Existence and Nonexistence of Positive Solutions for Fractional Boundary Value Problems with Lidstone-Inspired Fractional Conditions

Abstract

This paper investigates the existence and nonexistence of positive solutions for a class of nonlinear Riemann–Liouville fractional boundary value problems of order $\alpha +2n$, where $\alpha \in (m-1,m]$ with $m\ge 3$ and $m,n\in \mathbb{N}$. The conjugate fractional boundary conditions are inspired by Lidstone conditions. The nonlinearity depends on a positive parameter on which we identify constraints that determine the existence or nonexistence of positive solutions. Our method involves constructing Green’s function by convolving the Green functions of a lower-order fractional boundary value problem and a conjugate boundary value problem and using properties of this Green function to apply the Guo–Krasnosel’skii fixed-point theorem. Illustrative examples are provided to demonstrate existence and nonexistence intervals.

1. Introduction

Let $m,n\in \mathbb{N}$, $m\ge 3$, with $\alpha \in (m-1,m]$ and $\beta \in [1,m-1]$. Consider a Riemann–Liouville fractional boundary value problem

$$D_{0^{+}}^{\alpha +2n}u\left(t\right)+{(-1)}^n\lambda g\left(t\right)f\left(u\right)=0,0<t<1,$$

(1)

subject to the Lidstone-inspired boundary conditions

$$\begin{array}{c}\hfill u^{\left(i\right)}\left(0\right)=0,i=0,1,\dots ,m-2,D_{0^{+}}^{\beta }u\left(1\right)=0,\\ \hfill D_{0^{+}}^{\alpha +2l}u\left(0\right)=D_{0^{+}}^{\alpha +2l}u\left(1\right)=0,l=0,1,\dots ,n-1.\end{array}$$

(2)

Here, $f:[0,\infty )\to [0,\infty )$ and $g:[0,1]\to [0,\infty )$ are continuous functions with $g\left(t\right)$ satisfying the condition ${\int }_0^{1}g\left(t\right)dt>0$, and $\lambda >0$ is a positive parameter. This paper is concerned with the existence and nonexistence of positive solutions to (1) and (2).

To address this, we follow the procedure of Eloe et al. in [1] of constructing the associated Green’s function for the given problem by convolving a lower-order Green’s function, $G_0(t,s)$, for the equation of a conjugate boundary value problem. We present properties of Green’s function, many of which can be found in [2,3]. Then, we deploy those in an application of the Guo–Krasnosel’skii fixed-point theorem.

Our method involves the analysis of the operator defined by

$$Tu\left(t\right)={(-1)}^n\lambda {\int }_0^{1}G(t,s)g\left(s\right)f\left(u\left(s\right)\right)ds,$$

which will be shown to have a fixed point under suitable conditions on parameter $\lambda$. This fixed point is a positive solution to (1) and (2). We will also give suitable conditions on $\lambda$ for the nonexistence of solutions to (1) and (2).

This study extends the existing literature on fractional boundary value problems that leverage Guo–Krasnosel’skii’s fixed-point theorem. The major impetus of this work is two papers [2,3]. In the former, Lyons and Neugebauer used the convolution of two Green functions to prove existence and nonexistence results for two-point fractional boundary value problems. Recently, the latter work by Neugebauer and Wingo investigated a way to move to an even higher-order two-point boundary value problem using convolution and induction. The novelty here is that repeated convolution with induction leads one to creating arbitrary $\alpha +2n$-order two-point boundary value problems.

Previous articles have applied various fixed-point theorems to demonstrate the existence of positive solutions for similar problems. For results obtained by employing the Guo–Krasnosel’skii fixed-point theorem similar to that realized in this paper, we cite [4,5,6,7]. One may find singular nonlinearity results in [8,9]. A quite recent application for the Guo–Krasnosel’skii fixed-point theorem to fractional boundary value problems was studied by Raghavendran et al. in [10]. Other recent applications of fixed-point theory to fractional boundary value problems were carried out by Zhang et al. in [11,12].

Here, we use the Guo–Krasnosel’skii fixed-point theorem to guarantee the existence of positive solutions by establishing two separate sizing conditions on parameter $\lambda$ based upon the liminfs and limsups of the nonlinearity. Additionally, we provide nonexistence results determined with the parameter. This approach is based on the properties of Green’s function, which plays a critical role in showing the existence of positive solutions.

Section 2 provides definitions for the Riemann–Liousville fractional derivative and suggestions for further study therein and states the Guo–Krasnosel’skii fixed-point theorem. The subsequent sections are devoted to the construction of Green’s function and its properties. Then, in Section 5 and Section 6, we establish intervals for $\lambda$ that yield the existence or nonexistence of positive solutions. Finally, we present two examples to illustrate the application of our results.

2. Preliminaries and the Fixed-Point Theorem

We begin by defining the Riemann–Liouville fractional integral, which is used to define the Riemann–Liouville fractional derivative used in this work. Both are widely adopted and commonly used. Then, we present Guo–Krasnosel’skii’s fixed-point theorem [13,14].

Definition 1. 

Let $\nu >0$. The Riemann–Liouville fractional integral of a function u of order ν, denoted $I_{0^{+}}^{\nu }u$, is defined as

$$I_{0^{+}}^{\nu }u\left(t\right)={\displaystyle \frac{1}{\Gamma \left(\nu \right)}}{\int }_0^t{(t-s)}^{\nu -1}u\left(s\right)ds,$$

provided the right-hand side exists.

Definition 2. 

Let n denote a positive integer and assume $n-1<\alpha \le n$. The Riemann–Liouville fractional derivative of order α of the function $u:[0,1]\to \mathbb{R}$, denoted $D_{0^{+}}^{\alpha }u$, is defined as

$$D_{0^{+}}^{\alpha }u\left(t\right)={\displaystyle \frac{1}{\Gamma (n-\alpha )}}{\displaystyle \frac{d^n}{d{t}^n}}{\int }_0^t{(t-s)}^{n-\alpha -1}u\left(s\right)ds=D^n{I}_{0^{+}}^{n-\alpha }u\left(t\right),$$

provided the right-hand side exists.

We refer to [15,16,17,18] for further study of fractional calculus and fractional differential equations.

Theorem 1 

(Guo–Krasnosel’skii fixed-point theorem). Let $\mathcal{B}$ be a Banach space, and let $\mathcal{P}\subset X$ be a cone in $\mathcal{P}$. Assume that ${\Omega }_1$ and ${\Omega }_2$ are open sets with $0\in {\Omega }_1,$ and ${\overline{\Omega }}_1\subset {\Omega }_2.$ Let $T:\mathcal{P}\cap ({\overline{\Omega }}_2\setminus {\Omega }_1)\to \mathcal{P}$ be a completely continuous operator such that either of the following holds:

1. 

$\parallel Tu\parallel \ge \parallel u\parallel ,u\in \mathcal{P}\cap \partial {\Omega }_1,and\parallel Tu\parallel \le \parallel u\parallel ,u\in \mathcal{P}\cap \partial {\Omega }_2;$

2. 

$\parallel Tu\parallel \le \parallel u\parallel ,u\in \mathcal{P}\cap \partial {\Omega }_1,and\parallel Tu\parallel \ge \parallel u\parallel ,u\in \mathcal{P}\cap \partial {\Omega }_2.$

Then, T has a fixed point in $\mathcal{P}\cap ({\overline{\Omega }}_2\setminus {\Omega }_1).$

3. Green’s Function

Now, we construct Green’s function, used for (1) and (2) utilizing induction with a convolution of a lower-order problem and a conjugate problem. The procedure is similar to that found in [3].

First, the conjugate boundary value problem

$$-u^{″}=0,0<t<1,u\left(0\right)=0,u\left(1\right)=0,$$

has a well-known Green’s function

$$G_{conj}(t,s)=\left\{\begin{array}{cc}s(1-t),\hfill & \hfill 0\le s<t\le 1,\\ t(1-s),\hfill & \hfill 0\le t<s\le 1.\end{array}\right.$$

Let $G_0(t,s)$ be Green’s function for

$$-D_{0^{+}}^{\alpha }au=0,0<t<1,u^{\left(i\right)}\left(0\right)=0,i=0,1,\dots ,m-2,D_{0^{+}}^{\beta }u\left(1\right)=0,$$

which is given by [19]

$$G_0(t,s)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left\{\begin{array}{cc}{t}^{\alpha -1}{(1-s)}^{\alpha -1-\beta }-{(t-s)}^{\alpha -1},\hfill & \hfill 0\le s<t\le 1,\\ \\ t^{\alpha -1}{(1-s)}^{\alpha -1-\beta },\hfill & \hfill 0\le t\le s<1.\end{array}\right.$$

For $k=1,\dots ,n-1$, recursively define $G_k(t,s)$ by

$$G_k(t,s)=-{\int }_0^1{G}_{k-1}(t,r)G_{conj}(r,s)dr.$$

Then,

$$G_n(t,s)=-{\int }_0^1{G}_{n-1}(t,r)G_{conj}(r,s)dr,$$

is Green’s function for

$$-D_{0^{+}}^{\alpha +2n}u\left(t\right)=0,0<t<1,$$

with boundary condition (2), and $G_{n-1}(t,s)$ is Green’s function for

$$-D_{0^{+}}^{\alpha +2(n-1)}u\left(t\right)=0,0<t<1,$$

with boundary conditions

$$u^{\left(i\right)}\left(0\right)=0,i=0,1,\dots ,m-2,D_{0^{+}}^{\beta }u\left(1\right)=0,$$

$$D_{0^{+}}^{\alpha +2l}u\left(0\right)=D_{0^{+}}^{\alpha +2l}u\left(1\right)=0,l=0,1,\dots ,n-2.$$

To see this, for the base case $k=1$, consider the linear differential equation

$$D_{0^{+}}^{\alpha +2}u\left(t\right)+h\left(t\right)=0,0<t<1,$$

satisfying the boundary conditions

$$u^{\left(i\right)}\left(0\right)=0,i=0,1,\dots ,m-2,D_{0^{+}}^{\beta }u\left(1\right)=0,$$

$$D_{0^{+}}^{\alpha }u\left(0\right)=0,D_{0^{+}}^{\alpha }u\left(1\right)=0.$$

Make the change of variable:

$$v\left(t\right)=D_{0^{+}}^{\alpha +2-2}u\left(t\right).$$

Then,

$$D^{2}v\left(t\right)=D^2{D}_{0^{+}}^{\alpha +2-2}u\left(t\right)=D_{0^{+}}^{\alpha +2}u\left(t\right)=-h\left(t\right),$$

and since $v\left(t\right)=D_{0^{+}}^{\alpha }u\left(t\right)$,

$$v\left(0\right)=D_{0^{+}}^{\alpha }u\left(0\right)=0\mathrm{and}v\left(1\right)=D_{0^{+}}^{\alpha }u\left(1\right)=0.$$

Thus, v satisfies the Dirichlet boundary value problem:

$$v^{″}+h\left(t\right)=0,0<t<1,$$

$$v\left(0\right)=0,v\left(1\right)=0.$$

Also, u now satisfies a lower-order boundary value problem:

$$D_{0^{+}}^{\alpha }u\left(t\right)=v\left(t\right),0<t<1,$$

$$u^{\left(i\right)}\left(0\right)=0,i=0,1,\dots ,m-2,D_{0^{+}}^{\beta }u\left(1\right)=0.$$

So,

$$\begin{array}{cc}\hfill u\left(t\right)& ={\int }_0^1{G}_0(t,s)(-v\left(s\right))ds\hfill \\ \hfill & ={\int }_0^1{G}_0(t,s)\left(-{\int }_0^1{G}_{conj}(s,r)h\left(r\right)ds\right)dr\hfill \\ \hfill & ={\int }_0^1\left({\int }_0^1-G_0(t,s)G_{conj}(s,r)ds\right)h\left(r\right)dr.\hfill \end{array}$$

Therefore,

$$u\left(t\right)={\int }_0^1{G}_1(t,s)h\left(s\right)ds,$$

where

$$G_1(t,s)=-{\int }_0^1{G}_0(t,r)G_{conj}(r,s)dr.$$

For the inductive step, the argument is similar. Assume that $k=n-1$ is true, and consider the linear differential equation

$$D_{0^{+}}^{\alpha +2n}u\left(t\right)+k\left(t\right)=0,0<t<1,$$

satisfying boundary condition (2).

Make the change of variables

$$v\left(t\right)=D_{0^{+}}^{\alpha +2(n-1)}u\left(t\right)$$

so that

$$D^{2}v\left(t\right)=D_{0^{+}}^{\alpha +2n}=-k\left(t\right)$$

and

$$v\left(0\right)=D_{0^{+}}^{\alpha +2(n-1)}u\left(0\right)=0\mathrm{and}v\left(1\right)=D_{0^{+}}^{\alpha +2(n-1)}v\left(1\right)=0.$$

Similarly to before, $v\left(t\right)$ satisfies the Dirichlet boundary value problem

$$v^{″}+k\left(t\right)=0,0<t<1,$$

$$v\left(0\right)=0,v\left(1\right)=0,$$

while $u\left(t\right)$ satisfies the lower-order problem

$$D_{0^{+}}^{\alpha +2(n-1)}u\left(t\right)=v\left(t\right),0<t<1,$$

$$u\left(0\right)=0,D_{0^{+}}^{\beta }u\left(1\right)=0,$$

$$D_{0^{+}}^{\alpha +2l}u\left(0\right)=D_{0^{+}}^{\alpha +2l}u\left(1\right)=0,l=0,1,\dots ,n-2.$$

By induction,

$$\begin{array}{cc}\hfill u\left(t\right)& ={\int }_0^1{G}_{n-1}(t,s)(-v\left(s\right))ds\hfill \\ \hfill & ={\int }_0^1\left(-{\int }_0^1{G}_{n-1}(t,s)G_{conj}(s,r)ds\right)k\left(r\right)dr\hfill \\ \hfill & ={\int }_0^1{G}_n(t,s)k\left(s\right)ds.\hfill \end{array}$$

Therefore,

$$u\left(t\right)={\int }_0^1{G}_n(t,s)k\left(s\right)ds,$$

where

$$G_n(t,s)=-{\int }_0^1{G}_{n-1}(t,r)G_{conj}(r,s)dr.$$

So, the unique solution to

$$D_{0^{+}}^{\alpha +2n}u\left(t\right)+k\left(t\right)=0,0<t<1,$$

satisfying boundary condition (2) is given by

$$u\left(t\right)={\int }_0^1{G}_n(t,s)k\left(s\right)ds.$$

4. Green’s Function Properties

We now discuss properties for $G_n(t,s)$ that are inherited from $G_0(t,s)$ and $G_{conj}(t,s)$. The results of the first lemma regarding $G_{conj}(t,s)$ are well known and easily verifiable.

Lemma 1. 

For $(t,s)\in [0,1]\times [0,1]$, $G_{conj}(t,s)\in C^{\left(1\right)}$ and $G_{conj}(t,s)\ge 0$.

The following lemma regarding $G_0(t,s)$ is Lemma 3.1 proved in [2].

Lemma 2. 

The following are true:

(1) 

For $(t,s)\in [0,1]\times [0,1)$, $G_0(t,s)\in C^{\left(1\right)}$.

(2) 

For $(t,s)\in (0,1)\times (0,1)$, $G_0(t,s)>0$ and ${\displaystyle \frac{\partial }{\partial t}}{G}_0(t,s)>0$.

(3) 

For $(t,s)\in [0,1]\times [0,1)$, $t^{\alpha -1}{G}_0(1,s)\le G_0(t,s)\le G_0(1,s)$.

Parts (1) and (2) of the following lemma regarding the convoluted Green’s function $G_n(t,s)$ are proved in Lemma 5.1 [3], and part (3) is proven here inductively.

Lemma 3. 

The following are true:

(1) 

For $(t,s)\in [0,1]\times [0,1)$, $G_n(t,s)\in C^{\left(1\right)}$.

(2) 

For $(t,s)\in (0,1)\times (0,1)$, ${(-1)}^n{G}_n(t,s)>0$ and ${(-1)}^n{\displaystyle \frac{\partial }{\partial t}}{G}_n(t,s)>0.$

(3) 

For $(t,s)\in [0,1]\times [0,1)$,

$${(-1)}^n{t}^{\alpha -1}{G}_n(1,s)\le {(-1)}^n{G}_n(t,s)\le {(-1)}^n{G}_n(1,s).$$

Proof. 

For part (3), we proceed inductively.

For the base case $k=1$, we use Lemma 2 (3) to find

$$\begin{array}{cc}\hfill {(-1)}^1{t}^{\alpha -1}{G}_1(1,s)& =-t^{\alpha -1}\left(-{\int }_0^1{G}_0(1,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & =-\left({\int }_0^1-t^{\alpha -1}{G}_0(1,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & \le -\left({\int }_0^1-G_0(t,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & =-\left(-{\int }_0^1{G}_0(t,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & ={(-1)}^1{G}_1(t,s),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {(-1)}^1{G}_1(t,s)& =-\left(-{\int }_0^1{G}_0(t,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & ={\int }_0^1{G}_0(t,r)G_{conj}(r,s)dr\hfill \\ \hfill & \le {\int }_0^1{G}_0(1,r)G_{conj}(r,s)dr\hfill \\ \hfill & =-\left(-{\int }_0^1{G}_0(1,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & ={(-1)}^1{G}_1(1,s).\hfill \end{array}$$

Now, assume that $k=n-1$ is true. Then,

$$\begin{array}{cc}\hfill {(-1)}^n{t}^{\alpha -1}{G}_n(1,s)& ={(-1)}^n{t}^{\alpha -1}\left(-{\int }_0^1{G}_{n-1}(1,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & ={(-1)}^2\left({\int }_0^1{(-1)}^{n-1}{t}^{\alpha -1}{G}_{n-1}(1,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & \le {(-1)}^2\left({\int }_0^1{(-1)}^{n-1}{G}_{n-1}(t,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & ={(-1)}^n\left(-{\int }_0^1{G}_{n-1}(t,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & ={(-1)}^n{G}_n(t,s),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {(-1)}^n{G}_n(t,s)& ={(-1)}^n\left(-{\int }_0^1{G}_{n-1}(t,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & ={(-1)}^2\left({\int }_0^1{(-1)}^{n-1}{G}_{n-1}(t,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & \le {(-1)}^2\left({\int }_0^1{(-1)}^{n-1}{G}_{n-1}(1,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & ={(-1)}^n\left(-{\int }_0^1{G}_{n-1}(1,r)G_{conj}(r,s)dr\right)\hfill \\ \hfill & ={(-1)}^n{G}_n(1,s).\hfill \end{array}$$

□

5. Existence of Solutions

We are now in a position to demonstrate the existence of positive solutions to (1) and (2) based upon the parameter $\lambda$ using the Guo–Krasnosel’skii fixed-point theorem and our constructed Green’s function and its properties.

Define the constants

$$A_{G_n}={\int }_0^1{(-1)}^n{s}^{\alpha -1}{G}_n(1,s)g\left(s\right)ds,B_{G_n}={\int }_0^1{(-1)}^n{G}_n(1,s)g\left(s\right)ds,$$

$$F_0=\underset{u\to 0^{+}}{lim\; sup}{\displaystyle \frac{f\left(u\right)}{u}},f_0=\underset{u\to 0^{+}}{lim\; inf}{\displaystyle \frac{f\left(u\right)}{u}},$$

$$F_{\infty }=\underset{u\to \infty }{lim\; sup}{\displaystyle \frac{f\left(u\right)}{u}},f_{\infty }=\underset{u\to \infty }{lim\; inf}{\displaystyle \frac{f\left(u\right)}{u}}.$$

Let $\mathcal{B}=C[0,1]$ be a Banach space with norm

$$\parallel u\parallel =\underset{t\in [0,1]}{max}\left|u\left(t\right)\right|.$$

Define the cone:

$$\begin{array}{cc}\hfill \mathcal{P}& =\left\{u\in \mathcal{B}:u\left(0\right)=0,u\left(t\right)\mathrm{is} \mathrm{n}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}, \mathrm{and}\right.\hfill \\ & \phantom{=}\left.t^{\alpha -1}u\left(1\right)\le u\left(t\right)\le u\left(1\right)\mathrm{on}[0,1]\right\}.\hfill \end{array}$$

Define the operator $T:\mathcal{P}\to \mathcal{B}$ by

$$Tu\left(t\right)={(-1)}^n\lambda {\int }_0^1{G}_n(t,s)g\left(s\right)f\left(u\left(s\right)\right)ds.$$

Lemma 4. 

Operator $T:\mathcal{P}\to \mathcal{P}$ is completely continuous.

Proof. 

Let $u\in \mathcal{P}$. Then, by definition,

$$Tu\left(0\right)={(-1)}^n\lambda {\int }_0^1{G}_n(0,s)g\left(s\right)f\left(u\left(s\right)\right)ds=0.$$

Also, for $t\in (0,1)$ and by Lemma 3 (2),

$${\displaystyle \frac{\partial }{\partial t}}\left[Tu\left(t\right)\right]={(-1)}^n\lambda {\int }_0^1{\displaystyle \frac{\partial }{\partial t}}{G}_n(t,s)g\left(s\right)f\left(u\left(s\right)\right)ds>0$$

which implies that $Tu\left(t\right)$ is nondecreasing.

Next, for $t\in [0,1]$ and by Lemma 3 (3),

$$\begin{array}{cc}\hfill t^{\alpha -1}Tu\left(1\right)& =t^{\alpha -1}{(-1)}^n\lambda {\int }_0^1{G}_n(1,s)g\left(s\right)f\left(u\left(s\right)\right)ds\hfill \\ \hfill & \le {(-1)}^n\lambda {\int }_0^1{G}_n(t,s)g\left(s\right)f\left(u\left(s\right)\right)ds\hfill \\ \hfill & =Tu\left(t\right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill Tu\left(t\right)& ={(-1)}^n\lambda {\int }_0^1{G}_n(t,s)g\left(s\right)f\left(u\left(s\right)\right)ds\hfill \\ \hfill & \le {(-1)}^n\lambda {\int }_0^1{G}_n(1,s)g\left(s\right)f\left(u\left(s\right)\right)ds\hfill \\ \hfill & =Tu\left(1\right).\hfill \end{array}$$

Therefore, $Tu\in \mathcal{P}$. A standard application of the Arzela–Ascoli Theorem yields the result that T is completely continuous. □

Theorem 2. 

If

$${\displaystyle \frac{1}{A_{G_n}{f}_{\infty }}}<\lambda <{\displaystyle \frac{1}{B_{G_n}{F}_0}},$$

then (1) and (2) have at least one positive solution.

Proof. 

Since $F_0\lambda B_{G_n}<1$, there exists an $ϵ>0$ such that

$$(F_0+ϵ)\lambda B_{G_n}\le 1.$$

Also, since

$$F_0=\underset{u\to 0^{+}}{lim\; sup}{\displaystyle \frac{f\left(u\right)}{u}},$$

there exists an $H_1>0$ such that

$$f\left(u\right)\le (F_0+ϵ)u\mathrm{for}u\in (0,H_1].$$

Define ${\Omega }_1=\{u\in \mathcal{B}:\parallel u\parallel <H_1\}$. If $u\in \mathcal{P}\cap \partial {\Omega }_1$, then $\parallel u\parallel =H_1$, and

$$\begin{array}{cc}\hfill \left|\right(Tu\left)\right(1\left)\right|& ={(-1)}^n\lambda {\int }_0^1{G}_n(1,s)g\left(s\right)f\left(u\left(s\right)\right)ds\hfill \\ \hfill & \le {(-1)}^n\lambda {\int }_0^1{G}_n(1,s)g\left(s\right)(F_0+ϵ)u\left(s\right)ds\hfill \\ \hfill & \le (F_0+ϵ)u\left(1\right)\lambda {\int }_0^1{(-1)}^n{G}_n(1,s)g\left(s\right)ds\hfill \\ \hfill & \le (F_0+ϵ)\parallel u\parallel \lambda B_{G_n}\hfill \\ \hfill & \le \parallel u\parallel .\hfill \end{array}$$

Since $Tu\in \mathcal{P}$, $\parallel Tu\parallel \le \parallel u\parallel$ for $u\in \mathcal{P}\cap \partial {\Omega }_1$.

Next, since $f_{\infty }\lambda >{\displaystyle \frac{1}{A_{G_n}}}$, there exists a $c\in (0,1)$ and an $ϵ>0$ such that

$$(f_{\infty }-ϵ)\lambda >{\left({(-1)}^n{\int }_c^1{G}_n(1,s)g\left(s\right)ds\right)}^{-1}.$$

Since

$$f_{\infty }=\underset{u\to \infty }{lim\; inf}{\displaystyle \frac{f\left(u\right)}{u}},$$

there exists an $H_3>0$ such that

$$f\left(u\right)\ge (f_{\infty }-ϵ)u\mathrm{for}u\in [H_3,\infty ).$$

Define

$$H_2=max\left\{{\displaystyle \frac{H_3}{c^{\alpha -1}}},2H_1\right\},$$

and define ${\Omega }_2=\{u\in \mathcal{B}:\parallel u\parallel <H_2\}$.

Let $u\in \mathcal{P}\cap \partial {\Omega }_2$. Then, $\parallel u\parallel =H_2$. Notice that for $t\in [c,1]$,

$$u\left(t\right)\ge t^{\alpha -1}u\left(1\right)\ge c^{\alpha -1}{H}_2\ge c^{\alpha -1}{\displaystyle \frac{H_3}{c^{\alpha -1}}}=H_3.$$

Therefore,

$$\begin{array}{cc}\hfill \left|\right(Tu\left)\right(1\left)\right|& \ge {(-1)}^n\lambda {\int }_c^1{G}_n(1,s)g\left(s\right)f\left(u\left(s\right)\right)ds\hfill \\ \hfill & \ge {(-1)}^n\lambda {\int }_c^1{G}_n(1,s)g\left(s\right)(f_{\infty }-ϵ)u\left(s\right)ds\hfill \\ \hfill & \ge \lambda (f_{\infty }-ϵ)u\left(1\right){\int }_c^1{(-1)}^n{s}^{\alpha -1}{G}_n(1,s)g\left(s\right)ds\hfill \\ \hfill & \ge \parallel u\parallel .\hfill \end{array}$$

Hence, $\parallel Tu\parallel \ge \parallel u\parallel$ for $u\in \mathcal{P}\cap {\partial }_2$. Notice that since $H_1<H_2$, we have ₁ ⊂ ₂. Thus, by Theorem 1 (1), T has a fixed point $u\in \mathcal{P}$. By the definition of T, this fixed point is a positive solution of (1) and (2). □

Theorem 3. 

If

$${\displaystyle \frac{1}{A_{G_n}{f}_0}}<\lambda <{\displaystyle \frac{1}{B_{G_n}{F}_{\infty }}},$$

then (1) and (2) have at least one positive solution.

Proof. 

Since $f_0\lambda A_{G_n}>1$, there exists an $ϵ>0$ such that

$$(f_0-ϵ)\lambda A_{G_n}\ge 1.$$

Then, since

$$f_0=\underset{u\to 0^{+}}{lim\; inf}{\displaystyle \frac{f\left(u\right)}{u}},$$

there exists an $H_1>0$ such that

$$f\left(u\right)\ge (f_0-ϵ)u,t\in (0,H_1].$$

Define ${\Omega }_1=\{u\in \mathcal{B}:\parallel u\parallel <H_1\}$. If $u\in \mathcal{P}\cap \partial {\Omega }_1$, then $u\left(t\right)\le H_1\mathrm{for}t\in [0,1]$. So,

$$\begin{array}{cc}\hfill \left|\right(Tu\left)\right(1\left)\right|& ={(-1)}^n\lambda {\int }_0^1{G}_n(1,s)g\left(s\right)f\left(u\left(s\right)\right)ds\hfill \\ \hfill & \ge {(-1)}^n\lambda {\int }_0^1{G}_n(1,s)g\left(s\right)(f_0-ϵ)u\left(s\right)ds\hfill \\ \hfill & \ge \lambda (f_0-ϵ)u\left(1\right){\int }_0^1{(-1)}^n{s}^{\alpha -1}{G}_n(1,s)g\left(s\right)ds\hfill \\ \hfill & \ge \lambda (f_0-ϵ)\parallel u\parallel A_{G_n}\hfill \\ \hfill & \ge \parallel u\parallel .\hfill \end{array}$$

Thus, $\parallel Tu\parallel \ge \parallel u\parallel$ for $u\in \mathcal{P}\cap \partial {\Omega }_1$.

Next, since $F_{\infty }{B}_{G_n}\lambda <1$, there exists an $ϵ\in (0,1)$ such that

$$((F_{\infty }+ϵ)B_{G_n}+ϵ)\lambda \le 1.$$

Since

$$F_{\infty }=\underset{u\to \infty }{lim\; sup}{\displaystyle \frac{f\left(u\right)}{u}},$$

there exists an $H_3>0$ such that

$$f\left(u\right)\le (F_{\infty }+ϵ)u,u\in [H_3,\infty ).$$

Define

$$M=\underset{u\in [0,H_3]}{max}f\left(u\right).$$

Now, there exists a $k\in (0,1)$ with

$${(-1)}^n{\int }_0^k{G}_n(1,s)g\left(s\right)ds\le {\displaystyle \frac{ϵ}{M}}.$$

Let

$$H_2=max\left\{2H_1,{\displaystyle \frac{H_3}{k^{\alpha -1}}},1\right\},$$

and define ${\Omega }_2=\{u\in \mathcal{B}:\parallel u\parallel <H_2\}$. Let $u\in \mathcal{P}\cap \partial {\Omega }_2$. Then, $\parallel u\parallel =H_2$, and so

$$u\left(1\right)=H_2\ge {\displaystyle \frac{H_3}{k^{\alpha -1}}}>H_3.$$

Now, $u\left(0\right)=0$. So, by the Intermediate Value Theorem, there exists a $\gamma \in (0,1)$ with $u\left(\gamma \right)=H_3$. But for $t\in [k,1]$, we have

$$u\left(t\right)\ge t^{\alpha -1}u\left(1\right)=t^{\alpha -1}{H}_2\ge k^{\alpha -1}{\displaystyle \frac{H_3}{k^{\alpha -1}}}=H_3.$$

So, $\gamma \in (0,k]$. Moreover, since $u\left(t\right)$ is nondecreasing, this implies that

$$0\le u\left(t\right)\le H_3,t\in [0,\gamma )$$

and

$$u\left(t\right)\ge H_3,t\in (\gamma ,1].$$

Therefore,

$$\begin{array}{cc}\hfill \left|\right(Tu\left)\right(1\left)\right|& ={(-1)}^n\lambda {\int }_0^1{G}_n(1,s)g\left(s\right)f\left(u\left(s\right)\right)ds\hfill \\ \hfill & =\lambda \left({(-1)}^n{\int }_0^{\gamma }{G}_n(1,s)g\left(s\right)f\left(u\left(s\right)\right)ds+{(-1)}^n{\int }_{\gamma }^1{G}_n(1,s)g\left(s\right)f\left(u\left(s\right)\right)ds\right)\hfill \\ \hfill & \le \lambda \left(M{\int }_0^{\gamma }{(-1)}^n{G}_n(1,s)g\left(s\right)ds+{(-1)}^n{\int }_{\gamma }^1{G}_n(1,s)g\left(s\right)(F_{\infty }+ϵ)u\left(s\right)ds\right)\hfill \\ \hfill & \le \lambda \left(M{\displaystyle \frac{ϵ}{M}}+(F_{\infty }+ϵ)u\left(1\right){\int }_{\gamma }^1{(-1)}^n{G}_n(1,s)g\left(s\right)ds\right)\hfill \\ \hfill & \le \lambda (ϵ+(F_{\infty }+ϵ)\parallel u\parallel B_{G_n})\hfill \\ \hfill & \le \lambda (ϵ\parallel u\parallel +(F_{\infty }+ϵ)\parallel u\parallel B_{G_n})\hfill \\ \hfill & =\lambda \parallel u\parallel (ϵ+(F_{\infty }+ϵ)B_{G_n})\hfill \\ \hfill & \le \parallel u\parallel .\hfill \end{array}$$

Thus, $\parallel Tu\parallel \le \parallel u\parallel$ for $u\in \mathcal{P}\cap \partial {\Omega }_2$. Notice that since $H_1<H_2$, we have ${\overline{\Omega }}_1\subset {\Omega }_2$. Thus, by Theorem 1 (2), T has a fixed point $u\in \mathcal{P}$. By the definition of T, this fixed point is a positive solution of (1) and (2). □

6. Nonexistence Results

Penultimately, we provide two nonexistence results of positive solutions based on the size of parameter $\lambda$. First, we need the following lemma.

Lemma 5. 

Suppose $D_{0^{+}}^{\alpha +2n}u\in C[0,1]$. If ${(-1)}^n(-D_{0^{+}}^{\alpha +2n}u\left(t\right))\ge 0$ for all $t\in [0,1]$ and $u\left(t\right)$ satisfies (2). Then, we have the following:

(1) 

$u^{\prime }\left(t\right)\ge 0,0\le t\le 1$;

(2) 

$t^{\alpha -1}u\left(1\right)\le u\left(t\right)\le u\left(1\right),0\le t\le 1$.

Proof. 

Let $0\le t\le 1$.

For (1), by Lemma 3 (2),

$$\begin{array}{cc}\hfill u^{\prime }\left(t\right)& ={\int }_0^1{\displaystyle \frac{\partial }{\partial t}}{G}_n(t,s)(-D_{0^{+}}^{\alpha +2n}u\left(s\right))ds\hfill \\ \hfill & ={\int }_0^1{(-1)}^n{\displaystyle \frac{\partial }{\partial t}}{G}_n(t,s){(-1)}^n(-D_{0^{+}}^{\alpha +2n}u\left(s\right))ds\hfill \\ \hfill & >0.\hfill \end{array}$$

For (2), by Lemma 3 (3),

$$\begin{array}{cc}\hfill t^{\alpha -1}u\left(1\right)& =t^{\alpha -1}{\int }_0^1{G}_n(1,s)(-D_{0^{+}}^{\alpha +2n}u\left(s\right))ds\hfill \\ \hfill & ={\int }_0^1{(-1)}^n{t}^{\alpha -1}{G}_n(1,s){(-1)}^n(-D_{0^{+}}^{\alpha +2n}u\left(s\right))ds\hfill \\ \hfill & \le {\int }_0^1{(-1)}^n{G}_n(t,s){(-1)}^n(-D_{0^{+}}^{\alpha +2n}u\left(s\right))ds\hfill \\ \hfill & ={\int }_0^1{G}_n(t,s)(-D_{0^{+}}^{\alpha +2n}u\left(s\right))ds\hfill \\ \hfill & =u\left(t\right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill u\left(t\right)& ={\int }_0^1{G}_n(t,s)(-D_{0^{+}}^{\alpha +2n}u\left(s\right))ds\hfill \\ \hfill & ={\int }_0^1{(-1)}^n{G}_n(t,s){(-1)}^n(-D_{0^{+}}^{\alpha +2n}u\left(s\right))ds\hfill \\ \hfill & \le {\int }_0^1{(-1)}^n{G}_n(1,s){(-1)}^n(-D_{0^{+}}^{\alpha +2n}u\left(s\right))ds\hfill \\ \hfill & ={\int }_0^1{G}_n(1,s)(-D_{0^{+}}^{\alpha +2n}u\left(s\right))ds\hfill \\ \hfill & =u\left(1\right).\hfill \end{array}$$

□

Theorem 4. 

If

$$\lambda <{\displaystyle \frac{u}{B_{G_n}f\left(u\right)}}$$

for all $u\in (0,\infty )$, then no positive solution exists for (1) and (2).

Proof. 

For contradiction, suppose that $u\left(t\right)$ is a positive solution to (1) and (2). Then, ${(-1)}^n(-D_{0^{+}}^{\alpha +2n}u\left(t\right))=\lambda g\left(t\right)f\left(u\left(t\right)\right)\ge 0$. So, by Lemma 5,

$$\begin{array}{cc}\hfill u\left(1\right)& ={(-1)}^n\lambda {\int }_0^1{G}_n(1,s)g\left(s\right)f\left(u\left(s\right)\right)ds\hfill \\ \hfill & <{(-1)}^n{\left(B_{G_n}\right)}^{-1}{\int }_0^1{G}_n(1,s)g\left(s\right)u\left(s\right)ds\hfill \\ \hfill & \le u\left(1\right){\left(B_{G_n}\right)}^{-1}{\int }_0^1{(-1)}^n{G}_n(1,s)g\left(s\right)ds\hfill \\ \hfill & =u\left(1\right),\hfill \end{array}$$

which is a contradiction. □

Theorem 5. 

If

$$\lambda >{\displaystyle \frac{u}{A_{G_n}f\left(u\right)}}$$

for all $u\in (0,\infty )$, then no positive solution exists for (1) and (2).

Proof. 

For contradiction, suppose that $u\left(t\right)$ is a positive solution to (1) and (2). Then, ${(-1)}^n(-D_{0^{+}}^{\alpha +2n}u\left(t\right))=\lambda g\left(t\right)f\left(u\left(t\right)\right)\ge 0$. So, by Lemma 5,

$$\begin{array}{cc}\hfill u\left(1\right)& ={(-1)}^n\lambda {\int }_0^1{G}_n(1,s)g\left(s\right)f\left(u\left(s\right)\right)ds\hfill \\ \hfill & >{(-1)}^n{\left(A_{G_n}\right)}^{-1}{\int }_0^1{G}_n(1,s)g\left(s\right)u\left(s\right)ds\hfill \\ \hfill & \ge u\left(1\right){\left(A_{G_n}\right)}^{-1}{\int }_0^1{(-1)}^n{s}^{\alpha -1}{G}_n(1,s)g\left(s\right)ds\hfill \\ \hfill & =u\left(1\right),\hfill \end{array}$$

which is a contradiction. □

7. An Example

To conclude this paper, we provide an explicit example and calculate approximate bounds of the parameter $\lambda$ for the existence and nonexistence of positive solutions. We use Theorems 2, 4 and 5. Examples constructed using Theorems 3–5 are found and proved similarly.

Set $n=2$, $m=3$, $\alpha =2.5$, $\beta =1.5$, and $g\left(t\right)=t$. We note that $g\left(t\right)\ge 0$ is continuous for $0\le t\le 1$ and ${\int }_0^{1}g\left(t\right)dt>0$. Now, we have that

$$\begin{array}{cc}\hfill G_0(1,s)& ={\displaystyle \frac{1}{\Gamma \left(2.5\right)}}\left\{\begin{array}{cc}{1}^{1.5}{(1-s)}^0-{(1-s)}^{1.5},\hfill & 0\le s<t\le 1,\hfill \\ 1^{1.5}{(1-s)}^0,\hfill & 0\le t\le s<1\hfill \end{array}\right.\hfill \\ \hfill & ={\displaystyle \frac{1-{(1-s)}^{1.5}}{\Gamma \left(2.5\right)}},\hfill \end{array}$$

and we compute

$$\begin{array}{cc}\hfill A_{G_2}& ={\int }_0^1{(-1)}^2{s}^{1.5}{G}_2(1,s)\left(s\right)ds\hfill \\ \hfill & ={\int }_0^1\left[-{\int }_0^1{G}_1(1,r_1)G_{conj}(r_1,s)d{r}_1\right]s^{2.5}ds\hfill \\ \hfill & ={\int }_0^1\left[-{\int }_0^1\left({\int }_0^1-G_0(1,r_2)G_{conj}(r_2,r_1)d{r}_2\right)G_{conj}(r_1,s)d{r}_1\right]s^{2.5}ds\hfill \\ \hfill & \approx 0.00095454,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill B_{G_2}& ={\int }_0^1{(-1)}^2{G}_2(1,s)\left(s\right)ds\hfill \\ \hfill & ={\int }_0^1\left[-{\int }_0^1{G}_1(1,r_1)G_{conj}(r_1,s)d{r}_1\right]sds\hfill \\ \hfill & ={\int }_0^1\left[-{\int }_0^1\left({\int }_0^1-G_0(1,r_2)G_{conj}(r_2,r_1)d{r}_2\right)G_{conj}(r_1,s)d{r}_1\right]sds\hfill \\ \hfill & \approx 0.00197039.\hfill \end{array}$$

Now that we have $A_{G_2}$ and $B_{G_2}$, applying these Theorems is much simpler as they are based on the liminfs and limsups of choice of $f\left(u\right)$.

Example 1. 

We demonstrate an example for Theorems 2, 4, and 5. Set $f\left(u\right)=u(3u+1)/(u+1)$. We note that $f\left(u\right)\ge 0$ is continuous for $u\ge 0$. Thus, the fractional boundary value problem is

$$D_{0^{+}}^{6.5}u\left(t\right)+\lambda tu\left({\displaystyle \frac{3u+1}{u+1}}\right)=0,0<t<1,$$

(3)

subject to

$$\begin{array}{cc}\hfill u\left(0\right)=u^{\prime }\left(0\right)=0,& D_{0^{+}}^{1.5}\left(1\right)=0\hfill \\ \hfill D_{0^{+}}^{2.5}u\left(0\right)=D_{0^{+}}^{4.5}\left(0\right)=0,& D_{0^{+}}^{2.5}\left(1\right)=D_{0^{+}}^{4.5}\left(1\right)=0.\hfill \end{array}$$

(4)

We compute the liminfs and limsups for $f\left(u\right)/u=(3u+1)/(u+1)$.

$$\begin{array}{cccc}\hfill f_{\infty }& =\underset{u\to \infty }{lim\; inf}{\displaystyle \frac{3u+1}{u+1}}=3,\hfill & \hfill & F_0=\underset{u\to 0^{+}}{lim\; sup}{\displaystyle \frac{3u+1}{u+1}}=1\hfill \\ \hfill f_0& =\underset{u\to 0^{+}}{lim\; inf}{\displaystyle \frac{3u+1}{u+1}}=1,\hfill & \hfill & F_{\infty }=\underset{u\to \infty }{lim\; sup}{\displaystyle \frac{3u+1}{u+1}}=3\hfill \end{array}$$

Then, we have

$${\displaystyle \frac{1}{A_{G_2}{f}_{\infty }}}\approx {\displaystyle \frac{1}{0.00095454·3}}\approx 349.28$$

and

$${\displaystyle \frac{1}{B_{G_2}{F}_0}}\approx {\displaystyle \frac{1}{0.00197039·1}}\approx 507.61.$$

Next, for $u\in (0,\infty )$, we investigate

$${\displaystyle \frac{u}{B_{G_2}f\left(u\right)}}={\displaystyle \frac{u+1}{B_{G_2}(3u+1)}}.$$

We calculate

$$\underset{u\in (0,\infty )}{inf}{\displaystyle \frac{u+1}{B_{G_2}(3u+1)}}={\displaystyle \frac{1}{B_{G_2}}}\underset{u\in (0,\infty )}{inf}{\displaystyle \frac{u+1}{(3u+1)}}\approx {\displaystyle \frac{1}{0.00197039}}\left({\displaystyle \frac{1}{3}}\right)\approx 169.15.$$

Finally, for $u\in (0,\infty )$, we investigate

$${\displaystyle \frac{u}{A_{G_2}f\left(u\right)}}={\displaystyle \frac{u+1}{A_{G_2}(3u+1)}}.$$

We calculate

$$\underset{u\in (0,\infty )}{sup}{\displaystyle \frac{u+1}{A_{G_2}(3u+1)}}={\displaystyle \frac{1}{A_{G_2}}}\underset{u\in (0,\infty )}{sup}{\displaystyle \frac{u+1}{(3u+1)}}\approx {\displaystyle \frac{1}{0.00095454}}\left(1\right)\approx 1047.17.$$

Therefore, by Theorems 2, 4, and 5, if $349.28<\lambda <507.61$, then (3) and (4) have at least one positive solution, and if $\lambda <169.15$ or $\lambda >1047.17$, then (3) and (4) do not have a positive solution.

8. Conclusions

A Riemann–Liouville fractional derivative with fractional boundary conditions including Lidstone-inspired conditions was studied. With the use of Green’s function, convolution, induction, and fixed-point theory, at least one positive solution was proven to exist if parameter $\lambda$ was within certain bounds. Subsequently, no positive solutions were shown to exist if $\lambda$ satisfied other bounds. An explicit example was constructed to demonstrate how to utilize the presented theorems.

Future work would aim to investigate the convolution of the fractional boundary value problem with other types of Green functions such as those for right-focal or multipoint problems. Additionally, one could use the convolution with induction approach to find positive solutions for fractional boundary value problems that contain a singularity.