Positive Solutions for Tempered Riemann–Liouville Fractional Equations with Signed Measure and Sign-Changing Perturbations via Spectral Analysis

Abstract

In this paper, we investigate the existence of positive solutions for a class of p-Laplacian tempered Riemann–Liouville fractional equations involving signed measures and sign-changing perturbations with Riemann–Stieltjes boundary conditions. By performing a spectral analysis of the associated linear operator and applying Gelfand’s formula, we establish several fundamental properties of the principal eigenvalue for the corresponding linear fractional-order differential equations. Furthermore, under suitable growth conditions on the nonlinear terms, we employ the fixed point index theory to derive new existence results for positive solutions of the proposed equations.

1. Introduction

In this paper, we focus on the existence of positive solutions for a class of tempered Riemann–Liouville fractional equations involving signed measures and sign-changing perturbations. The system also involves the p-Laplacian operator and Riemann–Stieltjes integral boundary conditions:

$$\left\{\begin{array}{c}{\phi }_p(-{}_0{}^R\mathbb{D}_t^{\beta ,\lambda }x\left(t\right)-g(t,x\left(t\right)))=f(t,x\left(t\right)),\hfill \\ x\left(0\right)=x^{\prime }\left(0\right)=0,x\left(1\right)={\int }_0^{1}x\left(t\right)dA\left(t\right),\hfill \end{array}\right.$$

(1)

where $2<\beta \le 3$, and $\lambda >0$ is a constant, the operator ${\phi }_p\left(s\right)={\left|s\right|}^{p-2}s$ is the p-Laplacian with $p,q>1$ and $\frac{1}{p}+\frac{1}{q}=1$. ${}_0{}^R\mathbb{D}_t^{\beta ,\lambda }$ is the tempered Riemann–Liouville fractional derivative. The term ${\int }_0^{1}x\left(t\right)dA\left(t\right)$ is a Riemann–Stieltjes integral, where A is a function of bounded variation, which allows $dA\left(t\right)$ to be signed measures. The nonlinear function $f:[0,1]\times [0,+\infty )\to [0,+\infty )$ is continuous. The sign-changing perturbation $g:[0,1]\times (-\infty ,+\infty )\to (-\infty ,+\infty )$ satisfies the Carathéodory conditions, and may tend to negative infinity at certain singular points in $(0,1)$.

Fractional differential equations have attracted considerable attention because of their remarkable ability to characterize memory and hereditary effects arising in viscoelastic mechanics [1], chemotaxis–convection models and systems [2,3,4,5,6], diffusion processes [7,8,9], neural networks [10,11], biomedical engineering [12], and chemical graph theory [13]. Motivated by these broad applications, fractional calculus has undergone rapid theoretical development over the past decades [14,15,16,17,18]. Meanwhile, efficient numerical approaches, including finite element methods, finite difference schemes, and domain decomposition algorithms, have been extensively developed for both classical and fractional differential equations [19,20,21,22,23,24,25,26,27,28,29,30,31]. These advances have significantly improved the numerical simulation and analysis of complex diffusion, transport, and anomalous dynamic phenomena.

Considerable efforts have also been devoted to the theoretical analysis of tempered fractional equations. Zhang et al. [32] investigated a class of p-Laplacian tempered Riemann–Liouville fractional sub-diffusion equations with Riemann–Stieltjes integral boundary conditions and established existence results by applying fixed point theory. Zhang et al. [33] further studied singular tempered sub-diffusion equations involving sign-changing perturbations and obtained the existence of positive solutions. In addition, the study of tempered fractional equations has developed in several complementary directions. Obeidat and Bentil [34] developed new theoretical frameworks for tempered fractional differential equations and explored their applications to various nonlinear models, enriching the theoretical foundation of tempered fractional calculus. Alvarez et al. [35] characterized Hölder regularity and established maximal regularity results for tempered fractional abstract equations through resolvent estimates, providing deeper insights into the regularity theory of nonlocal evolution equations. From the perspective of numerical computation, Ounamane et al. [36] proposed an extended truncated exponential method for tempered fractional variational problems with integral constraints and demonstrated its computational efficiency and accuracy. Moreover, Verma and Tiwari [37] investigated a two-dimensional time–space tempered fractional diffusion-wave equation, where existence, uniqueness, and Ulam–Hyers stability results were obtained, together with analytical solutions derived by the two-step Adomian decomposition method. Further related results on tempered fractional equations can be found in [38,39,40]. Although these studies have substantially advanced the theory of tempered fractional equations, they mainly focus on abstract evolution equations, diffusion-wave models, variational problems, or equations with standard nonlinear structures. However, tempered fractional boundary value problems involving sign-changing perturbations and signed Riemann–Stieltjes integral boundary conditions remain largely unexplored. This motivates further investigation of positive solutions for more general tempered fractional systems with nonlocal boundary interactions.

It is worth emphasizing that Riemann–Stieltjes integral boundary conditions provide a unified framework that includes multi-point, integral, and measure-type boundary conditions as special cases. Recently, Almaghamsi [41] investigated a coupled Langevin system involving generalized proportional fractional derivatives under Riemann–Stieltjes integral boundary conditions, while Yang et al. [42] established existence and multiplicity results for positive solutions by employing fixed point index theory. Due to their flexibility in describing nonlocal interactions, such boundary conditions [43,44] have been widely applied in fractional boundary value problems.

Nevertheless, the simultaneous presence of sign-changing perturbations and signed measures in Riemann–Stieltjes boundary conditions introduces essential analytical difficulties, mainly because these factors may prevent the associated Green function and the corresponding integral operator from preserving positivity [45]. Consequently, many existing analytical techniques cannot be directly applied such as iterative technique [46], finite volume method [47,48,49,50], fixed point theorem [51], upper-lower solution method [52,53] and finite difference method [54]. Motivated by these challenges, this paper investigates a class of tempered Riemann–Liouville fractional p-Laplacian equations with sign-changing perturbations and signed Riemann–Stieltjes boundary conditions. By combining spectral analysis, the Krein–Rutman theorem, and fixed point index theory, we establish new existence and multiplicity results for positive solutions.

2. Preliminaries and Lemmas

In this section, we recall some basic definitions concerning the Riemann–Liouville fractional operators and Carathéodory functions. We also introduce several assumptions and preliminary lemmas that will be used throughout the paper.

Definition 1

([55]). The Riemann–Liouville fractional integral of order $\alpha >0$ of a function $u:(0,+\infty )\to \mathbb{R}$ is given by

$$I^{\alpha }u\left(t\right)=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_0^t{(t-s)}^{\alpha -1}u\left(s\right)ds$$

provided that the right-hand side is well defined for every $t\in (0,+\infty )$.

Definition 2

([55]). The Riemann–Liouville fractional derivative of order $\alpha >0$ of a function $u:(0,+\infty )\to \mathbb{R}$ is given by

$${\mathcal{D}}^{\alpha }u\left(t\right)=\frac{1}{\mathsf{\Gamma }(n-\alpha )}{\left(\frac{d}{dt}\right)}^n{\int }_0^t{(t-s)}^{n-\alpha -1}u\left(s\right)ds,$$

where $n=\left[\alpha \right]+1$, $\left[\alpha \right]$ denotes the greatest integer part of the number α, provided that the right-hand side is well defined for every $t\in (0,+\infty )$.

Definition 3

([56]). For a real order $\alpha >0$ and an exponential tempering parameter $\lambda >0$, the left-sided tempered Riemann–Liouville fractional derivative of a function $u:(0,+\infty )\to \mathbb{R}$ is defined as

$${}_0{}^R\mathbb{D}_t^{\alpha ,\lambda }u\left(t\right)=e^{-\lambda t}\frac{1}{\mathsf{\Gamma }(n-\alpha )}{\left(\frac{d}{dt}\right)}^n{\int }_0^t{(t-s)}^{n-\alpha -1}{e}^{\lambda s}u\left(s\right)ds.$$

(2)

Remark 1.

Letting $v\left(t\right)=e^{\lambda t}u\left(t\right)$, it follows from Definition 2 that

$${\mathcal{D}}^{\alpha }v\left(t\right)=\frac{1}{\mathsf{\Gamma }(n-\alpha )}{\left(\frac{d}{dt}\right)}^n{\int }_0^t{(t-s)}^{n-\alpha -1}{e}^{\lambda s}u\left(s\right)ds.$$

Multiplying both sides by $e^{-\lambda t}$ immediately yields

$${\mathbb{D}}_t^{\alpha ,\lambda }u\left(t\right)=e^{-\lambda t}{\mathcal{D}}^{\alpha }\left(e^{\lambda t}u\left(t\right)\right).$$

(3)

So a tempered Riemann–Liouville fractional derivative ${}_0{}^R\mathbb{D}_t^{\alpha ,\lambda }$ can be regarded as an exponentially weighted Riemann–Liouville fractional derivative.

Definition 4.

A mapping $g:[0,1]\times \mathbb{R}\to \mathbb{R}$ is called a Carathéodory function if the following conditions hold:

(a) 

for each $u\in \mathbb{R}$, the mapping $t\mapsto g(t,u)$ is Lebesgue measurable on $[0,1]$;

(b) 

for almost every $t\in [0,1]$, the mapping $u\mapsto g(t,u)$ is continuous on $\mathbb{R}$;

(c) 

there exists a function $h\in L^1([0,1],[0,+\infty ))$ such that

$$\left|g\right(t,u\left)\right|\le h\left(t\right),\mathit{for}\mathit{almost}\mathit{every}t\in [0,1],u\in \mathbb{R}.$$

Throughout the paper, we impose the following assumptions on the nonlinear term f and the perturbation term g.

• $\left({\mathbf{I}}_{\mathbf{1}}\right)$ The function $f:[0,1]\times [0,+\infty )\to [0,+\infty )$ is continuous.
• $\left({\mathbf{I}}_{\mathbf{2}}\right)$ The function $g:[0,1]\times (-\infty ,+\infty )\to (-\infty ,+\infty )$ satisfies the Carathéodory condition.

The following basic properties of the Riemann–Liouville fractional operators can be found in ([55], Lemmas 2.3–2.5, pp. 73–75).

Lemma 1.

Let $x\left(t\right)\in C[0,1]\cap L^1[0,1]$, $\alpha >\gamma >0$, and let $m=\left[\gamma \right]+1$. Then the following statements hold:

(i) 

$${\mathcal{I}}_t^{\gamma }{\mathcal{D}}_t^{\gamma }x\left(t\right)=x\left(t\right)+a_1{t}^{\gamma -1}+a_2{t}^{\gamma -2}+\dots +a_m{t}^{\gamma -m},$$

where $a_i\in \mathbb{R}$ for $i=1,2,\dots ,m$.

(ii) 

$${\mathcal{I}}_t^{\alpha }{\mathcal{I}}_t^{\gamma }x\left(t\right)={\mathcal{I}}_t^{\alpha +\gamma }x\left(t\right),{\mathcal{D}}_t^{\gamma }{\mathcal{I}}_t^{\alpha }x\left(t\right)={\mathcal{I}}_t^{\alpha -\gamma }x\left(t\right),{\mathcal{D}}_t^{\gamma }{\mathcal{I}}_t^{\gamma }x\left(t\right)=x\left(t\right).$$

Throughout this paper, we assume that

$$2<\beta \le 3,$$

as specified in Equation (1).

Lemma 2.

Let $h:[0,1]\to [0,+\infty )$ be the function given in Definition 4. Then the tempered Riemann–Liouville fractional boundary value problem

$$\left\{\begin{array}{cc}-{}_0{}^R\mathbb{D}_t^{\beta ,\lambda }x\left(t\right)=h\left(t\right),\hfill & 0<t<1,\hfill \\ x\left(0\right)=x^{\prime }\left(0\right)=0,x\left(1\right)=0,\hfill & \end{array}\right.$$

(4)

has a unique positive solution given by

$$x\left(t\right)={\int }_0^{1}H(t,s)h\left(s\right)ds,$$

(5)

where the Green’s function $H(t,s)$ is defined by

$$H(t,s)=\left\{\begin{array}{cc}\frac{t^{\beta -1}{(1-s)}^{\beta -1}-{(t-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s},\hfill & 0\le s\le t\le 1,\hfill \\ \frac{t^{\beta -1}{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s},\hfill & 0\le t\le s\le 1,\hfill \end{array}\right.$$

(6)

which is continuous on $[0,1]\times [0,1]$. Furthermore, H satisfies the estimate

$$\begin{array}{ll}\frac{t^{\beta -1}(1-t)s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}& e^{-\lambda t}{e}^{\lambda s}\le H(t,s)\\ & \le \min \left\{\frac{(\beta -1)t^{\beta -1}(1-t)e^{\lambda (1-t)}}{\mathsf{\Gamma }\left(\beta \right)},\frac{(\beta -1){(1-s)}^{\beta -1}s{e}^{\lambda s}}{\mathsf{\Gamma }\left(\beta \right)}\right\}.\end{array}$$

(7)

Proof. 

In fact, by $2<\beta \le 3$, Lemma 1 and (2), one has

$$e^{\lambda t}x\left(t\right)=-\frac{1}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^t{(t-s)}^{\beta -1}{e}^{\lambda s}h\left(s\right)ds+b_1{t}^{\beta -1}+b_2{t}^{\beta -2}+b_3{t}^{\beta -3},t\in [0,1].$$

Since $x\left(0\right)=x^{\prime }\left(0\right)=0$ and $x\left(1\right)=0$, we have $b_2=b_3=0$ and

$$b_1=\frac{1}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^1{(1-s)}^{\beta -1}{e}^{\lambda s}h\left(s\right)ds.$$

Thus

$$\begin{array}{l}x\left(t\right)=\frac{1}{\mathsf{\Gamma }\left(\beta \right)}\left[t^{\beta -1}{\int }_0^1{(1-s)}^{\beta -1}{e}^{-\lambda t}{e}^{\lambda s}h\left(s\right)ds-{\int }_0^t{(t-s)}^{\beta -1}{e}^{-\lambda t}{e}^{\lambda s}h\left(s\right)ds\right]\\ =\frac{1}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^t\left[t^{\beta -1}{(1-s)}^{\beta -1}-{(t-s)}^{\beta -1}\right]e^{-\lambda t}{e}^{\lambda s}h\left(s\right)ds\\ +\frac{1}{\mathsf{\Gamma }\left(\beta \right)}{\int }_t^1{t}^{\beta -1}{(1-s)}^{\beta -1}{e}^{-\lambda t}{e}^{\lambda s}h\left(s\right)ds\\ ={\int }_0^{1}H(t,s)h\left(s\right)ds,t\in [0,1].\end{array}$$

Since $2<\beta \le 3$, we have $\beta -1>0$, and therefore

$$\underset{t\to s^{+}}{\lim }{(t-s)}^{\beta -1}=0.$$

Consequently,

$$\underset{t\to s^{-}}{\lim }H(t,s)=\underset{t\to s^{+}}{\lim }H(t,s),$$

which implies that $H(t,s)$ is continuous on $[0,1]\times [0,1]$.

Next we prove the estimate (7).

Case 1. Assume that $s\le t$. Since $1-s\ge 1-t,$ we have

$$\begin{array}{cc}\hfill H(t,s)& =\frac{t^{\beta -1}{(1-s)}^{\beta -1}-{(t-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & =\frac{(\beta -1){\int }_{t-s}^{t-ts}{x}^{\beta -2}dx}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & \le \frac{(\beta -1){(t-ts)}^{\beta -2}\left((t-ts)-(t-s)\right)}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & =\frac{(\beta -1)t^{\beta -2}{(1-s)}^{\beta -2}(1-t)s}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & \le \min \left\{\frac{e^{\lambda }(\beta -1)t^{\beta -1}(1-t)e^{-\lambda t}}{\mathsf{\Gamma }\left(\beta \right)},\frac{(\beta -1){(1-s)}^{\beta -1}s{e}^{\lambda s}}{\mathsf{\Gamma }\left(\beta \right)}\right\}.\hfill \end{array}$$

Moreover, since

$$t(1-s)\ge t-s$$

and $\beta -2>0$, we have

$${(t-s)}^{\beta -2}\le {(t-ts)}^{\beta -2}.$$

Therefore,

$${(t-s)}^{\beta -1}={(t-s)}^{\beta -2}(t-s)\le {(t-ts)}^{\beta -2}(t-s).$$

Consequently,

$$\begin{array}{cc}\hfill H(t,s)& =\frac{t^{\beta -1}{(1-s)}^{\beta -1}-{(t-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & \ge \frac{{(t-ts)}^{\beta -2}(t-ts)-{(t-ts)}^{\beta -2}(t-s)}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & =\frac{t^{\beta -2}{(1-s)}^{\beta -2}(1-t)s}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & \ge \frac{t^{\beta -1}(1-t)s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}.\hfill \end{array}$$

Case 2. Assume that $t\le s$. Since $2<\beta \le 3$, we obtain

$$\begin{array}{cc}\hfill H(t,s)& =\frac{t^{\beta -1}{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & \le \frac{(\beta -1)t^{\beta -2}t{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & \le \frac{(\beta -1)t^{\beta -2}s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & \le \frac{(\beta -1)s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{\lambda s},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill H(t,s)& =\frac{{(1-s)}^{\beta -1}{t}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & \le \frac{(\beta -1)t^{\beta -1}(1-s){(1-s)}^{\beta -2}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & \le \frac{e^{\lambda }(\beta -1)t^{\beta -1}(1-t)e^{-\lambda t}}{\mathsf{\Gamma }\left(\beta \right)},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill H(t,s)& =\frac{t^{\beta -1}{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}\hfill \\ & \ge \frac{t^{\beta -1}(1-t)s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}.\hfill \end{array}$$

The proof is complete. □

Remark 2.

Since $0<t,s<1$ and $2<\beta \le 3$, it follows immediately that

$$t^{\beta -1}(1-t)s{(1-s)}^{\beta -1}>0.$$

Hence, the lower bound in Equation (7) is strictly positive on $(0,1)\times (0,1)$.

Define

$$\begin{array}{l}\mathcal{A}={\int }_0^1{e}^{\lambda (1-t)}{t}^{\beta -1}dA\left(t\right),\\ \overline{\mathcal{A}}={\int }_0^1{e}^{\lambda (1-t)}{t}^{\beta -1}(1-t)dA\left(t\right),\\ {\mathcal{H}}_A\left(s\right)={\int }_0^{1}H(t,s)dA\left(t\right).\end{array}$$

(8)

To guarantee the nonnegativity of the Green’s function, we impose the following basic assumption on $\mathcal{A}$.

• (A₀) A is a function of bounded variation such that ${\mathcal{H}}_A\left(s\right)\ge 0$ for all $s\in [0,1]$, and

  $$0<\overline{\mathcal{A}}\le \mathcal{A}<1.$$

Remark 3.

Throughout this paper, the boundary condition is understood in the Riemann–Stieltjes sense. In particular, the measure $dA\left(t\right)$ is allowed to contain atomic (Dirac delta) components. Consequently, the nonlocal boundary condition includes the classical multi-point boundary condition as a special case.

Remark 4.

In condition $\left({\mathbf{A}}_{\mathbf{0}}\right)$, let $dA=d{A}^{+}-d{A}^{-}$ be the Jordan decomposition of the signed measure induced by the bounded variation function A, and if for every $s\in [0,1]$,

$${\int }_0^{1}H(t,s)d{A}^{+}\left(t\right)\ge {\int }_0^{1}H(t,s)d{A}^{-}\left(t\right),$$

then

$${\mathcal{H}}_A\left(s\right)\ge 0,s\in [0,1].$$

Lemma 3.

Suppose that $\left({\mathbf{A}}_{\mathbf{0}}\right)$ holds. Then the function ${\mathcal{H}}_A\left(s\right)$ defined by (8) satisfies the following properties:

$$0\le \frac{\overline{\mathcal{A}}}{\mathsf{\Gamma }\left(\beta \right)}s{(1-s)}^{\beta -1}{e}^{\lambda s}\le {\mathcal{H}}_A\left(s\right)\le \frac{\overline{\mathcal{A}}(\beta -1)}{\mathsf{\Gamma }\left(\beta \right)}.$$

(9)

Proof. 

It follows from (7) and $\left({\mathbf{A}}_{\mathbf{0}}\right)$ that

$$\begin{array}{cc}& {\mathcal{H}}_A\left(s\right)={\int }_0^{1}H(t,s)dA\left(t\right)\hfill \\ & \ge {\int }_0^1\frac{t^{\beta -1}(1-t)s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s}dA\left(t\right)\hfill \\ & =\frac{s{(1-s)}^{\beta -1}{e}^{\lambda s}}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^1{t}^{\beta -1}(1-t)e^{-\lambda t}dA\left(t\right)\hfill \\ & =\frac{\overline{\mathcal{A}}}{\mathsf{\Gamma }\left(\beta \right)}s{(1-s)}^{\beta -1}{e}^{\lambda s}\ge 0.\hfill \end{array}$$

On the other hand, by using the right side of (7), we have

$$\begin{array}{cc}& {\mathcal{H}}_A\left(s\right)={\int }_0^{1}H(t,s)dA\left(t\right)\hfill \\ & \le {\int }_0^1\frac{(\beta -1)t^{\beta -1}(1-t)e^{\lambda (1-t)}}{\mathsf{\Gamma }\left(\beta \right)}dA\left(t\right)\hfill \\ & =\frac{(\beta -1)}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^1{t}^{\beta -1}(1-t)e^{\lambda (1-t)}dA\left(t\right)\hfill \\ & =\frac{\overline{\mathcal{A}}(\beta -1)}{\mathsf{\Gamma }\left(\beta \right)}.\hfill \end{array}$$

The proof is complete. □

Lemma 4.

Let h be defined in Definition 4. Then the tempered Riemann–Liouville fractional boundary value problem

$$\left\{\begin{array}{cc}-{}_0{}^R\mathbb{D}_t^{\beta ,\lambda }x\left(t\right)=h\left(t\right),\hfill & 0<t<1,\hfill \\ x\left(0\right)=x^{\prime }\left(0\right)=0,x\left(1\right)={\int }_0^{1}x\left(t\right)dA\left(t\right),\hfill & \end{array}\right.$$

(10)

has the unique positive solution

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

where

$$G(t,s)=\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\mathcal{H}}_A\left(s\right)+H(t,s).$$

(11)

is the Green function of the boundary value problem (10). Moreover, $\xi \left(t\right)$ satisfies the following estimate

$$\xi \left(t\right)\le \mathsf{\Lambda }\mathsf{\Phi }\left(t\right){\int }_0^{1}h\left(s\right)ds,$$

(12)

where

$$\mathsf{\Phi }\left(t\right)=(1-\mathcal{A})t^{\beta -1}(1-t)e^{-\lambda t}+\overline{\mathcal{A}}{e}^{-\lambda t}{t}^{\beta -1}\le 1.$$

(13)

$$\mathsf{\Lambda }=\frac{(\beta -1)e^{\lambda }}{\mathsf{\Gamma }\left(\beta \right)\left(1-\mathcal{A}\right)}.$$

(14)

$$0\le G(t,s)\le \vartheta =:\frac{(\beta -1)e^{\lambda }}{\mathsf{\Gamma }\left(\beta \right)}\left(1+\frac{\overline{\mathcal{A}}}{(1-\mathcal{A})}\right).$$

(15)

Proof. 

Applying Lemma 1 to the equation

$$-{}_0{}^R\mathbb{D}_t^{\beta ,\lambda }x\left(t\right)=h\left(t\right),$$

we obtain

$$e^{\lambda t}x\left(t\right)=-\frac{1}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^t{(t-s)}^{\beta -1}{e}^{\lambda s}h\left(s\right)ds+c_1{t}^{\beta -1}+c_2{t}^{\beta -2}+c_3{t}^{\beta -3},$$

where $c_1,c_2,c_3\in \mathbb{R}$.

Since

$$x\left(0\right)=x^{\prime }\left(0\right)=0,$$

it follows that

$$c_2=c_3=0.$$

Hence

$$x\left(t\right)=-\frac{e^{-\lambda t}}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^t{(t-s)}^{\beta -1}{e}^{\lambda s}h\left(s\right)ds+c_1{t}^{\beta -1}{e}^{-\lambda t}.$$

(16)

Consequently, one has

$$x\left(1\right)=-\frac{e^{-\lambda }}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^1{(1-s)}^{\beta -1}{e}^{\lambda s}h\left(s\right)ds+c_1{e}^{-\lambda }.$$

On the other hand,

$$\begin{array}{cc}\hfill {\int }_0^{1}x\left(t\right)dA\left(t\right)& =-\frac{1}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^1{e}^{-\lambda t}\left({\int }_0^t{(t-s)}^{\beta -1}{e}^{\lambda s}h\left(s\right)ds\right)dA\left(t\right)\hfill \\ & +c_1{\int }_0^1{t}^{\beta -1}{e}^{-\lambda t}dA\left(t\right).\hfill \end{array}$$

Interchanging the order of integration yields

$$\begin{array}{c}\hfill {\int }_0^{1}x\left(t\right)dA\left(t\right)=-{\int }_0^1{\mathcal{H}}_A\left(s\right)h\left(s\right)ds+c_1{e}^{-\lambda }\mathcal{A},\end{array}$$

where

$${\mathcal{H}}_A\left(s\right)={\int }_0^{1}H(t,s)dA\left(t\right),\mathcal{A}={\int }_0^1{t}^{\beta -1}{e}^{\lambda (1-t)}dA\left(t\right).$$

Using the boundary condition

$$x\left(1\right)={\int }_0^{1}x\left(t\right)dA\left(t\right),$$

we obtain

$$\begin{array}{c}\hfill c_1{e}^{-\lambda }(1-\mathcal{A})={\int }_0^1\left(\frac{{(1-s)}^{\beta -1}{e}^{\lambda s}}{\mathsf{\Gamma }\left(\beta \right)}-{\mathcal{H}}_A\left(s\right)\right)h\left(s\right)ds,\end{array}$$

that is,

$$c_1=\frac{e^{\lambda }}{1-\mathcal{A}}{\int }_0^1\left(\frac{{(1-s)}^{\beta -1}{e}^{\lambda s}}{\mathsf{\Gamma }\left(\beta \right)}-{\mathcal{H}}_A\left(s\right)\right)h\left(s\right)ds.$$

Substituting this expression into (16) gives

$$\begin{array}{cc}\hfill x\left(t\right)& =-{\int }_0^t\frac{e^{-\lambda t}{(t-s)}^{\beta -1}{e}^{\lambda s}}{\mathsf{\Gamma }\left(\beta \right)}h\left(s\right)ds+{\int }_0^1\frac{t^{\beta -1}{e}^{\lambda (1-t)}{(1-s)}^{\beta -1}{e}^{\lambda s}}{\mathsf{\Gamma }\left(\beta \right)(1-\mathcal{A})}h\left(s\right)ds\hfill \\ & -\frac{t^{\beta -1}{e}^{\lambda (1-t)}}{1-\mathcal{A}}{\int }_0^1{\mathcal{H}}_A\left(s\right)h\left(s\right)ds.\hfill \end{array}$$

Since

$$H(t,s)=\left\{\begin{array}{cc}\frac{t^{\beta -1}{(1-s)}^{\beta -1}-{(t-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s},\hfill & s\le t,\hfill \\ \frac{t^{\beta -1}{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s},\hfill & t\le s,\hfill \end{array}\right.$$

the above expression can be rewritten as

$$x\left(t\right)={\int }_0^1\left(H(t,s)+\frac{t^{\beta -1}{e}^{\lambda (1-t)}}{1-\mathcal{A}}{\mathcal{H}}_A\left(s\right)\right)h\left(s\right)ds.$$

(17)

Therefore, the Green function of problem (10) is

$$G(t,s)=H(t,s)+\frac{t^{\beta -1}{e}^{\lambda (1-t)}}{1-\mathcal{A}}{\mathcal{H}}_A\left(s\right).$$

In the end, it follows from (7), (9), (17) and $\left({\mathbf{A}}_{\mathbf{0}}\right)$ that

$$\begin{array}{l}x\left(t\right)={\int }_0^1\left(H(t,s)+\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\mathcal{H}}_A\left(s\right)\right)h\left(s\right)ds\\ \le (\beta -1)\left(\frac{t^{\beta -1}(1-t)e^{\lambda (1-t)}}{\mathsf{\Gamma }\left(\beta \right)}+\frac{\overline{\mathcal{A}}{e}^{\lambda (1-t)}{t}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)(1-\mathcal{A})}\right){\int }_0^{1}h\left(s\right)ds\\ =\frac{(\beta -1)e^{\lambda }\left((1-\mathcal{A})t^{\beta -1}(1-t)e^{-\lambda t}+\overline{\mathcal{A}}{e}^{-\lambda t}{t}^{\beta -1}\right)}{\mathsf{\Gamma }\left(\beta \right)(1-\mathcal{A})}{\int }_0^{1}h\left(s\right)ds\\ =\mathsf{\Lambda }\mathsf{\Phi }\left(t\right){\int }_0^{1}h\left(s\right)ds,\end{array}$$

where

$$\begin{array}{ll}\mathsf{\Phi }\left(t\right)& =(1-\mathcal{A})t^{\beta -1}(1-t)e^{-\lambda t}+\overline{\mathcal{A}}{e}^{-\lambda t}{t}^{\beta -1}\\ & \le (1-\mathcal{A})t^{\beta -1}{e}^{-\lambda t}+\overline{\mathcal{A}}{e}^{-\lambda t}{t}^{\beta -1}=(1-\mathcal{A}+\overline{\mathcal{A}})t^{\beta -1}{e}^{-\lambda t}\\ & \le 1,\end{array}$$

(18)

and

$$\mathsf{\Lambda }=\frac{(\beta -1)e^{\lambda }}{\mathsf{\Gamma }\left(\beta \right)\left(1-\mathcal{A}\right)}.$$

From (7) and (9) in Lemma 3, it is obvious that

$$\begin{array}{ll}\hfill 0\le G(t,s)& =H(t,s)+\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\mathcal{H}}_A\left(s\right)\\ & \le \frac{(\beta -1){(1-s)}^{\beta -1}s{e}^{\lambda s}}{\mathsf{\Gamma }\left(\beta \right)}+\frac{(\beta -1)\overline{\mathcal{A}}{e}^{\lambda }}{\mathsf{\Gamma }\left(\beta \right)(1-\mathcal{A})}\\ & \le \frac{(\beta -1)e^{\lambda }}{\mathsf{\Gamma }\left(\beta \right)}+\frac{(\beta -1)\overline{\mathcal{A}}{e}^{\lambda }}{\mathsf{\Gamma }\left(\beta \right)(1-\mathcal{A})}\\ & =\vartheta ,\end{array}$$

holds. The proof is completed. □

Next, for $\mathcal{F}\in C[0,1]$, define the nonnegative part of $\mathcal{F}$ by

$${\left[\mathcal{F}\left(t\right)\right]}^{*}=\max \{\mathcal{F}\left(t\right),0\}.$$

(19)

We then consider the following modified problem:

$$\left\{\begin{array}{c}{\phi }_p\left({-}_0^R{\mathbb{D}}_t^{\beta ,\lambda }u\left(t\right)-g\left(t,{[u\left(t\right)-\xi \left(t\right)]}^{*}\right)-h\left(t\right)\right)=f\left(t,{[u\left(t\right)-\xi \left(t\right)]}^{*}\right),\hfill \\ u\left(0\right)=u^{\prime }\left(0\right)=0,u\left(1\right)={\int }_0^{1}u\left(t\right)dA\left(t\right),\hfill \end{array}\right.$$

(20)

where $\xi \left(t\right)$ is the unique solution of (10).

In what follows, we introduce a transformation to establish the relationship between Equations (20) and (1).

Lemma 5.

Suppose that $\left({\mathbf{I}}_{\mathbf{1}}\right)$–$\left({\mathbf{I}}_{\mathbf{2}}\right)$ and $\left({\mathbf{A}}_{\mathbf{0}}\right)$ hold. Let

$$x\left(t\right)=u\left(t\right)-\xi \left(t\right),$$

(21)

then the following statements are true.

(i) 

If the modified tempered Riemann–Liouville fractional Equation (20) admits a solution $u\left(t\right)$ satisfying $u\left(t\right)\ge \xi \left(t\right)$ for all $t\in [0,1]$, then $x\left(t\right)=u\left(t\right)-\xi \left(t\right)$ is a positive solution of Equation (1).

(ii) 

Any solution of the modified tempered Riemann–Liouville fractional Equation (20) can be represented by the integral equation

$$u\left(t\right)={\int }_0^{1}G(t,s)\left[f^{q-1}(s,{[u\left(s\right)-\xi \left(s\right)]}^{*})+g(s,{[u\left(s\right)-\xi \left(s\right)]}^{*})+h\left(s\right)\right]ds.$$

(22)

Proof. 

(i) Assume that u is a solution of the modified tempered Riemann–Liouville fractional Equation (20) such that $u\left(t\right)\ge \xi \left(t\right)$ for all $t\in [0,1]$. By the definition of the positive part function, Equation (20) reduces to

$$\left\{\begin{array}{c}{\phi }_p\left({-}_0^R{\mathbb{D}}_t^{\beta ,\lambda }u\left(t\right)-g(t,u\left(t\right)-\xi \left(t\right)-h\left(t\right)\right)=f(t,u\left(t\right)-\xi \left(t\right)),\hfill \\ u\left(0\right)=u^{\prime }\left(0\right)=0,u\left(1\right)={\int }_0^{1}u\left(t\right)dA\left(t\right).\hfill \end{array}\right.$$

(23)

Since $\xi$ is the unique solution of Equation (10), it satisfies

$${-}_0^R{\mathbb{D}}_t^{\beta ,\lambda }\xi \left(t\right)=h\left(t\right)$$

which will be used in the following proof.

From $\left({\mathbf{I}}_{\mathbf{2}}\right)$ and Equations (10) and (23), we have

$$\xi \left(0\right)={\xi }^{\prime }\left(0\right)=0,\xi \left(1\right)={\int }_0^1\xi \left(t\right)dA\left(t\right),$$

and

$$u\left(0\right)=u^{\prime }\left(0\right)=0,u\left(1\right)={\int }_0^{1}u\left(t\right)dA\left(t\right).$$

Therefore,

$$x\left(0\right)=u\left(0\right)-\xi \left(0\right)=0,x^{\prime }\left(0\right)=u^{\prime }\left(0\right)-{\xi }^{\prime }\left(0\right)=0,$$

and

$$x\left(1\right)=u\left(1\right)-\xi \left(1\right)={\int }_0^1[u\left(t\right)-\xi \left(t\right)]dA\left(t\right)={\int }_0^{1}x\left(t\right)dA\left(t\right).$$

On the other hand, since

$${}_0{}^R\mathbb{D}_t^{\beta ,\lambda }x\left(t\right){=}_0^R{\mathbb{D}}_t^{\beta ,\lambda }(u\left(t\right)-\xi \left(t\right)){=}_0^R{\mathbb{D}}_t^{\beta ,\lambda }u\left(t\right){-}_0^R{\mathbb{D}}_t^{\beta ,\lambda }\xi \left(t\right){=}_0^R{\mathbb{D}}_t^{\beta ,\lambda }u\left(t\right)+h\left(t\right),$$

we obtain

$$\begin{array}{cc}\hfill f(t,x(t\left)\right)& =f(t,u\left(t\right)-\xi \left(t\right))={\phi }_p\left({-}_0^R{\mathbb{D}}_t^{\beta ,\lambda }u\left(t\right)-g(t,u\left(t\right)-\xi \left(t\right)-h\left(t\right)\right)\hfill \\ & ={\phi }_p\left({-}_0^R{\mathbb{D}}_t^{\beta ,\lambda }x\left(t\right)-g(t,x\left(t\right))\right)\hfill \end{array}$$

Thus $x\left(t\right)=u\left(t\right)-\xi \left(t\right)\ge 0$ is a solution of problem (1). Hence, Equation (1) admits a positive solution.

(ii) By Lemma 4 and Assumptions $\left({\mathbf{I}}_{\mathbf{1}}\right)$–$\left({\mathbf{I}}_{\mathbf{2}}\right)$, the solution of the modified tempered Riemann–Liouville fractional Equation (20) can be written in the integral form (22). This completes the proof. □

Let $E=C\left(\right[0,1],\mathbb{R})$ be endowed with the norm

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

Define the cone

$$P=\{u\in E:u(t)\ge 0,t\in [0,1\left]\right\},$$

and introduce the subcone

$$K=\left\{u\in P:u\left(t\right)\ge \frac{1}{2(\beta -1)}\mathsf{\Phi }\left(t\right)\parallel u\parallel ,t\in [0,1]\right\}.$$

For any $r>0$, set

$$K_r=\{u\in K:\parallel u\parallel <r\},\mathsf{\partial }{K}_r=\{u\in K:\parallel u\parallel =r\},{\overline{K}}_r=\{u\in K:\parallel u\parallel \le r\}.$$

Define the nonlinear operator $T:K\setminus \left\{0\right\}\to P$ and the linear operator $L:E\to E$ by

$$\left(Tu\right)\left(t\right)={\int }_0^{1}G(t,s)\left[f^{q-1}(s,{[u\left(s\right)-\xi \left(s\right)]}^{*})+g(s,{[u\left(s\right)-\xi \left(s\right)]}^{*})+h\left(s\right)\right]ds,$$

(24)

and

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

(25)

Lemma 6

(Krein–Rutman [57]). Let $L:E\to E$ be a continuous linear operator, and let K be a cone in E such that $L\left(K\right)\subset K$. If there exist $\phi \in E\setminus (-K)$ and a constant $c>0$ satisfying

$$cL\left(\phi \right)\ge \phi ,$$

then the spectral radius $r\left(L\right)\ne 0$, and L admits a positive eigenfunction corresponding to its first eigenvalue ${\delta }_1=r{\left(L\right)}^{-1}$.

Lemma 7

(Gelfand’s formula [57]). Let L be a bounded linear operator on a Banach space with operator norm $\parallel ·\parallel$. Then the spectral radius of L satisfies

$$r\left(L\right)=\underset{n\to +\infty }{\lim }{\parallel L^n\parallel }^{\frac{1}{n}}.$$

Lemma 8.

Assume that $\left({\mathbf{A}}_{\mathbf{0}}\right)$ holds. Then the operator $L:E\to E$ defined by (25) is a completely continuous linear operator. Moreover, its spectral radius satisfies $r\left(L\right)\ne 0$, and L admits a positive eigenfunction φ corresponding to its first eigenvalue ${\delta }_1=r{\left(L\right)}^{-1}$.

Proof. 

For any $u\in K$, it follows from (7) of Lemma 2 that

$$\begin{array}{cc}\hfill \parallel Lu\parallel & =\underset{t\in [0,1]}{\max }{\int }_0^{1}G(t,s)u\left(s\right)ds\hfill \\ \hfill & =\underset{t\in [0,1]}{\max }{\int }_0^1\left(H(t,s)+\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\mathcal{H}}_A\left(s\right)\right)u\left(s\right)ds\hfill \\ & \le {\int }_0^1\frac{(\beta -1){(1-s)}^{\beta -1}s{e}^{\lambda s}}{\mathsf{\Gamma }\left(\beta \right)}u\left(s\right)ds+\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\hfill \\ & \le (\beta -1)\left({\int }_0^1\frac{{(1-s)}^{\beta -1}s{e}^{\lambda s}}{\mathsf{\Gamma }\left(\beta \right)}u\left(s\right)ds+\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\right).\hfill \end{array}$$

(26)

On the other hand, notice that

$$0<1-\mathcal{A}<1,$$

and

$$\mathsf{\Phi }\left(t\right)=(1-\mathcal{A})t^{\beta -1}(1-t)e^{-\lambda t}+\overline{\mathcal{A}}{e}^{-\lambda t}{t}^{\beta -1}\le 1,$$

by (7), (9) and (26), we also have

$$\begin{array}{l}\left(Lu\right)\left(t\right)\ge t^{\beta -1}(1-t)e^{-\lambda t}{\int }_0^1\frac{s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{\lambda s}u\left(s\right)ds+\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\\ \ge \frac{1}{2}{t}^{\beta -1}(1-t)e^{-\lambda t}{\int }_0^1\frac{s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{\lambda s}u\left(s\right)ds+\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\\ =\frac{1}{2}\left\{t^{\beta -1}(1-t)e^{-\lambda t}{\int }_0^1\frac{s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{\lambda s}u\left(s\right)ds+\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\right\}\\ +\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{2(1-\mathcal{A})}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\\ \ge \frac{1}{2}\left\{t^{\beta -1}(1-t)e^{-\lambda t}{\int }_0^1\frac{s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{\lambda s}u\left(s\right)ds+\frac{\overline{\mathcal{A}}{e}^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\int }_0^1\frac{s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{\lambda s}u\left(s\right)ds\right\}\\ +\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{2(1-\mathcal{A})}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\\ =\frac{1}{2(1-\mathcal{A})}\left((1-\mathcal{A})t^{\beta -1}(1-t)e^{-\lambda t}+\overline{\mathcal{A}}{e}^{\lambda (1-t)}{t}^{\beta -1}\right){\int }_0^1\frac{s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{\lambda s}u\left(s\right)ds\\ +\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{2(1-\mathcal{A})}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\end{array}$$

$$\begin{array}{l}\ge \frac{1}{2}\left((1-\mathcal{A})t^{\beta -1}(1-t)e^{-\lambda t}+\overline{\mathcal{A}}{e}^{\lambda (1-t)}{t}^{\beta -1}\right){\int }_0^1\frac{s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{\lambda s}u\left(s\right)ds\\ +\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{2(1-\mathcal{A})}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\\ \ge \frac{1}{2}\mathsf{\Phi }\left(t\right){\int }_0^1\frac{s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{\lambda s}u\left(s\right)ds+\frac{\mathsf{\Phi }\left(t\right)e^{\lambda (1-t)}{t}^{\beta -1}}{2(1-\mathcal{A})}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\\ =\frac{1}{2}\mathsf{\Phi }\left(t\right)\left({\int }_0^1\frac{s{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{\lambda s}u\left(s\right)ds+\frac{e^{\lambda (1-t)}{t}^{\beta -1}}{1-\mathcal{A}}{\int }_0^1{\mathcal{H}}_A\left(s\right)u\left(s\right)ds\right)\\ \ge \frac{1}{2(\beta -1)}\mathsf{\Phi }\left(t\right)\left|\right|Lu\left|\right|,\end{array}$$

(27)

which yields that $L\left(K\right)\subset K.$

Moreover, since $G(t,s)$ is uniformly continuous on $[0,1]\times [0,1]$, it follows that the operator $L:E\to E$ is completely continuous.

On the other hand, choose $u\in K$ such that $u\left(t_0\right)>0$. Since u is continuous, there exists an interval $[a,b]\subset (0,1)$ such that $t_0\in [a,b]$ and $u\left(t\right)>0$ for all $t\in [a,b]$. Moreover from Lemma 4, we have $G(t,s)>0$ for all $t,s\in [a,b]\subset (0,1)$. Therefore, for any $t\in [a,b]$, we have

$$\begin{array}{cc}\hfill \left(Lu\right)\left(t\right)& ={\int }_0^{1}G(t,s)u\left(s\right)ds\hfill \\ & \ge {\int }_a^{b}G(t,s)u\left(s\right)ds>0.\hfill \end{array}$$

Consequently, there exists a constant $c>0$ such that $c\left(Lu\right)\left(t\right)\ge u\left(t\right)$ for all $t\in [0,1]$.

By the Krein–Rutman theorem, the spectral radius $r\left(L\right)\ne 0$, and L admits a positive eigenfunction corresponding to its first eigenvalue ${\delta }_1=r{\left(L\right)}^{-1}$, that is,

$${\delta }_{1}L\phi =\phi .$$

This completes the proof. □

Lemma 9.

Assume that $\left({\mathbf{I}}_{\mathbf{1}}\right)$–$\left({\mathbf{I}}_{\mathbf{2}}\right)$ and $\left({\mathbf{A}}_{\mathbf{0}}\right)$ holds. Then the operator $T:{\overline{K}}_R\to K$ is a completely continuous operator.

Proof. 

For any $t\in [0,1]$ and $u\in {\overline{K}}_R$, there exist constants $R>r>0$ such that $\parallel u\parallel \le R$. Hence,

$$0\le {[u\left(t\right)-\xi \left(t\right)]}^{*}\le u\left(t\right)\le \parallel u\parallel \le R.$$

Define

$$S=\underset{(t,u)\in [0,1]\times [0,R]}{\max }f(t,u)\ge 0.$$

By (15) of Lemma 4, we obtain

$$\begin{array}{cc}\hfill \parallel Tu\parallel & =\underset{t\in [0,1]}{\max }{\int }_0^{1}G(t,s)\left[f^{q-1}(s,{[u\left(s\right)-\xi \left(s\right)]}^{*})+g(s,{[u\left(s\right)-\xi \left(s\right)]}^{*})+h\left(s\right)\right]ds\hfill \\ \hfill & \le {\int }_0^1\vartheta \left(s\right)\left[S^{q-1}+2h\left(s\right)\right]ds=:M<+\infty .\hfill \end{array}$$

Hence $T:{\overline{K}}_R\to K$ is well defined and maps bounded sets into bounded sets.

Next, we verify that $T\left({\overline{K}}_R\right)\subset K$. For any $u\in {\overline{K}}_R$ and $t\in [0,1]$, arguing exactly as in the proof of (26) and (27), we also have

$$\begin{array}{c}\hfill \left(Tu\right)\left(t\right)\ge \frac{1}{2(\beta -1)}\mathsf{\Phi }\left(t\right)\parallel Tu\parallel .\end{array}$$

Hence, $T\left({\overline{K}}_R\right)\subset K$.

Finally, for any $t_1,t_2\in [0,1]$ and $u\in {\overline{K}}_R$, we have

$$\begin{array}{cc}\hfill |\left(Tu\right)\left(t_1\right)-\left(Tu\right)\left(t_2\right)|& \le {\int }_0^1|G(t_1,s)-G(t_2,s)|\left[S^{q-1}+2h\left(s\right)\right]ds.\hfill \end{array}$$

Since $G(t,s)$ is uniformly continuous on $[0,1]\times [0,1]$ and $S^{q-1}+2h\in L^1(0,1)$, it follows that $T\left({\overline{K}}_R\right)$ is equicontinuous. Together with the boundedness proved above, the Arzelà–Ascoli theorem implies that $T:{\overline{K}}_R\to K$ is completely continuous. □

Lemma 10

([57]). Let E be a real Banach space and K a cone in E. Suppose that $T:{\overline{K}}_r\to K$ is a completely continuous operator. If there exists $u_0\in K\setminus \left\{\theta \right\}$ such that

$$u-Tu\ne \mu u_0,u\in \mathsf{\partial }{K}_r,\mu \ge 0,$$

then the fixed point index satisfies $i(T,K_r,K)=0$.

Lemma 11

([57]). Let E be a real Banach space and K a cone in E. Suppose that $T:{\overline{K}}_r\to K$ is a completely continuous operator. If

$$Tu\ne \mu u,u\in \mathsf{\partial }{K}_r,\mu \ge 1,$$

then the fixed point index satisfies $i(T,K_r,K)=1$.

3. Main Results

For the next step, we introduce the following assumptions:

(A1)

For every $t\in [0,1]$, the function $s\mapsto g(t,s)$ is nondecreasing on $[0,\infty )$, and there exists a constant $r>0$ such that

$$\underset{t\in [0,1]}{\inf }\left(g(t,0)+h\left(t\right)\right)\ge {\delta }_{1}r,{\int }_0^{1}h\left(s\right)ds\le \frac{r}{2(\beta -1)\mathsf{\Lambda }},$$

where ${\delta }_1$ denotes the first eigenvalue of the operator L.

(A2)

There exist constants $R_1>r$ and $0<\rho \le 1$ such that the function f is nondecreasing with respect to its second variable on $[0,1]\times [0,R_1]$, and for any $u\ge R_1$,

$$f(t,u)<{\left({\delta }_1\rho u\right)}^{\frac{1}{q-1}}.$$

Remark 5.

Although $\left(\mathbf{A}\mathbf{1}\right)$ does not cover all nonlinearities, it is satisfied by a broad class of functions arising in applications, and is sufficient for establishing our existence result.

In this section, we establish our main result of this paper.

Theorem 1.

Suppose that $\left({\mathbf{I}}_{\mathbf{1}}\right)$,$\left({\mathbf{I}}_{\mathbf{2}}\right)$ and $\left({\mathbf{A}}_{\mathbf{0}}\right)$, $\left(\mathbf{A}\mathbf{1}\right)$, $\left(\mathbf{A}\mathbf{2}\right)$ hold. Then the boundary value problem (1) admits at least one positive solution.

Proof. 

By Assumption $\left(\mathbf{A}\mathbf{1}\right)$, for all $(t,u)\in [0,1]\times [0,r]$, one has

$$g\left(t,{[u-\xi \left(t\right)]}^{*}\right)+h\left(t\right)\ge {\delta }_{1}u,$$

(28)

where ${[u-\xi \left(t\right)]}^{*}=\max \{u-\xi \left(t\right),0\}$.

Indeed, fix arbitrary $t\in [0,1]$ and $u\in [0,r]$. If $u\le \xi \left(t\right)$, then ${[u-\xi \left(t\right)]}^{*}=0$, and hence

$$g\left(t,{[u-\xi \left(t\right)]}^{*}\right)+h\left(t\right)=g(t,0)+h\left(t\right)\ge {\delta }_{1}r\ge {\delta }_{1}u.$$

If $u>\xi \left(t\right)$, let $s:=u-\xi \left(t\right)>0$. Since $g(t,·)$ is nondecreasing on $[0,\infty )$, we again obtain

$$g\left(t,{[u-\xi \left(t\right)]}^{*}\right)+h\left(t\right)\ge g(t,0)+h\left(t\right)\ge {\delta }_{1}r\ge {\delta }_{1}u.$$

On the other hand, for any $u\in \mathsf{\partial }{K}_r$, one has

$${[u\left(\tau \right)-\xi \left(\tau \right)]}^{*}\le u\left(\tau \right)\le \parallel u\parallel \le r,$$

(29)

and thus, by (28) and (29),

$$\begin{array}{cc}\hfill \left(Tu\right)\left(t\right)& ={\int }_0^{1}G(t,s)\left[f^{q-1}(s,{[u\left(s\right)-\xi \left(s\right)]}^{*})+g(s,{[u\left(s\right)-\xi \left(s\right)]}^{*})+h\left(s\right)\right]ds\hfill \\ \hfill & \ge {\int }_0^{1}G(t,s)\left[g(s,{[u\left(s\right)-\xi \left(s\right)]}^{*})+h\left(s\right)\right]ds\hfill \\ \hfill & \ge {\delta }_1{\int }_0^{1}G(t,s)u\left(s\right)ds\hfill \\ \hfill & ={\delta }_1\left(Lu\right)\left(t\right).\hfill \end{array}$$

(30)

By Lemma 8, the operator L admits a positive eigenfunction $\phi$ corresponding to ${\delta }_1$, such that $\phi ={\delta }_{1}L\phi$. In the following, we shall show that

$$u-Tu\ne \mu \phi ,u\in \mathsf{\partial }{K}_r,\mu \ge 0.$$

(31)

Indeed, if (31) were false, then there would exist $u_0\in \mathsf{\partial }{K}_r$ and ${\mu }_0\ge 0$ such that

$$u_0-T{u}_0={\mu }_0\phi .$$

If ${\mu }_0=0$, the conclusion holds trivially. Thus we may assume that ${\mu }_0>0$, which yields

$$u_0=T{u}_0+{\mu }_0\phi \ge {\mu }_0\phi .$$

Define

$$\overline{\mu }=\sup \{\mu >0:u_0\ge \mu \phi \}.$$

Then

$$\overline{\mu }\ge {\mu }_0,u_0\ge \overline{\mu }\phi ,{\delta }_{1}L{u}_0\ge {\delta }_1\overline{\mu }L\phi =\overline{\mu }\phi .$$

Combining the above inequalities with (30), we obtain

$$u_0=T{u}_0+{\mu }_0\phi \ge {\delta }_{1}L{u}_0+{\mu }_0\phi \ge \overline{\mu }\phi +{\mu }_0\phi =(\overline{\mu }+{\mu }_0)\phi ,$$

(32)

which contradicts the definition of $\overline{\mu }$. Therefore, (31) holds, and by Lemma 10 we conclude that

$$i(T,K_r,K)=0.$$

(33)

Next, define a new linear operator

$$\overline{L}u=\rho {\delta }_{1}Lu,$$

where $\rho$ is given in Assumption $\left(\mathbf{A}\mathbf{2}\right)$. Then $\overline{L}:E\to E$ is a bounded linear operator satisfying $\overline{L}\left(K\right)\subset K$. Since L admits a positive eigenfunction $\phi$ corresponding to the first eigenvalue ${\delta }_1=r{\left(L\right)}^{-1}$, that is,

$${\delta }_{1}L\phi =\phi ,$$

it follows that

$$\overline{L}\phi =\rho {\delta }_{1}L\phi =\rho \phi .$$

Hence, the spectral radius of $\overline{L}$ is $r\left(\overline{L}\right)=\rho$, and $\overline{L}$ admits the first eigenvalue $r{\left(\overline{L}\right)}^{-1}={\rho }^{-1}>1$. By Gelfand’s formula,

$$\rho =\underset{n\to +\infty }{\lim }{\parallel {\overline{L}}^n\parallel }^{\frac{1}{n}},$$

(34)

where $\parallel \overline{L}\parallel$ denotes the operator norm induced by the original norm

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

on $E=C[0,1]$, namely,

$$\parallel \overline{L}\parallel =\underset{\parallel u\parallel =1}{\sup }\parallel \overline{L}u\parallel .$$

Choose $\epsilon =\frac{3(1-\rho )}{5}>0$. Then there exists a sufficiently large integer N such that, for all $n>N$,

$$\left|\parallel {\overline{L}}^n{\parallel }^{\frac{1}{n}}-\rho \right|\le \epsilon ,\mathrm{that}\mathrm{is},\parallel {\overline{L}}^n\parallel \le {(\rho +\epsilon )}^n.$$

For any $u\in E$, define

$${\parallel u\parallel }^{*}={\displaystyle \sum _{i=1}^N}{(\rho +\epsilon )}^{N-i}\parallel {\overline{L}}^{i-1}u\parallel ,$$

(35)

where ${\overline{L}}^0=I$ denotes the identity operator. Since

$${(\rho +\epsilon )}^{N-1}\parallel u\parallel \le {\parallel u\parallel }^{*}\le \left({\displaystyle \sum _{i=1}^N}{(\rho +\epsilon )}^{N-i}{\parallel \overline{L}\parallel }^{i-1}\right)\parallel u\parallel ,$$

the norm ${\parallel ·\parallel }^{*}$ is equivalent to the original norm on E.

For later use, we introduce the constant and the function

$$S^{*}={\int }_0^1\vartheta \left(s\right)f^{q-1}(s,R_1)ds,$$

and

$$Q\left(t\right)=S^{*}+2{\int }_0^{1}G(t,s)h\left(s\right)ds,$$

where $R_1$ is given in Assumption $\left(\mathbf{A}\mathbf{2}\right)$. Let $Q^{*}={\parallel Q\parallel }^{*}$ and choose

$$R=\max \left\{R_1,\frac{3}{\epsilon {(\rho +\epsilon )}^{N-1}}{Q}^{*}\right\}.$$

Next, we show that

$$Tu\ne \mu u,u\in \mathsf{\partial }{K}_R,\mu \ge 1.$$

(36)

Suppose, on the contrary, that (36) does not hold. Then there exist $u_1\in \mathsf{\partial }{K}_R$ and ${\kappa }_1\ge 1$ such that

$$T{u}_1={\kappa }_1{u}_1.$$

Let

$$D\left(u_1\right)=\{t\in [0,1]:{[u_1\left(t\right)-\xi \left(t\right)]}^{*}>R_1\}.$$

By Assumption $\left(\mathbf{A}\mathbf{2}\right)$, for any $u\ge R_1>r$,

$$f(t,u)\le {\left(\rho {\delta }_{1}u\right)}^{\frac{1}{q-1}}.$$

Thus for any $t\in D\left(u_1\right)$, noting that ${[u_1\left(t\right)-\xi \left(t\right)]}^{*}>R_1$, as $u_1\left(t\right)\ge {[u_1\left(t\right)-\xi \left(t\right)]}^{*}$, we have

$$f(t,{[u_1\left(t\right)-\xi \left(t\right)]}^{*})\le {\left(\rho {\delta }_1{[u_1\left(t\right)-\xi \left(t\right)]}^{*}\right)}^{\frac{1}{q-1}}\le {\left(\rho {\delta }_1{u}_1\left(t\right)\right)}^{\frac{1}{q-1}}.$$

(37)

On the other hand, for any $t\in [0,1]\setminus D\left(u_1\right)$, we have ${[u_1\left(t\right)-\xi \left(t\right)]}^{*}\le R_1$. By $\left(\mathbf{A}\mathbf{2}\right)$, we know that the function f is nondecreasing with respect to its second variable on $[0,1]\times [0,R_1]$, which implies that, for any $t\in [0,1]\setminus D\left(u_1\right)$,

$$f(t,{[u_1\left(t\right)-\xi \left(t\right)]}^{*})\le f(t,R_1).$$

(38)

Consequently, it follows from Lemma 3 that (37), (38) that

$$\begin{array}{cc}\hfill & \left(T{u}_1\right)\left(t\right)={\int }_0^{1}G(t,s)\left[f^{q-1}(s,{[u_1\left(s\right)-\xi \left(s\right)]}^{*})+g(s,{[u_1\left(s\right)-\xi \left(s\right)]}^{*})+h\left(s\right)\right]ds\hfill \\ & \le {\int }_0^{1}G(t,s)\left[f^{q-1}(s,{[u_1\left(s\right)-\xi \left(s\right)]}^{*})+2h\left(s\right)\right]ds\hfill \\ & ={\int }_{D\left(u_1\right)}G(t,s)\left[f^{q-1}(s,{[u_1\left(s\right)-\xi \left(s\right)]}^{*})+2h\left(s\right)\right]ds\hfill \\ & +{\int }_{[0,1]\setminus D\left(u_1\right)}G(t,s)\left[f^{q-1}(s,{[u_1\left(s\right)-\xi \left(s\right)]}^{*})+2h\left(s\right)\right]ds\hfill \\ & \le {\int }_{D\left(u_1\right)}G(t,s)f^{q-1}(s,{[u_1\left(s\right)-\xi \left(s\right)]}^{*})ds\hfill \\ & +{\int }_{[0,1]\setminus D\left(u_1\right)}G(t,s)f^{q-1}(s,R_1)ds+2{\int }_0^{1}G(t,s)h\left(s\right)ds\hfill \\ & \le {\int }_{D\left(u_1\right)}G(t,s)f^{q-1}(s,{[u_1\left(s\right)-\xi \left(s\right)]}^{*})ds\hfill \\ & +{\int }_0^1\vartheta \left(s\right)f^{q-1}(s,R_1)ds+2{\int }_0^{1}G(t,s)h\left(s\right)ds\hfill \\ & ={\int }_{D\left(u_1\right)}G(t,s)f^{q-1}(s,{[u_1\left(s\right)-\xi \left(s\right)]}^{*})ds+\left[S^{*}+2{\int }_0^{1}G(t,s)h\left(s\right)ds\right]\hfill \\ & \le {\int }_{D\left(u_1\right)}G(t,s){\left({\left(\rho {\delta }_1{u}_1\left(s\right)\right)}^{\frac{1}{q-1}}\right)}^{q-1}ds+\left[S^{*}+2{\int }_0^{1}G(t,s)h\left(s\right)ds\right]\hfill \\ & \le \rho {\delta }_1{\int }_{D\left(u_1\right)}G(t,s)u_1\left(s\right)ds+Q\left(t\right)\le \rho {\delta }_1\left(L{u}_1\right)\left(t\right)+Q\left(t\right)\hfill \\ & =\left(\overline{L}{u}_1\right)\left(t\right)+Q\left(t\right),\hfill \end{array}$$

(39)

where $\overline{L}u=\rho {\delta }_{1}Lu$.

Since $\overline{L}:K\to K$ is a bounded linear operator, it follows from (39) that

$$0\le \left({\overline{L}}^j\left(T{u}_1\right)\right)\left(t\right)\le \left({\overline{L}}^j(\overline{L}{u}_1+Q)\right)\left(t\right),j=0,1,\dots ,N-1,$$

(40)

which implies

$$\parallel {\overline{L}}^j\left(T{u}_1\right)\parallel \le \parallel {\overline{L}}^j(\overline{L}{u}_1+Q)\parallel ,j=0,1,\dots ,N-1.$$

Therefore,

$$\begin{array}{ll}\parallel T{u}_1{\parallel }^{*}& ={\displaystyle \sum _{i=1}^N}{(\rho +\epsilon )}^{N-i}\parallel {\overline{L}}^{i-1}\left(T{u}_1\right)\parallel \\ & \le {\displaystyle \sum _{i=1}^N}{(\rho +\epsilon )}^{N-i}\parallel {\overline{L}}^{i-1}(\overline{L}{u}_1+Q)\parallel \\ & =\parallel \overline{L}{u}_1+{Q\parallel }^{*}.\end{array}$$

(41)

Since $u_1\in \mathsf{\partial }{K}_R$ and $\parallel u_1\parallel =R$, by (35), we have

$$\parallel u_1{\parallel }^{*}>{(\rho +\epsilon )}^{N-1}R>\frac{3}{\epsilon }{Q}^{*},$$

which yields

$$Q^{*}<\frac{\epsilon }{3}{\parallel u_1\parallel }^{*}.$$

(42)

Combining (35), (41), (42) and $\epsilon =\frac{3(1-\rho )}{5}$, we obtain

$$\begin{array}{ll}{\kappa }_1\parallel u_1{\parallel }^{*}& =\parallel T{u}_1{\parallel }^{*}\le \parallel \overline{L}{u}_1{\parallel }^{*}+Q^{*}={\displaystyle \sum _{i=1}^N}{[\rho +\epsilon ]}^{N-i}\parallel {\overline{L}}^i{u}_1\parallel +Q^{*}\\ & =[\rho +\epsilon ]{\displaystyle \sum _{i=1}^{N-1}}{[\rho +\epsilon ]}^{N-i-1}\parallel {\overline{L}}^i{u}_1\parallel +\parallel {\overline{L}}^N{u}_1\parallel +Q^{*}\\ & \le [\rho +\epsilon ]{\displaystyle \sum _{i=1}^{N-1}}{[\rho +\epsilon ]}^{N-i-1}\parallel {\overline{L}}^i{u}_1\parallel +\parallel {\overline{L}}^N\parallel \parallel u_1\parallel +Q^{*}\\ & \le [\rho +\epsilon ]{\displaystyle \sum _{i=1}^{N-1}}{[\rho +\epsilon ]}^{N-i-1}\parallel {\overline{L}}^i{u}_1{\parallel +[\rho +\epsilon ]}^N\parallel u_1\parallel +Q^{*}\\ & =[\rho +\epsilon ]{\displaystyle \sum _{i=1}^N}{[\rho +\epsilon ]}^{N-i}\parallel {\overline{L}}^{i-1}{u}_1\parallel +Q^{*}=[\rho +\epsilon ]\parallel u_1{\parallel }^{*}+Q^{*}\\ & \le \left(\rho +\epsilon +\frac{\epsilon }{3}\right)\parallel u_1{\parallel }^{*}=\left(\frac{4}{5}+\frac{\rho }{5}\right){\parallel u_1\parallel }^{*}.\end{array}$$

(43)

Since ${\kappa }_1\ge 1$, it follows that $\frac{4}{5}+\frac{\rho }{5}\ge 1$, which implies $\rho \ge 1$, contradicting $0<\rho <1$. Hence, (36) holds. By Lemma 11,

$$i(T,K_R,K)=1.$$

(44)

Combining (33) and (44), we conclude that

$$i(T,K_R\setminus {\overline{K}}_r,K)=i(T,K_R,K)-i(T,K_r,K)=1.$$

(45)

Therefore, T admits at least one fixed point $u\left(t\right)$ in $K_R\setminus {\overline{K}}_r$ with $r\le \left|\right|u\left|\right|\le R$.

Next we prove that $u\left(t\right)\ge \xi \left(t\right),t\in [0,1].$ In fact, for any $t\in [0,1]$, by definition of cone K, (12) of Lemma 4 and $\left(\mathbf{A}\mathbf{1}\right)$, we have

$$\begin{array}{cc}\hfill u\left(t\right)-\xi \left(t\right)& \ge u\left(t\right)-\mathsf{\Lambda }\mathsf{\Phi }\left(t\right){\int }_0^{1}h\left(s\right)ds\hfill \\ \hfill & \ge \frac{r}{2(\beta -1)}\mathsf{\Phi }\left(t\right)-\mathsf{\Lambda }\mathsf{\Phi }\left(t\right){\int }_0^{1}h\left(s\right)ds\hfill \\ \hfill & =\mathsf{\Phi }\left(t\right)\left(\frac{r}{2(\beta -1)}-\mathsf{\Lambda }{\int }_0^{1}h\left(s\right)ds\right)\hfill \\ \hfill & \ge 0,\hfill \end{array}$$

(46)

which implies that $u\left(t\right)\ge \xi \left(t\right)$ for any $t\in [0,1]$.

Consequently, the boundary value problem (20) has at least one solution $u\left(t\right)\ge \xi \left(t\right)$ for any $t\in [0,1]$, and hence it follows from Lemma 5 that the boundary value problem (1) admits at least one positive solution. □

In the following, we study the case of boundary degeneration of the boundary value problem (1):

$$\left\{\begin{array}{c}{\phi }_p\left(-{}_0{}^R\mathbb{D}_t^{\beta ,\lambda }x\left(t\right)-g(t,x\left(t\right))\right)=f(t,x\left(t\right)),\hfill \\ x\left(0\right)=x^{\prime }\left(0\right)=0,x\left(1\right)=0.\hfill \end{array}\right.$$

(47)

Remark 6.

In this case, the Green function corresponding to the degenerate boundary condition is given by $H(t,s)$:

$$H(t,s)=\left\{\begin{array}{cc}\frac{t^{\beta -1}{(1-s)}^{\beta -1}-{(t-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s},\hfill & 0\le s\le t\le 1,\hfill \\ \frac{t^{\beta -1}{(1-s)}^{\beta -1}}{\mathsf{\Gamma }\left(\beta \right)}{e}^{-\lambda t}{e}^{\lambda s},\hfill & 0\le t\le s\le 1,\hfill \end{array}\right.$$

(48)

which has an estimate (7).

For the degenerate boundary value problem (47), we can obtain the following existence result of positive solutions.

Corollary 1.

Suppose that $\left({\mathbf{I}}_{\mathbf{1}}\right)$,$\left({\mathbf{I}}_{\mathbf{2}}\right)$ and $\left(\mathbf{A}\mathbf{2}\right)$ hold. If

(A*1)

For every $t\in [0,1]$, the function $s\mapsto g(t,s)$ is nondecreasing on $[0,\infty )$, and there exists a constant

$$0<{\int }_0^{1}h\left(s\right)ds<r$$

such that

$$\underset{t\in [0,1]}{\inf }\left(g(t,0)+h\left(t\right)\right)\ge {\delta }_{1}r,$$

where ${\delta }_1$ denotes the first eigenvalue of the operator L.

Then the boundary value problem (47) admits at least one positive solution.

Proof. 

Firstly, introduce a subcone of P

$$K^{*}=\left\{u\in P:u\left(t\right)\ge \frac{(\beta -1)t^{\beta -1}(1-t)e^{\lambda (1-t)}}{\mathsf{\Gamma }\left(\beta \right)}\parallel u\parallel ,t\in [0,1]\right\},$$

it follows from (7) that the unique positive solution of the fractional boundary value problem satisfies

$$\xi \left(t\right)={\int }_0^{1}H(t,s)h\left(s\right)ds\le \frac{(\beta -1)t^{\beta -1}(1-t)e^{\lambda (1-t)}}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^{1}h\left(s\right)ds.$$

(49)

Thus by Theorem 1, T admits at least one fixed point in $K_R\setminus {\overline{K}}_r$. Consequently, the boundary value problem (47) has at least one solution. In what follows, we show this solution is also positive.

In fact, since $r\le \left|\right|u\left|\right|\le R,$ it follows from (49) and the property of cone that

$$\begin{array}{ll}u\left(t\right)-\xi \left(t\right)& \ge \frac{(\beta -1)t^{\beta -1}(1-t)e^{\lambda (1-t)}}{\mathsf{\Gamma }\left(\beta \right)}\left|\right|u\left|\right|-\frac{(\beta -1)t^{\beta -1}(1-t)e^{\lambda (1-t)}}{\mathsf{\Gamma }\left(\beta \right)}{\int }_0^{1}h\left(s\right)ds\\ & \ge (\beta -1)\left(r-{\int }_0^{1}h\left(s\right)ds\right)\frac{t^{\beta -1}(1-t)e^{\lambda (1-t)}}{\mathsf{\Gamma }\left(\beta \right)}.\end{array}$$

(50)

Since

$$0<{\int }_0^{1}h\left(s\right)ds<r,$$

we have

$$r-{\int }_0^{1}h\left(s\right)ds>0.$$

Moreover,

$$\frac{t^{\beta -1}(1-t)e^{\lambda (1-t)}}{\mathsf{\Gamma }\left(\beta \right)}>0,t\in (0,1).$$

Hence,

$$u\left(t\right)-\xi \left(t\right)>0,t\in (0,1).$$

By Lemma 5, the solution $x\left(t\right)=u\left(t\right)-\xi \left(t\right)$ of Equation (47) is positive.

□

4. Example

Example 1.

We consider the following 3-Laplacian fractional diffusion equation with a sign-changing perturbation:

$$\left\{\begin{array}{c}{\phi }_3\left({-}_0^R{\mathbb{D}}_t^{\frac{7}{3},\frac{1}{1000}}x\left(t\right)-g(t,x\left(t\right))\right)=f(t,x\left(t\right)),\hfill \\ x\left(0\right)=x^{\prime }\left(0\right)=0,x\left(1\right)={\int }_0^{1}x\left(t\right)dA\left(t\right),\hfill \end{array}\right.$$

(51)

where

$$f(t,x)=\left\{\begin{array}{cc}\frac{1}{100}(1+\sin t)x,\hfill & 0\le x\le 1,\hfill \\ \frac{1}{100}(1+\sin t),\hfill & x\ge 1,\hfill \end{array}\right.$$

and

$$g(t,x)=1+\frac{1}{10}\arctan x-\frac{1}{10000|t-\frac{1}{2}{|}^{1/2}},$$

and

$$A\left(t\right)=\left\{\begin{array}{cc}0,\hfill & 0\le t<\frac{1}{4},\hfill \\ -\frac{1}{100},\hfill & \frac{1}{4}\le t<\frac{3}{4},\hfill \\ \frac{1}{20},\hfill & \frac{3}{4}\le t\le 1.\hfill \end{array}\right.$$

Then the 3-Laplacian fractional-order diffusion equation with a sign-changing perturbation admits at least one positive solution.

Remark 7.

The function $g(t,x)$ becomes singular at $t=\frac{1}{2}$ and changes sign in the variable t due to the singular negative perturbation term. Moreover, A is a function of bounded variation and the induced measure $dA$ is a signed measure.

Proof. 

Clearly,

$$f(t,x)\ge 0$$

and f is continuous on $[0,1]\times [0,+\infty )$. Hence $\left({\mathbf{I}}_{\mathbf{1}}\right)$ holds.

Take

$$p=3,q=\frac{3}{2},\beta =\frac{7}{3},\lambda =\frac{1}{1000}.$$

Define

$$h\left(t\right)=1+\frac{\pi }{20}+\frac{1}{10000|t-\frac{1}{2}{|}^{1/2}}.$$

Since

$$0\le \frac{1}{10}\arctan x\le \frac{\pi }{20},$$

we have

$$-h\left(t\right)\le g(t,x)\le h\left(t\right).$$

Moreover,

$$\begin{array}{cc}\hfill {\int }_0^{1}h\left(t\right)dt& =1+\frac{\pi }{20}+\frac{1}{10000}{\int }_0^1{|t-\frac{1}{2}|}^{-1/2}dt\hfill \\ & =1+\frac{\pi }{20}+\frac{2\sqrt{2}}{10000}\hfill \\ & \approx 1.15736.\hfill \end{array}$$

Thus $\left({\mathbf{I}}_{\mathbf{2}}\right)$ is satisfied.

For

$$A\left(t\right)=\left\{\begin{array}{cc}0,\hfill & 0\le t<\frac{1}{4},\hfill \\ -\frac{1}{100},\hfill & \frac{1}{4}\le t<\frac{3}{4},\hfill \\ \frac{1}{20},\hfill & \frac{3}{4}\le t\le 1,\hfill \end{array}\right.$$

take

$$p=3,q=\frac{3}{2},\beta =\frac{7}{3},\lambda =\frac{1}{1000}.$$

Since $A\left(t\right)$ has jumps

$$\Delta A\left(\frac{1}{4}\right)=-\frac{1}{100},\Delta A\left(\frac{3}{4}\right)=\frac{6}{100},$$

the Riemann–Stieltjes integrals are given by

$$\begin{array}{cc}\hfill \mathcal{A}& ={\int }_0^1{e}^{\lambda (1-t)}{t}^{\beta -1}dA\left(t\right)\hfill \\ & =-\frac{1}{100}{e}^{\frac{3}{4000}}{\left(\frac{1}{4}\right)}^{\frac{4}{3}}+\frac{6}{100}{e}^{\frac{1}{4000}}{\left(\frac{3}{4}\right)}^{\frac{4}{3}}\hfill \\ & \approx 0.039319353,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \overline{\mathcal{A}}& ={\int }_0^1{e}^{\lambda (1-t)}{t}^{\beta -1}(1-t)dA\left(t\right)\hfill \\ & =-\frac{3}{400}{e}^{\frac{3}{4000}}{\left(\frac{1}{4}\right)}^{\frac{4}{3}}+\frac{3}{200}{e}^{\frac{1}{4000}}{\left(\frac{3}{4}\right)}^{\frac{4}{3}}\hfill \\ & \approx 0.009041797.\hfill \end{array}$$

Hence,

$$0<\overline{\mathcal{A}}<\mathcal{A}<1.$$

Furthermore,

$${\mathcal{H}}_A\left(s\right)=\frac{1}{100}\left(6H(\frac{3}{4},s)-H(\frac{1}{4},s)\right).$$

Using a uniform mesh with $N={10}^5$ points, we obtain

$$\underset{s\in [0,1]}{\min }{\mathcal{H}}_A\left(s\right)\approx 0,$$

where the minimum is attained at $s=0$. Thus the numerical verification shows that

$${\mathcal{H}}_A\left(s\right)\ge 0,s\in [0,1].$$

Therefore, $\left({\mathbf{A}}_{\mathbf{0}}\right)$ holds.

Next, we verify $\left(\mathbf{A}\mathbf{1}\right)$. We have

$$g_x(t,x)=\frac{1}{10(1+x^2)}>0,$$

so g is nondecreasing with respect to x.

Moreover,

$$g(t,0)+h\left(t\right)=2+\frac{\pi }{20}.$$

The spectral radius of the positive compact operator L is computed by discretizing the Green operator with the composite trapezoidal rule and applying the power iteration method:

$$r\left(L\right)\approx 2.160.$$

Hence

$${\delta }_1=\frac{1}{r\left(L\right)}\approx 0.463.$$

Choose

$$r=4.$$

Then

$${\delta }_{1}r=0.463\times 4=1.852,$$

and

$$2+\frac{\pi }{20}=2.15708>1.852.$$

Thus

$$\underset{t\in [0,1]}{\inf }(g(t,0)+h\left(t\right))\ge {\delta }_{1}r.$$

Furthermore,

$$\mathsf{\Lambda }=\frac{(\beta -1)e^{\lambda }}{\mathsf{\Gamma }\left(\beta \right)(1-\mathcal{A})}\approx 1.167.$$

Therefore,

$$\frac{r}{2(\beta -1)\mathsf{\Lambda }}=\frac{4}{2\times \frac{4}{3}\times 1.167}\approx 1.285.$$

Since

$${\int }_0^{1}h\left(t\right)dt\approx 1.15736<1.285,$$

we obtain

$${\int }_0^{1}h\left(t\right)dt<\frac{r}{2(\beta -1)\mathsf{\Lambda }}.$$

Hence $\left(\mathbf{A}\mathbf{1}\right)$ is fulfilled.

Finally, we verify $\left(\mathbf{A}\mathbf{2}\right)$.

Take

$$R_1=5,\rho =0.8.$$

Clearly,

$$R_1>r.$$

Moreover, f is nondecreasing with respect to x on $[0,1]\times [0,R_1]$.

For $u\ge R_1=5$,

$$f(t,u)\le 0.02.$$

Since

$$\frac{1}{q-1}=2,$$

we have

$${\left({\delta }_1\rho u\right)}^{\frac{1}{q-1}}={\left({\delta }_1\rho u\right)}^2.$$

Thus,

$${\left({\delta }_1\rho u\right)}^2\ge {(0.463\times 0.8\times 5)}^2>3.43.$$

Consequently,

$$f(t,u)<{\left({\delta }_1\rho u\right)}^{\frac{1}{q-1}},u\ge 5.$$

Therefore, $\left(\mathbf{A}\mathbf{2}\right)$ holds.

Therefore, Assumptions $\left({\mathbf{I}}_{\mathbf{1}}\right)$,$\left({\mathbf{I}}_{\mathbf{2}}\right)$, $\left({\mathbf{A}}_{\mathbf{0}}\right)$, $\left(\mathbf{A}\mathbf{1}\right)$ and $\left(\mathbf{A}\mathbf{2}\right)$ all hold; hence, by Theorem 1, the tempered Riemann–Liouville fractional p-Laplacian boundary value problem (51) admits at least one positive solution. □

Example 2.

Consider the degenerate p-Laplacian tempered Riemann–Liouville fractional diffusion boundary value problem

$$\left\{\begin{array}{c}{\phi }_2\left({-}_0^R{\mathbb{D}}_t^{\frac{7}{3},\frac{1}{1000}}x\left(t\right)-g(t,x\left(t\right))\right)=f(t,x\left(t\right)),\hfill \\ x\left(0\right)=x^{\prime }\left(0\right)=0,x\left(1\right)=0,\hfill \end{array}\right.$$

(52)

where

$$p=2,q=2,\beta =\frac{7}{3},\lambda =\frac{1}{1000},$$

and

$$f(t,x)=\left\{\begin{array}{cc}(1+t)x,\hfill & 0\le x\le 2,\hfill \\ 2(1+t),\hfill & x\ge 2,\hfill \end{array}\right.g(t,x)=\frac{1}{2}\arctan x-\frac{1}{1000|t-\frac{1}{3}{|}^{1/3}}.$$

Then problem (52) admits at least one positive solution.

Proof. 

Clearly,

$$f(t,x)\ge 0$$

and f is continuous on $[0,1]\times [0,+\infty )$. Hence condition $\left({\mathbf{I}}_{\mathbf{1}}\right)$ holds.

Define

$$h\left(t\right)=\frac{\pi }{4}+\frac{1}{1000|t-\frac{1}{3}{|}^{1/3}}.$$

Since

$$-\frac{\pi }{4}\le \frac{1}{2}\arctan x\le \frac{\pi }{4},$$

we have

$$-h\left(t\right)\le g(t,x)\le h\left(t\right).$$

Moreover,

$$\begin{array}{cc}\hfill {\int }_0^{1}h\left(t\right)dt& =\frac{\pi }{4}+\frac{1}{1000}{\int }_0^1{|t-\frac{1}{3}|}^{-1/3}dt\hfill \\ & =\frac{\pi }{4}+\frac{1}{1000}\frac{3}{2}\left[{\left(\frac{1}{3}\right)}^{2/3}+{\left(\frac{2}{3}\right)}^{2/3}\right]\hfill \\ & \approx 0.7869.\hfill \end{array}$$

Therefore, condition $\left({\mathbf{I}}_{\mathbf{2}}\right)$ holds.

Since the boundary condition is the degenerate case of the Riemann–Stieltjes boundary condition with

$$A\left(t\right)\equiv 0,$$

the corresponding Green operator is

$$\left(Lu\right)\left(t\right)={\int }_0^{1}H(t,s)u\left(s\right)ds,$$

where $H(t,s)$ is given by (48).

The spectral radius of the positive compact operator L is computed by the composite trapezoidal rule and the power iteration method. We obtain

$$r\left(L\right)\approx 2.160.$$

Consequently, the first eigenvalue satisfies

$${\delta }_1=\frac{1}{r\left(L\right)}\approx 0.463.$$

Next, we verify condition $\left({\mathbf{A}}^{*}\mathbf{1}\right)$. Since

$$g_x(t,x)=\frac{1}{2(1+x^2)}>0,$$

the function $g(t,x)$ is nondecreasing with respect to $x\in [0,+\infty )$.

Furthermore,

$$g(t,0)+h\left(t\right)=\frac{\pi }{4},$$

and hence

$$\underset{t\in [0,1]}{\inf }(g(t,0)+h\left(t\right))=\frac{\pi }{4}=0.7854.$$

Choose

$$r=1.5.$$

Then

$${\int }_0^{1}h\left(t\right)dt\approx 0.7869<1.5,$$

and

$${\delta }_{1}r=0.463\times 1.5=0.6945<0.7854=\underset{t\in [0,1]}{\inf }(g(t,0)+h\left(t\right)).$$

Therefore, $\left({\mathbf{A}}^{*}\mathbf{1}\right)$ is satisfied.

Finally, we verify $\left(\mathbf{A}\mathbf{2}\right)$. Choose

$$\rho =0.8,R_1=11.$$

For $u\ge 11$, we have

$$f(t,u)=2(1+t)\le 4.$$

Since

$$q=2,$$

we have

$$\frac{1}{q-1}=1.$$

Thus,

$${\left({\delta }_1\rho u\right)}^{\frac{1}{q-1}}=0.463\times 0.8u.$$

For $u\ge 11$,

$$0.463\times 0.8u\ge 4.0744>4.$$

Therefore,

$$f(t,u)<{\left({\delta }_1\rho u\right)}^{\frac{1}{q-1}},u\ge R_1.$$

Hence condition $\left(\mathbf{A}\mathbf{2}\right)$ holds.

In summary, Assumptions $\left({\mathbf{I}}_{\mathbf{1}}\right)$,$\left({\mathbf{I}}_{\mathbf{2}}\right)$, $\left({\mathbf{A}}^{*}\mathbf{1}\right)$ and $\left(\mathbf{A}\mathbf{2}\right)$ hold; hence, by Corollary 1, the degenerate tempered Riemann–Liouville fractional p-Laplacian boundary (52) admits at least one positive solution $x\in C[0,1]$. □

5. Conclusions

In this paper, we present a comprehensive study of positive solutions to tempered Riemann–Liouville fractional equations with sign-changing perturbations, Riemann–Stieltjes integral boundary conditions, and p-Laplacian operators, using spectral analysis as the primary tool. Motivated by the pivotal role of tempered Riemann–Liouville fractional diffusion equations in advanced applications and the limitations of classical integer-order models in describing complex phenomena, this work addresses the mathematical difficulties arising from the coexistence of sign-changing perturbations and signed measures. By combining functional analysis, fixed point theory, fractional calculus, and spectral analysis of the associated linear operator, we establish new existence results for positive solutions under suitable assumptions.

It should be pointed out that the present approach has certain limitations. Our analysis relies essentially on the spectral properties of the associated compact operator, the Krein–Rutman theorem, and the construction of an appropriate Green function. These techniques are developed for one-dimensional tempered Riemann–Liouville fractional boundary value problems and cannot be directly extended to higher-dimensional fractional models or coupled systems. In particular, the loss of positivity caused by signed measures and sign-changing perturbations makes the construction of suitable Green functions and the spectral analysis of the corresponding operators substantially more difficult.

Future work will focus on extending the present framework to more general boundary conditions, higher-dimensional fractional problems, and coupled systems of tempered Riemann–Liouville fractional equations. Developing new analytical techniques to overcome the lack of positivity and to handle more general nonlinear couplings remains an interesting and challenging direction for future research. Such extensions are expected to broaden the applicability of the proposed approach in mathematical physics, engineering, and related fields.