilde{Q}:P\cap \overline{\mathsf{\Omega }}\to P$ be a condensing mapping. If $\tilde{Q}u\ne u$ for every $u\in P\cap \overline{\mathsf{\Omega }}$, then the fixed point index $i(\tilde{Q},P\cap \mathsf{\Omega },P)$ is well defined. One important fact is that if $i(\tilde{Q},P\cap \mathsf{\Omega },P)\ne 0$, then $\tilde{Q}$ has a fixed point in $P\cap \mathsf{\Omega }$. As the fixed point index theory of condensing mapping is similar to the fixed point index theory of completely continuous mapping, we give following some properties of the $i(\tilde{Q},P\cap \mathsf{\Omega },P)$.

We will use the fixed point index theory of condensing mapping to prove our main results. In the following, we state the fixed point index theory of condensing mapping.

Lemma 9 

([23]). Let $D\subset E$ and $Q:P\cap \overline{D}\to P$ be a condensing mapping. If $x\ne \mu Qx$ for every $x\in P\cap \partial D,0<\mu \le 1$, then $i(Q,P\cap D,P)=1$.

Lemma 10 

([23]). Let $D\subset E$ with $\theta \in D$ and $Q:P\cap \overline{D}\to P$ be a condensing map. If $\exists w_0\in P\backslash \left\{\theta \right\}$, such that $x-Qx\ne \delta w_0,\forall x\in P\cap \partial D,\delta \ge 0$, then $i(Q,P\cap D,P)=0$.

Lemma 11 

([23]). Let $D\subset E$, and $Q,Q_1:P\cap \overline{D}\to P$ be two completely continuous maps. If $(1-s)Qx+s{Q}_{1}x\ne x$, for $x\in P\cap \partial D$, $0\le s\le 1$, then $i(Q,P\cap D,P)=i(Q_1,P\cap D,P)$.

Lemma 12 

([23]). Let P be a reproducing cone in E, $Q:E\to E$ completely continuous linear operator with $Q\left(P\right)\subset P$. If $r\left(Q\right)>0$, then $\exists \sigma \in P\backslash \left\{\theta \right\}$, $\theta \in E$, such that $Q\sigma =r\left(Q\right)\sigma$.

Remark 2. 

Obviously, $\tilde{T}$ defined by (3) is continuous linear operator and $\tilde{T}\left(P\right)\subset \tilde{T}$. For more details, see the proof of Lemma 4.3 and Lemma 4.4 in [1].

The proofs of Theorems 1 and 2 are based on the fixed point index theory of condensing mapping in cones. We will choose a special cone P in the space $C(J,E)$ and convert Problem (1) to a fixed point problem of a cone mapping $\tilde{Q}:P\to P$, then find the fixed point of $\tilde{Q}$ by the theory of the fixed point index of condensing mapping in cones. The proofs will be given in Section 3.

In next section, we will use the three lemmas to prove Theorems 1 and 2.

3. Main Results

Denote $P_r=P\cap \tilde{B}(\theta ,r)$ for $r>0$. In the following, we assume that E is an ordered Banach spaces and whose positive cone P is normal with normal constant N.

Theorem 1. 

If f is continuous and satisfies (F0) and the following conditions hold:

(F1) 

There exist constants $0<a<{\displaystyle \frac{1}{M_1}}$, $\delta >0$,

$$f(t,x)\le ax,t\in J,x\in P_{\delta };$$

(F2) 

There exist constant $b>{\lambda }_1$, $m_1\in C(J,K)$,

$$f(t,x)\ge bx-h_1\left(t\right),t\in J,x\in P$$

then the problem (1) has at least one positive solution.

Proof. 

We choose the working space $E=C(J,E)$. Let $P\subset C(J,E)$ be the closed convex cone and $\tilde{Q}:P\to P$ defined by (5). Then, the positive solution of the problem (1) is equivalent to the nontrivial fixed point of $\tilde{Q}$. Let $0<r<R<\infty$, set

$$D_1={\{x\in C(J,E)|\parallel x\parallel }_C<r\},D_2={\{x\in C(J,E)|\parallel x\parallel }_C<R\},$$

and

$$\partial D_1={\{x\in C(J,E)|\parallel x\parallel }_C=r\},\partial D_2={\{x\in C(J,E)|\parallel x\parallel }_C=R\},$$

which are the relative boundary of $D_1,D_2$ in P, respectively. We prove that $\tilde{Q}$ has a fixed point in $P\cap (D_2\backslash {\overline{D}}_1)$ when r is small enough and R large enough.

Let $r\in (0,\delta )$, where $\delta$ is the positive constant in the condition (F1). We prove that $\tilde{Q}$ satisfies the condition of Lemma 9 in $P\cap \partial D_1$. Next, we verify

$$\begin{array}{c}\hfill x-\mu \tilde{Q}x\ne \theta ,\forall x\in P\cap \partial D_1,0<\mu \le 1.\end{array}$$

(6)

If this fact does not hold, then there exists $x_0\in P\cap \partial D_1$, $0<{\mu }_0\le 1$ such that $x_0={\mu }_0\tilde{Q}{x}_0$. Since $x_0\in P\cap \partial D_1$, by P and $D_1$, we obtain

$$0\le x_0\left(t\right)\le {\parallel x_0\parallel }_C=r<\delta ,t\in J.$$

For $t\in J$, by (F1), we obtain

$$\begin{array}{cc}\hfill x_0\left(t\right)& ={\mu }_0\tilde{Q}{x}_0\left(t\right)\le {\int }_0^1\tilde{G}(t,s)f(s,x_0\left(s\right))ds\hfill \\ \hfill & \le a{\int }_0^1\tilde{G}(t,s)x_0\left(s\right)ds=a\left(\tilde{T}{x}_0\right)\left(t\right).\hfill \end{array}$$

(7)

Moreover, we obtain

$$\begin{array}{c}\hfill x_0\left(t\right)\le a\left(\tilde{T}{x}_0\right)\left(t\right)\le \cdots \le a^n\left({\tilde{T}}^n{x}_0\right)\left(t\right),t\in J,n\in \mathbb{N}.\end{array}$$

By Lemma 3, we obtain

$$\begin{array}{c}\hfill \parallel x_0{\parallel }_C\le N{a}^n\parallel {\tilde{T}}^n{x}_0{\parallel }_C\le N{\left(a{M}_1\right)}^n{\parallel x_0\parallel }_C\to 0\end{array}$$

as $n\to \infty$. Thus, $\parallel u_0{\parallel }_C\equiv 0$, which contradicts the $x_0\in P\cap \partial D_1$. Thus, (6) holds. We have

$$\begin{array}{c}\hfill i(\tilde{Q},P\cap D_1,P)=1.\end{array}$$

(8)

On the other hand, we verify that when R is large and sufficient,

$$\begin{array}{c}\hfill i(\tilde{Q},P\cap D_2,P)=0.\end{array}$$

(9)

Then, when $R>\delta$ is large and sufficient, $\forall x\in P\cap \partial D_2,\sigma \ge 0$, we obtain

$$\begin{array}{c}\hfill x-\tilde{Q}x\ne \sigma x^{*},\end{array}$$

(10)

where $x^{*}$ is a eigenfunction and the eigenvalue ${\lambda }_1={\left(r\left(\tilde{T}\right)\right)}^{-1}$. If (10) does not hold, there exists $x_1\in P\cap \partial D_2$, ${\sigma }_0\ge 0$ such that $x_1-\tilde{Q}{x}_1={\sigma }_0{x}^{*}$. For $t\in J$, by (F2), we obtain

$$\begin{array}{cc}\hfill x_1\left(t\right)& =\tilde{Q}{x}_1\left(t\right)+{\sigma }_0{x}^{*}\left(t\right)\hfill \\ \hfill & ={\int }_0^1\tilde{G}(t,s)f(s,x_1\left(s\right))ds+{\sigma }_0{x}^{*}\left(t\right)\hfill \\ \hfill & \ge b{\int }_0^1\tilde{G}(t,s)x_1\left(s\right)-{\int }_0^1\tilde{G}(t,s)m_1\left(s\right)d\left(s\right)+{\sigma }_0{x}^{*}\left(t\right)\hfill \\ \hfill & =b\tilde{T}{x}_1\left(t\right)-{\int }_0^1\tilde{G}(t,s)m_1\left(s\right)d\left(s\right)+{\sigma }_0{x}^{*}\left(t\right),\hfill \end{array}$$

which implies that

$$(b\tilde{T}-I)x_1\left(t\right)\le {\int }_0^1\tilde{G}(t,s)m_1\left(s\right)ds-{\sigma }_0{x}^{*}\left(t\right)\le {\int }_0^1\tilde{G}(t,s)m_1\left(s\right)ds.$$

Since $b>{\lambda }_1$, then $(b\tilde{T}-I)$ is a positive, then there exists ${(b\tilde{T}-I)}^{-1}$. Thus,

$$\begin{array}{cc}\hfill \parallel x_1{\parallel }_C& \le N\parallel {(b\tilde{T}-I)}^{-1}\left(1\right){\parallel }_C\parallel {\int }_0^1\tilde{G}(t,s)m_1\left(s\right)ds{\parallel }_C\hfill \\ \hfill & \le M_{1}N\parallel {(b\tilde{T}-I)}^{-1}\left(1\right){\parallel }_C{\parallel m_1\parallel }_C.\hfill \end{array}$$

We choose $R>max\{\delta ,M_{1}N\parallel {(b\tilde{T}-I)}^{-1}\left(1\right){\parallel }_C\parallel m_1{\parallel }_C\}$. Then, (10) is satisfied, and by Lemma 10, we obtain

$$\begin{array}{c}\hfill i({\tilde{Q}}_1,P\cap D_2,P)=0.\end{array}$$

(11)

And, by (8) and (11), we have

$$i(\tilde{Q},P\cap (D_2\backslash {\overline{D}}_1),P)=i(\tilde{Q},P\cap D_2,P)-i(\tilde{Q},P\cap D_1,P)=-1\ne 0.$$

Hence, $\tilde{Q}$ has a fixed point in $P\cap ({\mathsf{\Omega }}_2\backslash {\overline{\mathsf{\Omega }}}_1)$, which is a positive solution of Problem (1). □

Theorem 2. 

If f is continuous and satisfies (F0) and the following conditions hold:

(F3) There exist $b>{\lambda }_1$, $\delta >0$,

$$f(t,x)\ge bx,t\in J,x\in P_{\delta },$$

where the eigenvalue ${\lambda }_1={\left(r\left(\tilde{T}\right)\right)}^{-1}$.

(F4) There exist constants $0<a<{\displaystyle \frac{1}{M_1}}$ and $m_2\in C(J,P)$,

$$f(t,x)\le ax+m_2\left(t\right),t\in J,x\in P,$$

then the problem (1) has at least one positive solution.

Proof. 

Let $D_1,D_2\subset C(J,E)$ be defined by Theorem 1. We prove that the mapping $\tilde{Q}:P\to P$ defined by (5) has a fixed point in $P\cap (D_2\backslash {\overline{D}}_1)$ if r is small enough and R large enough.

Let $r\in (0,\delta )$, where $\delta$ is the positive constant in the condition (F1). We prove that $\tilde{Q}$ satisfies the condition of Lemma 9 in $P\cap \partial D_1$. Next, $\forall x\in P\cap \partial D_1,\sigma \ge 0$, we test that

$$\begin{array}{c}\hfill x-\tilde{Q}x\ne \sigma x^{*},\end{array}$$

(12)

where $x^{*}$ is a eigenfunction and eigenvalue ${\lambda }_1={\left(r\left(\tilde{T}\right)\right)}^{-1}$. In fact, if (12) does not hold, there exist $x_2\in P\cap \partial D$ and ${\sigma }_1\ge 0$, such that $x_2-\tilde{Q}{x}_2={\sigma }_1{x}^{*}$, which implies that $x_2=\tilde{Q}{x}_2+{\sigma }_1{x}^{*}\ge {\sigma }_1{x}^{*}$. Since $x_2\in P\cap \partial D_1$, we have $0\le x_2\left(t\right),\forall t\in [0,1]$. For $t\in J$, by (F3), we obtain

$$\begin{array}{cc}\hfill \tilde{Q}{x}_2\left(t\right)& ={\int }_0^1\tilde{G}(t,s)f(s,x_2\left(s\right))ds\hfill \\ \hfill & \ge b{\int }_0^1\tilde{G}(t,s)x_2\left(s\right)ds\hfill \\ \hfill & =b\left(\tilde{T}{x}_2\right)\left(t\right),\hfill \end{array}$$

which implies that

$$x_2=\tilde{Q}{x}_2+{\sigma }_1{x}^{*}\ge b\tilde{T}{x}_2+{\sigma }_1{x}^{*}\ge {\sigma }_1{x}^{*}.$$

Let ${\sigma }^{*}=sup\left\{\sigma \right|x_2\ge \sigma x^{*},\sigma >0\}$; then, $0<{\sigma }_1\le {\sigma }^{*}<+\infty$ and $x_2\ge {\sigma }^{*}{x}^{*}$. By (3), we obtain

$${\lambda }_1\tilde{T}{x}_2\ge {\lambda }_1\tilde{T}\left({\sigma }^{*}{x}^{*}\right)={\sigma }^{*}{x}^{*}.$$

Thus,

$$\begin{array}{cc}\hfill u_2& \ge bT{u}_2+{\tau }_1{u}^{*}\hfill \\ \hfill & \ge {\tau }_1{u}^{*}+{\lambda }_{1}T{u}_2\hfill \\ \hfill & \ge {\tau }_1{u}^{*}+{\tau }^{*}{u}^{*}\hfill \\ \hfill & \ge ({\tau }^{*}+{\tau }_1)u^{*},\hfill \end{array}$$

which contradicts with ${\sigma }^{*}$. Thus, we obtain

$$\begin{array}{c}\hfill i(\tilde{Q},p\cap \partial D_1,P)=0.\end{array}$$

(13)

In addition, assuming that $R>\delta$ is sufficiently large, we prove that $\tilde{Q}$ in $P\cap \partial {\mathsf{\Omega }}_2$. By Lemma 9, we obtain

$$\begin{array}{c}\hfill \mu \tilde{Q}x\ne x,\forall x\in P\cap \partial D_2,0<\mu \le 1.\end{array}$$

(14)

If (14) does not hold, there exist $x_0\in P\cap \partial D_1$ with $0<{\mu }_0\le 1$, $x_0={\mu }_0\tilde{Q}{x}_0$. For $t\in J$, by (F4), we obtain

$$\begin{array}{cc}\hfill x_0\left(t\right)& ={\mu }_0\tilde{Q}{x}_0\left(t\right)\hfill \\ \hfill & \le {\int }_0^1\tilde{G}(t,s)f(s,x_0\left(s\right))ds\hfill \\ \hfill & \le {\int }_0^1\tilde{G}(t,s)[a{x}_0\left(s\right)+m_2\left(s\right)]ds\hfill \\ \hfill & =a\left(\tilde{T}{x}_0\right)\left(t\right)+{\int }_0^1\tilde{G}(t,s)m_2\left(s\right)ds.\hfill \end{array}$$

Therefore,

$$(I-a\tilde{T})x_0\left(t\right)\le {\int }_0^1\tilde{G}(t,s)m_2\left(s\right)ds=:v\left(t\right)$$

i.e., $(I-a\tilde{T})x_0\le v$, since

$$\parallel a\tilde{T}{\parallel }_C\le a{M}_1<1.$$

Thus, $(I-a\tilde{T})$ has a bounded inverse

$${(I-a\tilde{T})}^{-1}=\sum _{n=0}^{\infty }{\left(a\tilde{T}\right)}^n$$

Thus,

$$\theta \le x_0\le {(I-a\tilde{T})}^{-1}v.$$

Moreover, we obtain

$$\begin{array}{cc}\hfill \parallel x_0{\parallel }_C& \le N\parallel {(I-a\tilde{T})}^{-1}{v\parallel }_C\hfill \\ \hfill & \le N\sum _{n=0}^{\infty }\parallel {\left(a\tilde{T}\right)}^n{\parallel \parallel v\parallel }_C.\hfill \end{array}$$

From Lemma 3, we know that

$$\parallel x_0{\parallel }_C\le N{M}_1\parallel m_2{\parallel }_C\sum _{n=0}^{\infty }\parallel {\left(a\tilde{T}\right)}^n\parallel .$$

By (F4), $\exists ϵ>0$, $a+ϵ<{\displaystyle \frac{1}{M_1}}$, we obtain

$$\parallel {\left(a\tilde{T}\right)}^n{\parallel }_C={\parallel a^n{M}_1^n\parallel }_C\le {\left({\displaystyle \frac{a}{a+ϵ}}\right)}^n.$$

So, ${\sum }_{n=0}^{\infty }{\left({\displaystyle \frac{a}{a+ϵ}}\right)}^n$ is convergent, i.e., ${\sum }_{n=0}^{\infty }{a}^n{\parallel M_1^n\parallel }_C$ is convergent. Denote

$$\overline{M}=\sum _{n=0}^{\infty }{a}^n{\parallel M_1^n\parallel }_C<{\displaystyle \frac{1}{1-\parallel a{M}_1\parallel }}<+\infty ,$$

and it follows that

$$\parallel x_0{\parallel }_C\le N{M}_1\overline{M}{\parallel m_2\parallel }_C=:R_0,$$

Choose $R>max\{\delta ,R_0\}$. Since $u_0\in P\cap \partial {\mathsf{\Omega }}_2$, by $D_2$, we see that $\parallel x_0{\parallel }_C=R>R_0$, which contradicts with (14). By Lemma 9, we obtain

$$\begin{array}{c}\hfill i(\tilde{T},P\cap \partial D_2,P)=1.\end{array}$$

(15)

And by (13) and (15), we obtain

$$i(\tilde{Q},P\cap (D_2\backslash {\overline{D}}_1),P)=i(\tilde{Q},P\cap D_2,P)-i(\tilde{Q},P\cap D_1,P)=1\ne 0.$$

Hence, $\tilde{Q}$ has a fixed point in $P\cap (D_2\backslash {\overline{D}}_1)$, which is a positive solution of the problem (1). □

4. Conclusions

This article investigates the existence of positive solutions for a class of fractional order nonlinear boundary value problems in an ordered Banach spaces E with integral boundary conditions. The results we provide in this article improve and generalize many classical results in the real space $\mathbb{R}$.