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

1. Introduction

Let $m,n\in \mathbb{N}$, $m\ge 3$, with $\alpha \in (m-1,m]$ and $\beta \in [1,m-1]$. Consider the following 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

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

(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), (2).

To address this, we follow the procedure of Eloe et al. in [5] of constructing the associated Green’s function for the given problem by convoluting the Green’s function, $G_0(t,s)$, for equation $-D_{0^{+}}^{\alpha }u=0$ with boundary conditions (2) with that of a conjugate boundary value problem. We present properties of the Green’s function many of which can be found in [10] and [12]. Then, we deploy those in an application of the 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 the parameter $\lambda$. This fixed point is a positive solution to (1), (2). We will also give suitable conditions on $\lambda$ for the nonexistence of solutions to (1), (2).

This study extends the existing literature on fractional boundary value problems that leverage Krasnosel’skii’s fixed point theorem. Previous works have applied various fixed point theorems to demonstrate the existence of positive solutions for similar problems, such as those studied in [1,2,6,7,8,10,12,14,15]. Here, we use these results to guarantee the existence or nonexistence of a positive solutions by establishing two separate sizing conditions on the parameter $\lambda$ based upon liminfs and limsups of the nonlinearity. This approach is based on the properties of the Green’s function which plays a critical role in showing the existence of positive solutions.

Section two provides definitions on the Riemann-Liousville fractional derivative and suggestions for further study therein and states the Krasnosel’skii Fixed Point Theorem. The subsequent sections are devoted to the construction of the Green’s function and its properties. Then, in sections five and six, we establish intervals for $\lambda$ that yield 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. Then, we present Krasnosel’skii’s Fixed Point Theorem.

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 [3,9,11,13] for further study of fractional calculus and fractional differential equations.

Theorem 1

(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 ${\Omega }_1,{\Omega }_2$ 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

1. 

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

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. The Green’s Function

Now, we construct the Green’s function used for (1), (2) by utilizing induction with a convolution of a lower order problem and a conjugate problem. The procedure is similar to that found in [12].

First, the conjugate boundary value problem

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

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 the 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 ([4])

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

For $k=1,\cdots ,n-1$, 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 the Green’s function for

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

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

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

with boundary conditions

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

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

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

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

Thus, v satisfies the Dirichlet boundary value problem

$$v^{\prime \prime }+h\left(t\right)=0,0<t<1,$$

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

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 conditions (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

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

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

$$v^{\prime \prime }+k\left(t\right)=0,0<t<1,$$

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

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

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

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

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

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 conditions (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 [10].

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[12], 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 (2) 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), (2) based upon the parameter $\lambda$ using the Krasnosel’skii Fixed Point Theorem and our constructed Green’s function and 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

$$\begin{array}{cc}\hfill \mathcal{P}& =\left\{u\in \mathcal{B}:u\left(0\right)=0,u\left(t\right)isnondecreasing,and\right.\hfill \\ & \phantom{=}\left.t^{\alpha -1}u\left(1\right)\le u\left(t\right)\le u\left(1\right)on[0,1]\right\}.\hfill \end{array}$$

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.

The 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), (2) has 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

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

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 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 ${¯}_1{\subset }_2$. 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), (2). □

Theorem 3.

If

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

then (1), (2) has 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

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

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,

$$u\left(1\right)=H_2\ge {\displaystyle \frac{H_3}{k^{\alpha +2n-1}}}>H_3.$$

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

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

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), (2). □

6. Nonexistence Results

Penultimately, we provide two nonexistence of positive solutions results based on the size of the 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

(1) 

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

(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 to (1), (2).

Proof. 

For contradiction, suppose that $u\left(t\right)$ is a positive solution to (1), (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,

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

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 to (1), (2).

Proof. 

For contradiction, suppose that $u\left(t\right)$ is a positive solution to (1), (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,

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

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, 4, and 5 are proved similarly.

Set $n=2$, $m=3$, $\alpha =2.5$, $\beta =1.5$, and $g\left(t\right)=t$. We note that 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)

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

(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

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

and

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

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

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

We calculate

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

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

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

We calculate

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

Therefore, by Theorems 2, 4, and 5, if $349.28<\lambda <507.61$, then (3), (4) has at least one positive solution, and if $\lambda <169.15$ or $\lambda >1047.17$, then (3), (4) does 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 the Green’s function, convolution, induction, and fixed point theory, at least one positive solution was proven to exist if the 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.