A New Result Regarding Positive Solutions for Semipositone Boundary Value Problems of Fractional Differential Equations

Abstract

In this paper, we discuss the positive solutions to a class of semipositone boundary value problems of fractional differential equations. The nonlinearity $f(t,\mathfrak{x})$ may be singular at $t=0,1$ and satisfies $f(t,\mathfrak{x})\ge -\mathfrak{a}\left(t\right)\mathfrak{x}-\mathfrak{R}\left(t\right)$. We derive some new properties of the Green’s function of the auxiliary problems, and discover the multiplicity and existence of the positive solutions by utilizing the fixed point index theory. Two examples are illustrated to validate the main results.

1. Introduction

In recent decades, several scholars have probed into fractional differential equations (FDEs) due to their widespread applications in the realm of science. Many processes and phenomena in biology, control theory, electrochemistry, and other fields can be more accurately described by fractional-order models, see [1,2,3] and the references therein. An essential problem for FDEs is the existence of solutions. Many authors have studied this problem by using various tools of nonlinear analysis, such as the fixed point index theory [4], the monotone iterative method [5], the Leray–Schauder nonlinear alternative [6], Banach’s contraction principle [7,8], and various fixed point theorem [9,10,11,12]. Xu et al. [11] obtained positive solutions for fractional differential equation boundary value problems (FBVPs) by utilizing the Avery–Henderson fixed point theorem. Zhu et al. [12] obtained mild solutions to a integro-differential equation with fractional order derivative using the Schauder’s fixed point theorem. Considering the practical significance of solutions in most practical problems, positive solutions to FBVPs have drawn much attention. The usual method used to investigate the positive solutions is to convert the FBVP to a Fredholm operator and seek its positive fixed points. The nonlinearity needs to be non-negative in most of the papers that have considered positive solutions. For instance, Zhang and Zhong [13] considered the following FBVP:

$$\left\{\begin{array}{c}{D}_{0+}^{\gamma }\mathcal{V}\left(t\right)+f(t,\mathcal{V}\left(t\right))=0,0<t<1,n-1<\gamma <n,\hfill \\ \mathcal{V}\left(0\right)={\mathcal{V}}^{\prime }\left(0\right)=\cdots ={\mathcal{V}}^{(n-2)}\left(0\right)=0,D_{0+}^{\beta }\mathcal{V}\left(1\right)={\int }_0^{\eta }a\left(t\right)D_{0+}^{\beta }\mathcal{V}\left(t\right)dt,\hfill \end{array}\right.$$

where $D_{0+}^{\gamma }$ and $D_{0+}^{\beta }$ are the Riemann–Liouville (R-L) derivatives and $f\in C\left(\right(0,1)\times (0,+\infty \left)\right)$ is nonnegative. They established the multiplicity of the positive solutions by utilizing the Leggett–Williams theorem and Krasnoselskii’s fixed point theorem.

In [14], Cui obtained a uniqueness result to the following FBVP:

$$\left\{\begin{array}{c}{D}_{0+}^{\gamma }\mathcal{V}\left(t\right)+\mathfrak{p}\left(t\right)f(t,\mathcal{V}\left(t\right))+\mathfrak{q}\left(t\right)=0,0<t<1,\hfill \\ \mathcal{V}\left(0\right)=\mathcal{V}\left(1\right)={\mathcal{V}}^{\prime }\left(0\right)=0.\hfill \end{array}\right.$$

where $\gamma \in (2,3]$, $D_{0+}^{\gamma }$ is the R-L derivative and f is a Lipschitz continuous function. It should be noted that $f(t,\mathfrak{x})$ may change the sign, so the solution is not necessarily non-negative.

Semipositone problems depict some natural phenomena in the areas of engineering and science. Aris [15] introduced semipositone problems which occurred in the model for steady-state diffusion with the following reactions:

$$-{\mathcal{V}}^{″}\left(t\right)=\lambda f(t,\mathcal{V}\left(t\right))-\mathcal{M},$$

where $\mathcal{M}>0$ and $\lambda >0$ is the Thiele modulus. Models of fractional-order are proven to be more precise than integer-order models, since they permit more degrees of freedom. It is undeniable that there are a great many articles that handle positive solutions of FBVPs with non-negative nonlinearity, yet results considering semipositone cases are relatively scarce (see [16,17,18,19,20,21,22,23]). In [19], Yuan studied the positive solutions to the following FBVP:

$$\left\{\begin{array}{c}{D}_{0^{+}}^{\gamma }\mathcal{V}\left(t\right)+\lambda f(t,\mathcal{V}\left(t\right))=0,0<t<1,n-1<\gamma \le n,\hfill \\ {\mathcal{V}}^{\left(i\right)}=0,0\le i\le n-2,\hfill \\ \mathcal{V}\left(1\right)=0,\hfill \end{array}\right.$$

where $\lambda >0$ is a parameter, $n\ge 3$, $D_{0^{+}}^{\gamma }$ denotes the R-L derivative, and $f(t,\mathfrak{x})\ge -\mathfrak{R}\left(t\right)$ is a sign-changing continuous function, which, here, is as follows: $\mathfrak{R}\left(t\right)\in L^1([0,1],(0,+\infty ))$.

In [23], Henderson and Luca considered the following FBVP:

$$\left\{\begin{array}{c}{D}_{0+}^{\gamma }\mathcal{V}\left(t\right)+\lambda f(t,\mathcal{V}\left(t\right))=0,0<t<1,n-1<\gamma \le n,\hfill \\ \mathcal{V}\left(0\right)={\mathcal{V}}^{\prime }\left(0\right)=\cdots ={\mathcal{V}}^{(n-2)}\left(0\right)=0,D_{0+}^{{\beta }_1}\mathcal{V}\left(1\right)=\sum _{i=1}^m{\eta }_i{D}_{0+}^{{\beta }_2}\mathcal{V}\left({\xi }_i\right),\hfill \end{array}\right.$$

where $\lambda >0$ is a parameter, $n\ge 3$, ${\beta }_1\in [1,n-2],{\beta }_2\in [0,{\beta }_1]$, and $f(t,\mathfrak{x})\ge -\mathfrak{R}\left(t\right)$ is continuous. Also, $\mathfrak{R}\in L^1[0,1]\bigcap C(0,1)$ is non-negative. They established the existence results of positive solutions by utilizing the Guo–Krasnoselskii fixed point theorem.

Multi-term FDEs can be used to describe various types of visco-elastic damping. For instance, Elshehawey et.al [3] studied the following endolymph equation:

$${\mathcal{V}}^{″}\left(t\right)+a_1{\mathcal{V}}^{\prime }\left(t\right)+a_2{D}^{{\displaystyle \frac{1}{2}}}\mathcal{V}\left(t\right)+a_3\mathcal{V}\left(t\right)=-\mathfrak{y}\left(t\right),$$

which depicted the response of semicircular canals to angular acceleration. It is clear that the Green’s function plays a crucial part in seeking positive fixed points. However, due to the complexity caused by the extra term, the Green’s functions to such problems failed to be fully exploited. Based on the spectral theory, Graef [24] and Zou [25] studied the following FBVP:

$$\left\{\begin{array}{c}-D_{0+}^{\gamma }\mathcal{V}\left(t\right)+a\left(t\right)\mathcal{V}\left(t\right)=\mathfrak{q}\left(t\right)f\left(\mathcal{V}\left(t\right)\right),0<t<1,\hfill \\ \mathcal{V}\left(0\right)={\mathcal{V}}^{\prime }\left(0\right)=0,{\mathcal{V}}^{\prime }\left(1\right)=0,\hfill \end{array}\right.$$

where $\gamma \in (2,3)$, $a\in C[0,1]$. In [24], the authors deduced the associated Green’s function and obtained positive properties for it. Zou [25] obtained some new properties for the same Green’s function. In [4], we discussed the positive solutions to a resonant FDE, as follows:

$$-D_{0+}^{\gamma }\mathcal{V}\left(t\right)=f(t,\mathcal{V}\left(t\right)),$$

by converting it to the equivalent non-resonant FDE, as follows:

$$-D_{0+}^{\gamma }\mathcal{V}\left(t\right)+a_1\mathcal{V}\left(t\right)=f(t,\mathcal{V}\left(t\right))+a_1\mathcal{V}\left(t\right).$$

In view of the aforementioned, we now study the existence of positive solutions to the following semipositone FBVP:

$$\left\{\begin{array}{c}{D}_{0+}^{\alpha }\mathfrak{U}\left(t\right)+\lambda f(t,\mathfrak{U}\left(t\right))=0,t\in (0,1),\hfill \\ \mathfrak{U}\left(0\right)=\mathfrak{U}\left(1\right)={\mathfrak{U}}^{\prime }\left(0\right)=0.\hfill \end{array}\right.$$

(1)

where $\alpha \in (2,3]$, $\lambda >0$ is a parameter, $D_{0+}^{\alpha }$ is the R-L derivative, and $f\in C\left(\right(0,1)\times [0,+\infty \left)\right)$ satisfies the following hypothesis:

$\left(H_0\right)$ There exist the non-negative functions $\mathfrak{a},\mathfrak{R}\in C(0,1)\bigcap L^1[0,1]$, such that $f(t,\mathfrak{x})\ge -\mathfrak{a}\left(t\right)\mathfrak{x}-\mathfrak{R}\left(t\right)$, $(t,\mathfrak{x})\in (0,1)\times [0,+\infty )$.

The method used in previous papers [16,17,18,19,20,21,22,23] cannot be utilized under hypothesis $\left(H_0\right)$, since there is an extra term, specifically $\mathfrak{a}\left(t\right)\mathfrak{x}$. Therefore, we investigate the following linear auxiliary problem:

$$\left\{\begin{array}{c}-D_{0+}^{\alpha }\mathfrak{U}\left(t\right)+\lambda \mathfrak{a}\left(t\right)\mathfrak{U}\left(t\right)=y\left(t\right),t\in (0,1),\hfill \\ \mathfrak{U}\left(0\right)=\mathfrak{U}\left(1\right)={\mathfrak{U}}^{\prime }\left(0\right)=0.\hfill \end{array}\right.$$

(2)

We will prove some new properties of Green’s function of FBVP (2) and employ them to seek positive solutions for FBVP (1). The results presented in this article have some new features. First, we consider the existence and multiplicity of positive solutions to a semipositone FBVP, which is different from previous research [13,14]. Second, the coefficient a of the perturbation term in [24,25] is required to be continuous. Compared with [24,25], the linear auxiliary problem (2) possesses singularity; that is, $\mathfrak{a}$ may be singular at $t=0,1$. Third, the significant difference from the existing results [16,17,18,19,20,21,22,23] lies in the fact that we deal with more general cases; that is, the nonlinearity f satisfies the following: $f(t,\mathfrak{x})\ge -\mathfrak{a}\left(t\right)\mathfrak{x}-\mathfrak{R}\left(t\right)$. It is noted that the results obtained in the above papers fail to work under this condition. Furthermore, the tool used in this paper is the fixed point index theory. Therefore, our results enrich the study of semipositone FBVPs. In addition, the methods used in this paper can be applied to some other differential equations.

The paper is organized as follows. In Section 2, we introduce some necessary definitions and lemmas from fractional calculus theory, and deduce some new properties for Green’s function of the linear auxiliary problem (2). In Section 3, we establish the existence of positive solutions for the FBVP (1). In Section 4, we illustrate two examples to corroborate the main results.

2. Some Preliminaries and Lemmas

In this section, we will introduce some preliminary results which will be used in the proofs later.

Definition 1

([2]). The fractional integral of order $\beta (\beta >0)$ of a function $\mathfrak{u}:(0,+\infty )\to R$ is defined by the following:

$${\mathcal{J}}_{0+}^{\beta }\mathfrak{u}\left(x\right)={\displaystyle \frac{1}{\Gamma \left(\beta \right)}}{\mathit{\int }}_0^x{(x-\tau )}^{\beta -1}\mathfrak{u}\left(\tau \right)d\tau ,$$

in which $\Gamma \left(\beta \right)={\int }_0^{+\infty }{\tau }^{\beta -1}{e}^{-\tau }d\tau$ is the Euler gamma function.

Definition 2

([2]). The R-L fractional derivative of order $\beta (n-1<\beta <n)$ of a function $\mathfrak{U}:(0,+\infty )\to R$ is defined as follows:

$$D_{0+}^{\beta }\mathfrak{U}\left(x\right)={\displaystyle \frac{1}{\Gamma (n-\beta )}}{\left({\displaystyle \frac{d}{dx}}\right)}^n{\int }_0^x{(x-\tau )}^{n-\beta -1}\mathfrak{U}\left(\tau \right)d\tau ,$$

where $n=\left[\beta \right]+1$, $\left[\beta \right]$ is the integer part of β.

Lemma 1

([2]). Let $\beta ,\gamma >0$. If $y\left(\tau \right)\in L^1[0,1]$, then the following holds:

(1) 

$\left(D_{0+}^{\beta }{\mathcal{J}}_{0+}^{\beta }y\right)\left(\tau \right)=y\left(\tau \right).$

(2) 

$\left({\mathcal{J}}_{0+}^{\beta }{\mathcal{J}}_{0+}^{\gamma }y\right)\left(\tau \right)=\left({\mathcal{J}}_{0+}^{\beta +\gamma }y\right)\left(\tau \right)$.

Lemma 2

([2]). Let $\beta >0$ and $n=\left[\beta \right]+1$. $\left(D_{0+}^{\beta }y\right)\left(\tau \right)=0$ if and only if

$$y\left(\tau \right)=c_1{\tau }^{\beta -1}+c_2{\tau }^{\beta -2}+···+c_n{\tau }^{\beta -n},$$

where $c_i\in \mathbb{R}$, $i=1,2,...,n$.

First, we list the following hypotheses, which will be used in the rest of the paper.

Hypothesis 1.

$f(t,\mathfrak{x})$ is continuous on $(0,1)\times [0,+\infty )$, and there exist non-negative functions $\mathfrak{a},\mathfrak{R}\in C(0,1)\bigcap L^1[0,1]$, such that the following is true:

$$\begin{array}{c}\hfill f(t,\mathfrak{x})\ge -\mathfrak{a}\left(t\right)\mathfrak{x}-\mathfrak{R}\left(t\right),\forall (t,\mathfrak{x})\in (0,1)\times [0,+\infty ),\end{array}$$

with ${\int }_0^1\mathfrak{a}\left(\tau \right)d\tau >0$ and ${\int }_0^1\mathfrak{R}\left(\tau \right)d\tau >0$.

Hypothesis 2.

$f(t,\mathfrak{x})$ is continuous on $(0,1)\times [0,+\infty )$, and there exists a non-negative function $\mathfrak{a}\in C(0,1)\bigcap L^1[0,1]$, such that the following is true:

$$\begin{array}{c}\hfill f(t,\mathfrak{x})\ge -\mathfrak{a}\left(t\right)\mathfrak{x},\forall (t,\mathfrak{x})\in (0,1)\times [0,+\infty ),\end{array}$$

with ${\int }_0^1\mathfrak{a}\left(\tau \right)d\tau >0$.

Hypothesis 3.

There exist non-negative functions $\mathfrak{b}\in C(0,1)$ and $h\in C[0,+\infty )$, such that ${\int }_0^1\mathfrak{b}\left(\tau \right)\mathfrak{g}\left(\tau \right)d\tau <+\infty$ and that the following is true:

$$\begin{array}{c}\hfill f(t,\mathfrak{x})\le \mathfrak{b}\left(t\right)h\left(\mathfrak{x}\right),\forall (t,\mathfrak{x})\in (0,1)\times [0,+\infty ),\end{array}$$

where $\mathfrak{g}\left(\tau \right)=\tau {(1-\tau )}^{\alpha -1}$.

Hypothesis 4.

$\exists [c,d]\subset (0,1)$, such that the following is true:

$$\begin{array}{c}\hfill \underset{\mathfrak{x}\to +\infty }{lim\; inf}\underset{t\in [c,d]}{min}{\displaystyle \frac{f(t,\mathfrak{x})}{\mathfrak{x}}}=+\infty .\end{array}$$

For convenience, we introduce the following notations:

$$\begin{array}{cc}\hfill & {\mathcal{K}}_0(t,\tau )={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left\{\begin{array}{cc}{\left[t(1-\tau )\right]}^{\alpha -1},\hfill & 0\le t\le \tau \le 1,\hfill \\ {\left[t(1-\tau )\right]}^{\alpha -1}-{(t-\tau )}^{\alpha -1},\hfill & 0\le \tau \le t\le 1,\hfill \end{array}\right.\hfill \\ \hfill & {\mathcal{K}}_n(t,\tau )={\int }_0^1\lambda \mathfrak{a}\left(s\right){\mathcal{K}}_0(t,s){\mathcal{K}}_{n-1}(s,\tau )ds,n=1,2,\cdots .\hfill \\ \hfill & {\lambda }_1={\displaystyle \frac{\Gamma (\alpha -1)}{{\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }},{\lambda }_2={\displaystyle \frac{\Gamma (\alpha -1)}{{\int }_0^1\mathfrak{a}\left(\tau \right)d\tau }},\hfill \\ \hfill & {\lambda }_3={\displaystyle \frac{\Gamma (\alpha -1)}{3(\alpha -1){\int }_0^1\mathfrak{a}\left(\tau \right)d\tau +{\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }},\hfill \\ \hfill & \mathrm{\Lambda }={\displaystyle \frac{\lambda (\alpha -1){\int }_0^1\mathfrak{a}\left(\tau \right)d\tau }{({\lambda }_1-\lambda ){\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }},\hfill \end{array}$$

(3)

in which the function $\mathfrak{a}$ satisfies the assumption in Hypothesis 1 or Hypothesis 2, and $\mathfrak{g}$ is defined in Hypothesis 3.

Lemma 3

([20]). If $y\in C(0,1)\cap L^1[0,1]$, then the solution to the below problem

$$\left\{\begin{array}{c}-D_{0+}^{\alpha }\mathfrak{U}\left(t\right)=y\left(t\right),t\in (0,1),\hfill \\ \mathfrak{U}\left(0\right)=\mathfrak{U}\left(1\right)={\mathfrak{U}}^{\prime }\left(0\right)=0.\hfill \end{array}\right.$$

is $\mathfrak{U}\left(t\right)={\int }_0^1{\mathcal{K}}_0(t,\tau )y\left(\tau \right)d\tau .$ The function ${\mathcal{K}}_0(t,\tau )$ possesses the following properties:

(1) 

${\mathcal{K}}_0(t,\tau )={\mathcal{K}}_0(1-\tau ,1-t);$

(2) 

${\mathcal{K}}_0(t,\tau )>0,fort,\tau \in (0,1);$

(3) 

${\displaystyle \frac{\mathfrak{g}(1-t)\mathfrak{g}\left(\tau \right)}{\Gamma \left(\alpha \right)}}\le {\mathcal{K}}_0(t,\tau )\le {\displaystyle \frac{\mathfrak{g}\left(\tau \right)}{\Gamma (\alpha -1)}},fort,\tau \in [0,1];$

(4) 

${\displaystyle \frac{\mathfrak{g}(1-t)\mathfrak{g}\left(\tau \right)}{\Gamma \left(\alpha \right)}}\le {\mathcal{K}}_0(t,\tau )\le {\displaystyle \frac{\mathfrak{g}(1-t)}{\Gamma (\alpha -1)}},fort,\tau \in [0,1].$

Remark 1.

Green’s function of FBVP (1) is ${\mathcal{K}}_0(t,\tau )$.

Lemma 4.

Assume that the positive parameter $\lambda <{\lambda }_1$. Then, the series of functions

$$\sum _{n=0}^{+\infty }{(-1)}^n{\mathcal{K}}_n(t,\tau )$$

is uniformly convergent on $[0,1]\times [0,1]$.

Proof. 

It follows from Lemma 3 that the following is true:

$$\begin{array}{cc}\hfill {\mathcal{K}}_1(t,\tau )& ={\int }_0^1\lambda \mathfrak{a}\left(s\right){\mathcal{K}}_0(t,s){\mathcal{K}}_0(s,\tau )ds\le {\int }_0^1\lambda \mathfrak{a}\left(s\right){\displaystyle \frac{\mathfrak{g}(1-t)}{\Gamma (\alpha -1)}}·{\displaystyle \frac{\mathfrak{g}\left(\tau \right)}{\Gamma (\alpha -1)}}ds\hfill \\ \hfill & ={\displaystyle \frac{\lambda {\int }_0^1\mathfrak{a}\left(s\right)ds}{{(\Gamma (\alpha -1))}^2}}\mathfrak{g}(1-t)\mathfrak{g}\left(\tau \right),\hfill \end{array}$$

and that the following is also true:

$$\begin{array}{cc}\hfill {\mathcal{K}}_1(t,\tau )& ={\int }_0^1\lambda \mathfrak{a}\left(s\right){\mathcal{K}}_0(t,s){\mathcal{K}}_0(s,\tau )ds\ge {\int }_0^1\lambda \mathfrak{a}\left(s\right){\displaystyle \frac{\mathfrak{g}(1-t)\mathfrak{g}\left(s\right)}{\Gamma \left(\alpha \right)}}·{\displaystyle \frac{\mathfrak{g}(1-s)\mathfrak{g}\left(\tau \right)}{\Gamma \left(\alpha \right)}}ds\hfill \\ \hfill & ={\displaystyle \frac{\lambda {\int }_0^1\mathfrak{a}\left(s\right)\mathfrak{g}\left(s\right)\mathfrak{g}(1-s)ds}{{(\Gamma \left(\alpha \right))}^2}}\mathfrak{g}(1-t)\mathfrak{g}\left(\tau \right).\hfill \end{array}$$

Then, the following is also true:

$$\begin{array}{cc}\hfill {\mathcal{K}}_2(t,\tau )& ={\int }_0^1\lambda \mathfrak{a}\left(s\right){\mathcal{K}}_0(t,s){\mathcal{K}}_1(s,\tau )ds\le {\int }_0^1\lambda \mathfrak{a}\left(s\right){\displaystyle \frac{\mathfrak{g}(1-t)}{\Gamma (\alpha -1)}}·{\displaystyle \frac{\lambda {\int }_0^1\mathfrak{a}\left(s\right)ds}{{(\Gamma (\alpha -1))}^2}}\mathfrak{g}(1-s)\mathfrak{g}\left(\tau \right)ds\hfill \\ \hfill & ={\displaystyle \frac{{\lambda }^2{\int }_0^1\mathfrak{a}\left(s\right)ds{\int }_0^1\mathfrak{a}\left(s\right)\mathfrak{g}(1-s)ds}{{(\Gamma (\alpha -1))}^3}}\mathfrak{g}(1-t)\mathfrak{g}\left(\tau \right),\hfill \end{array}$$

and the following is true:

$$\begin{array}{cc}\hfill {\mathcal{K}}_2(t,\tau )& ={\int }_0^1\lambda \mathfrak{a}\left(s\right){\mathcal{K}}_0(t,s){\mathcal{K}}_1(s,\tau )ds\hfill \\ \hfill & \ge {\int }_0^1\lambda \mathfrak{a}\left(s\right){\displaystyle \frac{\mathfrak{g}(1-t)\mathfrak{g}\left(s\right)}{\Gamma \left(\alpha \right)}}·{\displaystyle \frac{\lambda {\int }_0^1\mathfrak{a}\left(s\right)\mathfrak{g}\left(s\right)\mathfrak{g}(1-s)ds}{{(\Gamma \left(\alpha \right))}^2}}\mathfrak{g}(1-s)\mathfrak{g}\left(\tau \right)ds\hfill \\ \hfill & ={\displaystyle \frac{{\lambda }^2{\left({\int }_0^1\mathfrak{a}\left(s\right)\mathfrak{g}\left(s\right)\mathfrak{g}(1-s)ds\right)}^2}{{(\Gamma \left(\alpha \right))}^3}}\mathfrak{g}(1-t)\mathfrak{g}\left(\tau \right).\hfill \end{array}$$

By induction, we obtain the following:

$${\mathcal{K}}_n(t,\tau )\le {\displaystyle \frac{{\lambda }^n{\int }_0^1\mathfrak{a}\left(s\right)ds{\left({\int }_0^1\mathfrak{a}\left(s\right)\mathfrak{g}(1-s)ds\right)}^{n-1}}{{(\Gamma (\alpha -1))}^{n+1}}}\mathfrak{g}(1-t)\mathfrak{g}\left(\tau \right),$$

(4)

and the following is also true:

$${\mathcal{K}}_n(t,\tau )\ge {\displaystyle \frac{{\lambda }^n{\left({\int }_0^1\mathfrak{a}\left(s\right)\mathfrak{g}\left(s\right)\mathfrak{g}(1-s)ds\right)}^n}{{(\Gamma \left(\alpha \right))}^{n+1}}}\mathfrak{g}(1-t)\mathfrak{g}\left(\tau \right),n=1,2,\cdots .$$

Note that the notation of ${\lambda }_1$ and $\mathfrak{g}\left(\tau \right)<1$. It follows from (4) that the following holds:

$${\mathcal{K}}_n(t,\tau )\le {\displaystyle \frac{\lambda {\int }_0^1\mathfrak{a}\left(s\right)ds}{{(\Gamma (\alpha -1))}^2}}{\left({\displaystyle \frac{\lambda }{{\lambda }_1}}\right)}^{n-1},n=1,2,\cdots ,$$

which implies that the following series of functions is uniformly convergent on $[0,1]\times [0,1]$:

$$\sum _{n=0}^{+\infty }{(-1)}^n{\mathcal{K}}_n(t,\tau ).$$

□

Denote a Banach space $E=C[0,1]$ using the maximum norm. Let the following hold:

$$\mathcal{K}(t,\tau )=\sum _{n=0}^{+\infty }{(-1)}^n{\mathcal{K}}_n(t,\tau ).$$

(5)

Lemma 5.

Assume that the positive parameter $\lambda <{\lambda }_2$ and $y\in C(0,1)\cap L(0,1)$. Then, the unique solution of FBVP (2) can be written as follows:

$$\mathfrak{U}\left(t\right)={\int }_0^1\mathcal{K}(t,\tau )y\left(\tau \right)d\tau .$$

Proof. 

It is clear that ${\lambda }_2<{\lambda }_1$ and ${\mathcal{K}}_n(t,\tau )$$(n=0,1,2,\cdots )$ are continuous on $[0,1]\times [0,1]$. From Lemma 4, we learn that $\mathcal{K}(t,\tau )$ is also continuous on $[0,1]\times [0,1]$.

Let $\mathfrak{U}\in E$ be a solution of FBVP (2). It follows from Lemma 3 that the following is true:

$$\mathfrak{U}\left(t\right)={\int }_0^1{\mathcal{K}}_0(t,\tau )(y\left(\tau \right)-\mathfrak{a}\left(\tau \right)\mathfrak{U}\left(\tau \right))d\tau ,$$

that is, the following is also true:

$$\mathfrak{U}\left(t\right)+{\int }_0^1{\mathcal{K}}_0(t,\tau )\mathfrak{a}\left(\tau \right)\mathfrak{U}\left(\tau \right)d\tau ={\int }_0^1{\mathcal{K}}_0(t,\tau )y\left(\tau \right)d\tau .$$

(6)

We denote the linear operators T and $L_{\lambda }:E\to E$ as follows:

$$T\mathfrak{U}\left(t\right)={\int }_0^1{\mathcal{K}}_0(t,\tau )\mathfrak{U}\left(\tau \right)d\tau ,$$

$$L_{\lambda }\mathfrak{U}\left(t\right)={\int }_0^1\lambda \mathfrak{a}\left(\tau \right){\mathcal{K}}_0(t,\tau )\mathfrak{U}\left(\tau \right)d\tau .$$

Since $\lambda <{\lambda }_2$, we obtain the following:

$$\parallel L_{\lambda }\parallel =\underset{\parallel \mathfrak{U}\parallel =1}{max}\parallel L_{\lambda }\mathfrak{U}\parallel <1.$$

Then, $(\mathcal{I}+L_{\lambda })$ is invertible and the following holds:

$${(\mathcal{I}+L_{\lambda })}^{-1}=\mathcal{I}-L_{\lambda }+L_{\lambda }^2+···+{(-1)}^n{L}_{\lambda }^n+···,$$

in which $\mathcal{I}$ denotes the identity operator. Therefore, (6) can be expressed as follows:

$$\mathfrak{U}={(\mathcal{I}+L_{\lambda })}^{-1}Ty=\sum _{n=0}^{+\infty }{(-L_{\lambda })}^{n}Ty.$$

(7)

Next, we prove the following:

$$\left({(-L_{\lambda })}^{n}Ty\right)\left(t\right)={\int }_0^1{(-1)}^n{\mathcal{K}}_n(t,\tau )y\left(\tau \right)d\tau ,n=0,1,2,\cdots .$$

(8)

It is obvious that (8) holds for $n=0$ according to the definition of T. We assume (8) holds for $n=m$. By the definition of $L_{\lambda }$ and notation of ${\mathcal{K}}_n$, the following also holds:

$$\begin{array}{cc}\hfill & \left({(-L_{\lambda })}^{m+1}Ty\right)\left(t\right)=(-L_{\lambda }\left({(-L_{\lambda })}^{m}Ty\right))\left(t\right)\hfill \\ \hfill & ={\int }_0^1{(-1)}^{m+1}\lambda \mathfrak{a}\left(s\right){\mathcal{K}}_0(t,s){\int }_0^1{\mathcal{K}}_m(s,\tau )y\left(\tau \right)d\tau ds\hfill \\ & ={\int }_0^1{(-1)}^{m+1}y\left(\tau \right){\int }_0^1\lambda \mathfrak{a}\left(s\right){\mathcal{K}}_0(t,s){\mathcal{K}}_m(s,\tau )dsd\tau \hfill \\ & ={\int }_0^1{(-1)}^{m+1}{\mathcal{K}}_{m+1}(t,\tau )y\left(\tau \right)d\tau .\hfill \end{array}$$

Thus, (8) holds.

Taking into account Equations (5), (7), and (8), as well as Lemma 4, we obtain the following:

$$\mathfrak{U}\left(t\right)={\int }_0^1\mathcal{K}(t,\tau )y\left(\tau \right)d\tau .$$

□

Lemma 6.

Assume the positive parameter $\lambda <{\lambda }_2$. $\mathcal{K}(t,\tau )$ satisfies the following properties:

$$(1-\Lambda ){\mathcal{K}}_0(t,\tau )\le \mathcal{K}(t,\tau )\le (1+\Lambda ){\mathcal{K}}_0(t,\tau ),$$

where Λ is defined in (3).

Proof. 

From (4) and Lemma 3, we obtain the following:

$$\begin{array}{cc}\hfill \sum _{n=1}^{+\infty }{\mathcal{K}}_n(t,\tau )\le & \mathfrak{g}(1-t)\mathfrak{g}\left(\tau \right)·\sum _{n=1}^{+\infty }{\displaystyle \frac{\lambda {\int }_0^1\mathfrak{a}\left(s\right)ds}{{(\Gamma (\alpha -1))}^2}}{\left({\displaystyle \frac{\lambda }{{\lambda }_1}}\right)}^{n-1}\hfill \\ \hfill =& {\displaystyle \frac{\lambda {\lambda }_1\mathfrak{g}(1-t)\mathfrak{g}\left(\tau \right){\int }_0^1\mathfrak{a}\left(s\right)ds}{({\lambda }_1-\lambda ){(\Gamma (\alpha -1))}^2}}\hfill \\ \hfill \le & {\displaystyle \frac{\lambda {\lambda }_1{\int }_0^1\mathfrak{a}\left(s\right)ds}{({\lambda }_1-\lambda ){(\Gamma (\alpha -1))}^2}}\times \Gamma \left(\alpha \right){\mathcal{K}}_0(t,\tau )\hfill \\ \hfill =& {\displaystyle \frac{\lambda {\lambda }_1(\alpha -1){\int }_0^{1}a\left(\tau \right)d\tau }{({\lambda }_1-\lambda )\Gamma (\alpha -1)}}{\mathcal{K}}_0(t,\tau )=\Lambda {\mathcal{K}}_0(t,\tau ).\hfill \end{array}$$

Thus, the following holds:

$$\mathcal{K}(t,\tau )\le {\mathcal{K}}_0(t,\tau )+\sum _{n=1}^{+\infty }{\mathcal{K}}_n(t,\tau )\le (1+\Lambda ){\mathcal{K}}_0(t,\tau ),$$

and the equation below also holds:

$$\mathcal{K}(t,\tau )\ge {\mathcal{K}}_0(t,\tau )-\sum _{n=1}^{+\infty }{\mathcal{K}}_n(t,\tau )\ge (1-\Lambda ){\mathcal{K}}_0(t,\tau ).$$

□

Remark 2.

As a function of λ, Λ is strictly increasing on $(0,{\lambda }_1)$. If $\lambda \in (0,{\lambda }_3)$, then $0<\Lambda <{\displaystyle \frac{1}{3}}$. Furthermore, the following holds:

$${\displaystyle \frac{2}{3}}{\mathcal{K}}_0(t,\tau )\le \mathcal{K}(t,\tau )\le {\displaystyle \frac{4}{3}}{\mathcal{K}}_0(t,\tau ).$$

Proof. 

Clearly, $0<{\lambda }_3<{\lambda }_2<{\lambda }_1$. $\forall \lambda \in (0,{\lambda }_1)$, as follows:

$$\Lambda ={\displaystyle \frac{\lambda (\alpha -1){\int }_0^1\mathfrak{a}\left(\tau \right)d\tau }{({\lambda }_1-\lambda ){\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }}={\displaystyle \frac{(\alpha -1){\int }_0^1\mathfrak{a}\left(\tau \right)d\tau }{\left(\frac{{\lambda }_1}{\lambda }-1\right){\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }},$$

This yields that $\Lambda$, as a function of $\lambda$, is strictly increasing on $(0,{\lambda }_1)$. If $\lambda \in (0,{\lambda }_3)$, then the following is true:

$${\displaystyle \frac{{\lambda }_1}{{\lambda }_3}}-1={\displaystyle \frac{3(\alpha -1){\int }_0^1\mathfrak{a}\left(\tau \right)d\tau +{\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }{{\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }}-1={\displaystyle \frac{3(\alpha -1){\int }_0^1\mathfrak{a}\left(\tau \right)d\tau }{{\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }}.$$

Therefore, the following is also true:

$$0<\Lambda <{\displaystyle \frac{(\alpha -1){\int }_0^1\mathfrak{a}\left(\tau \right)d\tau }{\left(\frac{{\lambda }_1}{{\lambda }_3}-1\right){\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }}={\displaystyle \frac{1}{3}}.$$

This inequality and Lemma 6 imply the following:

$${\displaystyle \frac{2}{3}}{\mathcal{K}}_0(t,\tau )\le \mathcal{K}(t,\tau )\le {\displaystyle \frac{4}{3}}{\mathcal{K}}_0(t,\tau ).$$

□

Lemma 7.

Assume the positive parameter $\lambda <{\lambda }_3$. Then, the solution for the FBVP, as shown below, is ${\omega }_{\lambda }\left(t\right)=\lambda {\int }_0^1\mathcal{K}(t,\tau )\mathfrak{R}\left(\tau \right)d\tau .$

$$\left\{\begin{array}{c}-D_{0+}^{\alpha }\mathfrak{U}\left(t\right)+\lambda \mathfrak{a}\left(t\right)\mathfrak{U}\left(t\right)=\lambda \mathfrak{R}\left(t\right),0<t<1,\hfill \\ \mathfrak{U}\left(0\right)=\mathfrak{U}\left(1\right)={\mathfrak{U}}^{\prime }\left(0\right)=0.\hfill \end{array}\right.$$

Moreover, the following holds:

$${\omega }_{\lambda }\left(t\right)\le {\displaystyle \frac{4\lambda {\int }_0^1\mathfrak{R}\left(\tau \right)d\tau }{3\Gamma (\alpha -1)}}\mathfrak{g}(1-t).$$

(9)

Proof. 

This Lemma can be deduced directly from Lemma 3, Lemma 5, and Remark 2, so we have omitted this Proof section. □

Lemma 8

([26]). We assume that E is a Banach space, $\mathcal{P}\subset E$ is a cone, the zero element $\theta \in \mathrm{\Omega }$ and $\mathrm{\Omega }\subset E$ is a bounded open set, and $A:$$\mathcal{P}\cap \overline{\mathrm{\Omega }}\to \mathcal{P}$ is completely continuous. If the following is true:

$$A\mathfrak{U}\ne \mu \mathfrak{U},\forall \mu \ge 1,\mathfrak{U}\in \mathcal{P}\cap \partial \mathrm{\Omega },$$

then $i(A,\mathcal{P}\cap \mathrm{\Omega },\mathcal{P})=1$.

Lemma 9

([26]). We assume that E is a Banach space, $\mathcal{P}\subset E$ is a cone, $\mathrm{\Omega }\subset E$ is a bounded open set, and $A:$$\mathcal{P}\cap \overline{\mathrm{\Omega }}\to \mathcal{P}$ is completely continuous. If there exists ${\mathfrak{U}}_0\in \mathcal{P}$ with ${\mathfrak{U}}_0\ne \theta$, such that the following holds:

$$\mathfrak{U}-A\mathfrak{U}\ne \mu {\mathfrak{U}}_0,\forall \mu \ge 0,\mathfrak{U}\in \mathcal{P}\cap \partial \mathrm{\Omega },$$

then $i(A,\mathcal{P}\cap \mathrm{\Omega },\mathcal{P})=0$.

3. Main Results

We define a cone as follows:

$$\mathcal{P}=\left\{\mathfrak{U}\in E:\mathfrak{U}\left(t\right)\ge {\displaystyle \frac{\mathfrak{g}(1-t)\parallel \mathfrak{U}\parallel }{2(\alpha -1)}},t\in [0,1]\right\}.$$

We also set the following:

$$h^{*}\left(z\right)=max\{h\left(x\right):0\le x\le z\}.$$

Additionally, we denote the following: $B_l=\{\mathfrak{U}\left(t\right)\in E:\parallel \mathfrak{U}\parallel <l\}$, ${\mathcal{P}}_l=\mathcal{P}\cap B_l$ and $\partial {\mathcal{P}}_l=\mathcal{P}\cap \partial B_l$.

Now, we study the following FBVP:

$$\left\{\begin{array}{c}-D_{0+}^{\alpha }\mathfrak{U}\left(t\right)+\lambda \mathfrak{a}\left(t\right)\mathfrak{U}\left(t\right)\hfill \\ =\lambda \left(f(t,{[\mathfrak{U}\left(t\right)-{\omega }_{\lambda }\left(t\right)]}^{+})+\mathfrak{a}\left(t\right){[\mathfrak{U}\left(t\right)-{\omega }_{\lambda }\left(t\right)]}^{+}+\mathfrak{R}\left(t\right)\right),t\in (0,1),\hfill \\ \mathfrak{U}\left(0\right)=\mathfrak{U}\left(1\right)={\mathfrak{U}}^{\prime }\left(0\right)=0.\hfill \end{array}\right.$$

where ${[\mathfrak{U}\left(t\right)-{\omega }_{\lambda }\left(t\right)]}^{+}=max\{0,\mathfrak{U}\left(t\right)-{\omega }_{\lambda }\left(t\right)\}.$

We define the operator A as follows:

$$A\mathfrak{U}\left(t\right)=\lambda {\int }_0^1\mathcal{K}(t,\tau )\left(f(\tau ,{[\mathfrak{U}\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})+\mathfrak{a}\left(\tau \right){[\mathfrak{U}\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}+\mathfrak{R}\left(\tau \right)\right)d\tau .$$

Theorem 1.

We assume that Hypothesis 1 and Hypothesis 3 hold. If $\lambda \in (0,{\lambda }_3)$, then $A:$$P\to P$ is a completely continuous operator.

Proof. 

Let $\mathfrak{U}\in \mathcal{P}$; it follows from Remark 2 and Lemma 3 that the following holds:

$$A\mathfrak{U}\left(t\right)\le {\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_0^1\mathfrak{g}\left(\tau \right)[\mathfrak{b}\left(\tau \right)h^{*}(\parallel \mathfrak{U}\parallel )+\mathfrak{a}\left(\tau \right)\parallel \mathfrak{U}\parallel +\mathfrak{R}\left(\tau \right)]d\tau <+\infty .$$

(10)

This implies the operator A is well-defined for $\mathcal{P}$.

Again, from Remark 2 and Lemma 3, we obtain the following:

$$A\mathfrak{U}\left(t\right)\le {\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_0^1\mathfrak{g}\left(\tau \right)\left(f(\tau ,{[\mathfrak{U}\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})+\mathfrak{a}\left(\tau \right){[\mathfrak{U}\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}+\mathfrak{R}\left(\tau \right)\right)d\tau ,$$

and

$$A\mathfrak{U}\left(t\right)\ge {\displaystyle \frac{2\lambda \mathfrak{g}(1-t)}{3\Gamma \left(\alpha \right)}}{\int }_0^1\mathfrak{g}\left(\tau \right)\left(f(\tau ,{[\mathfrak{U}\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})+\mathfrak{a}\left(\tau \right){[\mathfrak{U}\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}+\mathfrak{R}\left(\tau \right)\right)d\tau .$$

According to the above analysis, we obtain $A\mathfrak{U}\left(t\right)\ge {\displaystyle \frac{\mathfrak{g}(1-t)}{2(\alpha -1)}}\parallel A\mathfrak{U}\parallel$. Thus, $A:$$\mathcal{P}\to \mathcal{P}$.

Next, we will show the continuity of A.

We assume $\left\{{\mathfrak{U}}_n\right\}\subset \mathcal{P}$ and $\parallel {\mathfrak{U}}_n-{\mathfrak{U}}_0\parallel \to 0(n\to \infty )$. Then, there is a $M>0$, such that the following holds:

$$\parallel {\mathfrak{U}}_n\parallel +\parallel {\omega }_{\lambda }\parallel \le M,n=0,1,2,\cdots .$$

While $\forall ϵ>0$, by considering (10) and absolute continuity of integrals, there exists $\delta \in (0,{\displaystyle \frac{1}{2}})$, such that the following holds:

$${\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_0^{\delta }\mathfrak{g}\left(s\right)[\mathfrak{b}\left(s\right)h^{*}\left(M\right)+M·\mathfrak{a}\left(s\right)]ds<{\displaystyle \frac{ϵ}{6}},$$

and

$${\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_{1-\delta }^1\mathfrak{g}\left(s\right)[\mathfrak{b}\left(s\right)h^{*}\left(M\right)+M·\mathfrak{a}\left(s\right)]ds<{\displaystyle \frac{ϵ}{6}}.$$

Clearly, the following is true:

$$\left|{[{\mathfrak{U}}_n\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}-{[{\mathfrak{U}}_0\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}\right|\le |{\mathfrak{U}}_n\left(\tau \right)-{\mathfrak{U}}_0\left(\tau \right)|.$$

It then follows from $\parallel {\mathfrak{U}}_n-{\mathfrak{U}}_0\parallel \to 0$ that the following is true:

$$\parallel {[{\mathfrak{U}}_n\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}-{[{\mathfrak{U}}_0\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}\parallel \to 0.$$

Since $f(t,x)$ is uniformly continuous on $[\delta ,1-\delta ]\times [0,M]$, there is a $N>0$ for any $n>N$ as follows:

$${\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_{\delta }^{1-\delta }\mathfrak{g}\left(\tau \right)\left|f(\tau ,{[{\mathfrak{U}}_n\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})-f(\tau ,{[{\mathfrak{U}}_0\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})\right|d\tau <{\displaystyle \frac{ϵ}{6}},$$

and

$${\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_{\delta }^{1-\delta }\mathfrak{g}\left(\tau \right)\mathfrak{a}\left(\tau \right)\left|{[{\mathfrak{U}}_n\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}-{[{\mathfrak{U}}_0\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}\right|d\tau <{\displaystyle \frac{ϵ}{6}}.$$

Thus, the following holds:

$$\begin{array}{cc}\hfill & |A{\mathfrak{U}}_n-A{\mathfrak{U}}_0|\le {\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_0^1\mathfrak{g}\left(\tau \right)\mathfrak{a}\left(\tau \right)|{[{\mathfrak{U}}_n\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}-{[{\mathfrak{U}}_0\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}|d\tau \hfill \\ \hfill & +{\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_0^1\mathfrak{g}\left(\tau \right)\left|f(\tau ,{[{\mathfrak{U}}_n\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})-f(\tau ,{[{\mathfrak{U}}_0\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})\right|d\tau \hfill \\ \hfill <& {\displaystyle \frac{ϵ}{6}}+{\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_0^{\delta }\mathfrak{g}\left(\tau \right)\left[f(\tau ,{[{\mathfrak{U}}_n\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})+f(\tau ,{[{\mathfrak{U}}_0\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})\right]d\tau \hfill \\ \hfill & +{\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_{1-\delta }^1\mathfrak{g}\left(\tau \right)\left[f(\tau ,{[{\mathfrak{U}}_n\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})+f(\tau ,{[{\mathfrak{U}}_0\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})\right]d\tau \hfill \\ \hfill & +{\displaystyle \frac{ϵ}{6}}+{\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_0^{\delta }\mathfrak{g}\left(\tau \right)\mathfrak{a}\left(\tau \right)[{\mathfrak{U}}_n\left(\tau \right)+{\mathfrak{U}}_0\left(\tau \right)]d\tau \hfill \\ \hfill & +{\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_{1-\delta }^1\mathfrak{g}\left(\tau \right)\mathfrak{a}\left(\tau \right)[{\mathfrak{U}}_n\left(\tau \right)+{\mathfrak{U}}_0\left(\tau \right)]d\tau \hfill \\ \hfill <& {\displaystyle \frac{ϵ}{3}}+{\displaystyle \frac{8\lambda }{3\Gamma (\alpha -1)}}{\int }_0^{\delta }\mathfrak{g}\left(\tau \right)[\mathfrak{b}\left(\tau \right)h^{*}\left(M\right)+M·\mathfrak{a}\left(\tau \right)]d\tau \hfill \\ \hfill & +{\displaystyle \frac{8\lambda }{3\Gamma (\alpha -1)}}{\int }_{\delta }^1\mathfrak{g}\left(\tau \right)[\mathfrak{b}\left(\tau \right)h^{*}\left(M\right)+M·\mathfrak{a}\left(\tau \right)]d\tau \hfill \\ \hfill <& {\displaystyle \frac{ϵ}{3}}+{\displaystyle \frac{ϵ}{3}}+{\displaystyle \frac{ϵ}{3}}=ϵ,\hfill \end{array}$$

which implies that $\parallel A{\mathfrak{U}}_n-A{\mathfrak{U}}_0\parallel <ϵ$. Hence, we find that A is continuous.

Let $D\subset \mathcal{P}$ be a bounded set. According to (10) we understand that $A\left(D\right)$ is uniformly bounded. On the other hand, since $\mathcal{K}(t,\tau )$ is uniformly continuous on $[0,1]\times [0,1]$, we find that $A\left(D\right)$ is equicontinuous. According to the Arzela–Ascoli theorem, A is compact. □

Theorem 2.

If Hypotheses 1, 3, and 4 hold, then there is a ${\lambda }^{*}>0$, and when $\lambda \in (0,{\lambda }^{*})$, the FBVP (1) has at least one positive solution.

Proof. 

It is clear that, if $\mathfrak{U}\in E$ is a fixed point of A and $\mathfrak{U}\left(t\right)\ge {\omega }_{\lambda }\left(t\right)$ on $[0,1]$, then $\mathfrak{U}-{\omega }_{\lambda }$ is the solution of FBVP (1).

We denote the following:

$$r_1={\displaystyle \frac{{\int }_0^1\mathfrak{R}\left(\tau \right)d\tau }{{\int }_0^1\mathfrak{a}\left(\tau \right)d\tau }},$$

and

$${\lambda }^{*}=min\left\{{\lambda }_3,{\displaystyle \frac{3\Gamma (\alpha -1)r_1}{4{\int }_0^1\mathfrak{g}\left(\tau \right)[\mathfrak{b}\left(\tau \right)h^{*}\left(r_1\right)+r_1\mathfrak{a}\left(\tau \right)+\mathfrak{R}\left(\tau \right)]d\tau }}\right\}.$$

In the remainder of the proofing process, we assume the positive parameter $\lambda <{\lambda }^{*}$ always holds.

For $\forall \mathfrak{U}\in \mathcal{P}$ with $\parallel \mathfrak{U}\parallel \ge r_1$, we obtain the following:

$$\mathfrak{U}\left(t\right)\ge {\displaystyle \frac{\parallel \mathfrak{U}\parallel }{2(\alpha -1)}}\mathfrak{g}(1-t).$$

This inequality, together with (9), leads to the following:

$$\begin{array}{cc}\hfill \mathfrak{U}\left(t\right)-{\omega }_{\lambda }\left(t\right)& \ge \left({\displaystyle \frac{\parallel \mathfrak{U}\parallel }{2(\alpha -1)}}-{\displaystyle \frac{4\lambda {\int }_0^1\mathfrak{R}\left(\tau \right)d\tau }{3\Gamma (\alpha -1)}}\right)\mathfrak{g}(1-t)\hfill \\ \hfill & \ge \left({\displaystyle \frac{\parallel \mathfrak{U}\parallel }{2(\alpha -1)}}-{\displaystyle \frac{4{\lambda }_3{\int }_0^1\mathfrak{R}\left(\tau \right)d\tau }{3\Gamma (\alpha -1)}}\right)\mathfrak{g}(1-t).\hfill \end{array}$$

Notice that the following holds:

$$\begin{array}{cc}\hfill {\displaystyle \frac{4{\lambda }_3{\int }_0^1\mathfrak{R}\left(\tau \right)d\tau }{3\Gamma (\alpha -1)}}& ={\displaystyle \frac{\frac{4}{3}{\int }_0^1\mathfrak{R}\left(\tau \right)d\tau }{3(\alpha -1){\int }_0^1\mathfrak{a}\left(\tau \right)d\tau +{\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }}\hfill \\ & ={\displaystyle \frac{\frac{4r_1}{3}{\int }_0^1\mathfrak{a}\left(\tau \right)d\tau }{3(\alpha -1){\int }_0^1\mathfrak{a}\left(\tau \right)d\tau +{\int }_0^1\mathfrak{a}\left(\tau \right)\mathfrak{g}(1-\tau )d\tau }}\hfill \\ & <{\displaystyle \frac{4r_1}{9(\alpha -1)}}\hfill \\ & <{\displaystyle \frac{r_1}{2(\alpha -1)}}\hfill \end{array}$$

Thus, the following also holds:

$$\begin{array}{c}\hfill \mathfrak{U}\left(t\right)-{\omega }_{\lambda }\left(t\right)\ge {\displaystyle \frac{\parallel \mathfrak{U}\parallel -r_1}{2(\alpha -1)}}\mathfrak{g}(1-t)\ge 0.\end{array}$$

(11)

We assume that there are no fixed points of A on $\partial {\mathcal{P}}_{r_1}$ (otherwise, the proof would be complete). Next, we will show that the following are true:

$$A\mathfrak{U}\ne \mu \mathfrak{U},\forall \mu \ge 1,\mathfrak{U}\in \partial {\mathcal{P}}_{r_1}.$$

(12)

Alternatively, there exist ${\mathfrak{U}}_0\in \partial {\mathcal{P}}_{r_1}$ and ${\mu }_0>1$, such that the following is true:

$$A{\mathfrak{U}}_0={\mu }_0{\mathfrak{U}}_0.$$

(13)

Clearly, the following holds:

$$\begin{array}{cc}\hfill A{\mathfrak{U}}_0\left(t\right)& =\lambda {\int }_0^1\mathcal{K}(t,\tau )\left(f(\tau ,{[{\mathfrak{U}}_0\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})+\mathfrak{a}\left(\tau \right){[{\mathfrak{U}}_0\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}+\mathfrak{R}\left(\tau \right)\right)d\tau \hfill \\ \hfill & \le {\displaystyle \frac{4\lambda }{3\Gamma (\alpha -1)}}{\int }_0^1\mathfrak{g}\left(\tau \right)[\mathfrak{b}\left(\tau \right)h^{*}(\parallel {\mathfrak{U}}_0\parallel )+\mathfrak{a}\left(\tau \right)\parallel {\mathfrak{U}}_0\parallel +\mathfrak{R}\left(\tau \right)]d\tau \hfill \\ \hfill & \le {\displaystyle \frac{4{\lambda }^{*}}{3\Gamma (\alpha -1)}}{\int }_0^1\mathfrak{g}\left(\tau \right)[\mathfrak{b}\left(\tau \right)h^{*}(\parallel r_1\parallel )+\mathfrak{a}\left(\tau \right)\parallel r_1\parallel +\mathfrak{R}\left(\tau \right)]d\tau \hfill \\ \hfill & \le r_1=\parallel {\mathfrak{U}}_0\parallel .\hfill \end{array}$$

This contradicts with (13). Thus, (12) holds. According to Lemma 8, we obtain the following:

$$i(A,{\mathcal{P}}_{r_1},\mathcal{P})=1.$$

(14)

On the other hand, we denote the following:

$$G={\displaystyle \frac{6\Gamma \left(\alpha \right)(\alpha -1)}{\lambda m^2{\int }_c^d\mathfrak{g}\left(\tau \right)d\tau }},$$

in which the following holds:

$$m=\underset{t\in [c,d]}{min}\mathfrak{g}(1-t).$$

Hypothesis 4 implies there is an $X>0$, such that the following are true.

$$f(t,\mathfrak{x})\ge G\mathfrak{x},\forall \mathfrak{x}\ge X,c\le t\le d.$$

We denote the following:

$$r_2=max\left\{2r_1,{\displaystyle \frac{4(\alpha -1)X}{m}}\right\}.$$

We may suppose that there are no fixed points of A on $\partial P_{r_2}$. In the subsequent sections, we shall prove that the following are true:

$$\mathfrak{U}-A\mathfrak{U}\ne \mu \mathfrak{g}(1-t),\forall \mu \ge 0,\mathfrak{U}\in \partial {\mathcal{P}}_{r_2}.$$

(15)

Otherwise, $\exists {\mathfrak{U}}_1\in \partial {\mathcal{P}}_{r_2}$ and ${\mu }_1>0$, such that the following is true:

$${\mathfrak{U}}_1=A{\mathfrak{U}}_1+{\mu }_1\mathfrak{g}(1-t).$$

For any $\mathfrak{U}\in \partial {\mathcal{P}}_{r_2}$, (11) yields the following

$$\mathfrak{U}\left(t\right)-{\omega }_{\lambda }\left(t\right)\ge {\displaystyle \frac{r_2-r_1}{2(\alpha -1)}}\mathfrak{g}(1-t).$$

Then, the following holds:

$$\mathfrak{U}\left(t\right)-{\omega }_{\lambda }\left(t\right)\ge 0,t\in [0,1].$$

When $t\in [c,d]$, we also obtain the following:

$$\mathfrak{U}\left(t\right)-{\omega }_{\lambda }\left(t\right)\ge {\displaystyle \frac{m(r_2-r_1)}{2(\alpha -1)}}\ge {\displaystyle \frac{m{r}_2}{4(\alpha -1)}}.$$

By considering the denotation of $r_2$, we obtain the following:

$$\mathfrak{U}\left(t\right)-{\omega }_{\lambda }\left(t\right)\ge X,t\in [c,d].$$

Thus, the following is true:

$$\begin{array}{cc}\hfill A{\mathfrak{U}}_1\left(t\right)& =\lambda {\int }_0^1\mathcal{K}(t,\tau )\left(f(\tau ,{[{\mathfrak{U}}_1\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})+\mathfrak{a}\left(\tau \right){[{\mathfrak{U}}_1\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}+\mathfrak{R}\left(\tau \right)\right)d\tau \hfill \\ \hfill & \ge \lambda {\int }_c^d\mathcal{K}(t,\tau )\left(f(\tau ,{[{\mathfrak{U}}_1\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})+\mathfrak{a}\left(\tau \right){[{\mathfrak{U}}_1\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+}+\mathfrak{R}\left(\tau \right)\right)d\tau \hfill \\ \hfill & \ge \lambda {\int }_c^d\mathcal{K}(t,\tau )f(\tau ,{[{\mathfrak{U}}_1\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right)]}^{+})d\tau \hfill \\ \hfill & \ge \lambda G{\int }_c^d\mathcal{K}(t,\tau )({\mathfrak{U}}_1\left(\tau \right)-{\omega }_{\lambda }\left(\tau \right))d\tau \hfill \\ \hfill & \ge \lambda G{\int }_c^d{\displaystyle \frac{m{r}_2}{4(\alpha -1)}}\mathcal{K}(t,\tau )d\tau \hfill \\ \hfill & \ge {\displaystyle \frac{\lambda Gm{r}_2{\int }_c^d\mathfrak{g}\left(\tau \right)d\tau }{6(\alpha -1)\Gamma \left(\alpha \right)}}\mathfrak{g}(1-t).\hfill \end{array}$$

By considering the denotation of G, we obtain the following:

$${\mathfrak{U}}_1=A{\mathfrak{U}}_1+{\mu }_1\mathfrak{g}(1-t)\ge \left({\displaystyle \frac{r_2}{m}}+{\mu }_1\right)\mathfrak{g}(1-t).$$

Therefore, the following is true:

$$\parallel {\mathfrak{U}}_1\parallel \ge \left({\displaystyle \frac{r_2}{m}}+{\mu }_1\right)\underset{t\in [0,1]}{max}g(1-t)>r_2,$$

which contradicts with $\mathfrak{U}\in \partial {\mathcal{P}}_{r_2}$.

Hence, (15) holds, and we confirm, using Lemma 9, that the following is true:

$$i(A,{\mathcal{P}}_{r_2},\mathcal{P})=0.$$

(16)

From (14) and (16), through direct calculation, we obtain the following:

$$i(A,\mathcal{P}\cap (B_{r_2}\setminus \overline{B_{r_1}}),\mathcal{P})=-1.$$

Thus, A has a fixed point $\mathfrak{U}$ on $\mathcal{P}\cap (B_{r_2}\setminus \overline{B_{r_1}})$. It is clear that $\mathfrak{U}-{\omega }_{\lambda }$ is a positive solution to the semipositone FBVP (1). □

Remark 3.

Theorem 2 is valid if Hypothesis 1 is replaced by the following hypothesis:

(H): $f(t,\mathfrak{x})\in C\left(\right(0,1)\times [0,+\infty ),\mathbb{R})$, and there are non-negative functions $\mathfrak{R}\in L^1(0,1)\cap C(0,1)$ with ${\int }_0^1\mathfrak{R}\left(\tau \right)d\tau >0$, such that

$$\begin{array}{c}\hfill f(t,\mathfrak{x})\ge -\mathfrak{R}\left(t\right),(t,\mathfrak{x})\in (0,1)\times [0,+\infty ).\end{array}$$

Theorem 3.

We assume that Hypothesis 2, Hypothesis 3, and Hypothesis 4 hold, and that $f(t,0)\not\equiv 0$ in $(0,1)$. Then, there is a ${\lambda }^{\star }>0$, and, when $\lambda \in (0,{\lambda }^{\star })$, the FBVP (1) has at least two positive solutions.

Proof. 

Let the following be true:

$$\mathcal{A}\mathfrak{U}\left(t\right)=\lambda {\int }_0^1\mathcal{K}(t,\tau )\left(f\right(\tau ,\mathfrak{U}\left(\tau \right))+\mathfrak{a}(\tau \left)\mathfrak{U}\right(\tau \left)\right)d\tau .$$

Clearly, when $\lambda \in (0,{\lambda }_3)$, $\mathcal{A}:$$P\to P$ is a completely continuous operator.

We denote the following:

$${\lambda }^{\star }=min\left\{{\lambda }_3,{\displaystyle \frac{3\Gamma (\alpha -1)}{4{\int }_0^1\mathfrak{g}\left(s\right)[\mathfrak{b}\left(s\right)h^{*}\left(1\right)+\mathfrak{a}\left(s\right)]ds}}\right\}.$$

If $\mathcal{A}$ has no fixed point on $\overline{{\mathcal{P}}_1}$, we let $r_1=1$ and $\mathfrak{R}=\theta$; in a similar way to the proof for (14), we can obtain the following:

$$i(\mathcal{A},{\mathcal{P}}_1,\mathcal{P})=1.$$

It is obvious that the zero element is not the fixed point of $\mathcal{A}$, since $f(t,0)\not\equiv 0$ in $[0,1]$. Therefore, $\mathcal{A}$ has a positive fixed point ${\mathfrak{U}}_1\in \overline{{\mathcal{P}}_1}$.

We denote the following:

$$r_3=max\left\{2,{\displaystyle \frac{4(\alpha -1)X}{m}}\right\},$$

in which m and X are defined the same as those in the proof of Theorem 2. Next, we will prove that $\mathcal{A}$ has at least one positive fixed point on $\overline{{\mathcal{P}}_{r_3}}\setminus {\mathcal{P}}_1$, if there are no fixed points of $\mathcal{A}$ on $\partial {\mathcal{P}}_{r_3}$. From the proof of Theorem 2, we obtain the following:

$$i(\mathcal{A},{\mathcal{P}}_{r_3},\mathcal{P})=0.$$

Then, the following is true:

$$i(\mathcal{A},\mathcal{P}\cap (B_{r_3}\setminus \overline{B_1}),\mathcal{P})=-1.$$

Thus, $\mathcal{A}$ has a fixed point, ${\mathfrak{U}}_2$, on $\overline{{\mathcal{P}}_{r_3}}\setminus {\mathcal{P}}_1$. It is clear that ${\mathfrak{U}}_1$ and ${\mathfrak{U}}_2$ are positive solutions to FBVP (1). □

Corollary 1.

Assume that Hypothesis 2, Hypothesis 3, and Hypothesis 4 hold. The FBVP (1) has positive solutions if the positive parameter λ is small enough.

4. Examples

Example 1.

Consider the following semipositone FBVP:

$$\left\{\begin{array}{c}{D}_{0+}^{{\displaystyle \frac{5}{2}}}\mathfrak{U}\left(t\right)+\lambda f(t,\mathfrak{U}\left(t\right))=0,t\in (0,1),\hfill \\ \mathfrak{U}\left(0\right)=\mathfrak{U}\left(1\right)={\mathfrak{U}}^{\prime }\left(0\right)=0,\hfill \end{array}\right.$$

(17)

where the following is true:

$$f(t,\mathfrak{x})={\left[t(1-t)\right]}^{-{\displaystyle \frac{3}{2}}}ln(1+t){\mathfrak{x}}^{{\displaystyle \frac{3}{2}}}-t^{-{\displaystyle \frac{1}{3}}}{(1-t)}^{-{\displaystyle \frac{2}{3}}}{\mathfrak{x}}^{{\displaystyle \frac{1}{2}}}.$$

Let $\mathfrak{a}\left(\tau \right)=\mathfrak{R}\left(\tau \right)={\tau }^{-{\displaystyle \frac{1}{3}}}{(1-\tau )}^{-{\displaystyle \frac{2}{3}}}$, $\mathfrak{b}\left(\tau \right)={\left[\tau (1-\tau )\right]}^{-{\displaystyle \frac{3}{2}}}ln(1+\tau )$, $h\left(\mathfrak{x}\right)={\mathfrak{x}}^{{\displaystyle \frac{3}{2}}}$. Then, the following is true:

$$f(t,\mathfrak{x})\ge -\mathfrak{a}\left(t\right){\mathfrak{x}}^{{\displaystyle \frac{1}{2}}}\ge -\mathfrak{a}\left(t\right)(1+\mathfrak{x})=-\mathfrak{a}\left(t\right)\mathfrak{x}-\mathfrak{R}\left(t\right),$$

and the following is also true:

$$f(t,\mathfrak{x})\le {\left[t(1-t)\right]}^{-{\displaystyle \frac{3}{2}}}ln(1+t){\mathfrak{x}}^{{\displaystyle \frac{3}{2}}}=\mathfrak{b}\left(t\right)h\left(\mathfrak{x}\right).$$

Through calculations, the following is true:

$${\int }_0^1\mathfrak{a}\left(\tau \right)d\tau ={\int }_0^1\mathfrak{R}\left(\tau \right)d\tau ={\int }_0^1{\tau }^{-{\displaystyle \frac{1}{3}}}{(1-\tau )}^{-{\displaystyle \frac{2}{3}}}d\tau ={\displaystyle \frac{2\sqrt{3}\pi }{3}},$$

$${\int }_0^1\mathfrak{b}\left(\tau \right)\mathfrak{g}\left(\tau \right)d\tau ={\int }_0^1{\tau }^{-{\displaystyle \frac{1}{2}}}ln(1+\tau )d\tau =2ln2+\pi -4.$$

Clearly, Hypotheses 1 and 3 hold. Moreover, the following implies that Hypothesis 4 holds:

$$\begin{array}{c}\hfill \underset{\mathfrak{x}\to +\infty }{lim\; inf}\underset{t\in [{\displaystyle \frac{1}{4}},{\displaystyle \frac{3}{4}}]}{min}{\displaystyle \frac{f(t,\mathfrak{x})}{\mathfrak{x}}}=+\infty \end{array}$$

Therefore, Theorem 2 guarantees that the FBVP (17) has positive solutions, if the positive parameter λ is small enough.

Example 2.

Consider the FBVP, as follows:

$$\left\{\begin{array}{c}{D}_{0+}^{{\displaystyle \frac{7}{3}}}\mathfrak{U}\left(t\right)+\lambda f(t,\mathfrak{U}\left(t\right))=0,t\in (0,1),\hfill \\ \mathfrak{U}\left(0\right)=\mathfrak{U}\left(1\right)={\mathfrak{U}}^{\prime }\left(0\right)=0,\hfill \end{array}\right.$$

(18)

where the following is true:

$$f(t,\mathfrak{x})=t^{-{\displaystyle \frac{1}{3}}}{(1-t)}^{-2}+{(1-t)}^{-{\displaystyle \frac{4}{3}}}(sint){\mathfrak{x}}^2+lntln(1+\mathfrak{x}).$$

Let $\mathfrak{a}\left(\tau \right)=-ln\tau$, $\mathfrak{b}\left(\tau \right)={\tau }^{-{\displaystyle \frac{1}{3}}}{(1-\tau )}^{-2}+{(1-\tau )}^{-{\displaystyle \frac{4}{3}}}sin\tau$, $h\left(\mathfrak{x}\right)={\mathfrak{x}}^2+1$. Then, the following holds:

$${\int }_0^1\mathfrak{a}\left(\tau \right)d\tau ={\int }_0^1-ln\tau d\tau =1,$$

$${\int }_0^1\mathfrak{b}\left(\tau \right)\mathfrak{g}\left(\tau \right)d\tau ={\int }_0^1{\tau }^{{\displaystyle \frac{2}{3}}}{(1-\tau )}^{-{\displaystyle \frac{2}{3}}}d\tau +{\int }_0^1\tau sin\tau d\tau =sin1-cos1+{\displaystyle \frac{4\sqrt{3}\pi }{9}}.$$

Clearly, the following is true:

$$\begin{array}{c}\hfill f(t,\mathfrak{x})\le \mathfrak{b}\left(t\right)h\left(\mathfrak{x}\right),(t,\mathfrak{x})\in (0,1)\times [0,+\infty ).\end{array}$$

By considering the following:

$$0\le ln(1+\mathfrak{x})\le \mathfrak{x},\mathfrak{x}\ge 0,$$

we obtain the following:

$$\begin{array}{c}\hfill f(t,\mathfrak{x})\ge -\mathfrak{a}\left(t\right)\mathfrak{x},(t,\mathfrak{x})\in (0,1)\times [0,+\infty ).\end{array}$$

Thus, Hypotheses 2 and 3 hold, according to the above discussion.

Moreover, the following also holds:

$$\begin{array}{c}\hfill \underset{\mathfrak{x}\to +\infty }{lim\; inf}\underset{t\in [{\displaystyle \frac{1}{4}},{\displaystyle \frac{3}{4}}]}{min}{\displaystyle \frac{f(t,\mathfrak{x})}{\mathfrak{x}}}=+\infty ,\end{array}$$

This means that Hypothesis 4 also holds. Moreover, $f(t,0)=t^{-{\displaystyle \frac{1}{3}}}{(1-t)}^{-2}\ne 0$ in $(0,1)$. Theorem 3 guarantees that the FBVP (18) has no fewer than two positive solutions if the positive parameter λ is small enough.

5. Conclusions

This study investigates positive solutions to semipositone FBVPs. An essential assumption of nonlinearity is $f(t,\mathfrak{x})\ge -\mathfrak{a}\left(t\right)\mathfrak{x}-\mathfrak{R}\left(t\right)$, which contains $f(t,\mathfrak{x})\ge -\mathfrak{R}\left(t\right)$ as a special case. As far as we know, semipositone problems under such a condition have not yet been studied due to the difficulties caused by an extra term $\mathfrak{a}\left(t\right)\mathfrak{x}$. We derive some new properties of Green’s function for the auxiliary problems, and obtain both their existence and their multiplicity, by means of fixed point index theory. It is noted that the methods used in this article can be utilized for some other differential equations, provided the kernel demonstrates suitable behavior.