## 1 Introduction

Differential equations of fractional order have been recently proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, porous media, electromagnetism, etc. (see [15]). There has been a significant development in the study of fractional differential equations in recent years, see the monographs of Kilbas et al. [6], Lakshmikantham et al. [7], Podlubny [4], Samko et al. [8], and the survey by Agarwal et al. [9].

For some recent contributions on fractional differential equations, see for example, [1028] and the references therein. Especially, in [15], by means of Guo-Krasnosel'skiĭ's fixed point theorem, Zhao et al. investigated the existence of positive solutions for the nonlinear fractional boundary value problem (BVP for short)

$\left\{\begin{array}{cc}{D}_{0+}^{\alpha }u\left(t\right)=\lambda f\left(u\left(t\right)\right),\hfill & t\in \left(0,1\right),\hfill \\ u\left(0\right)+{u}^{\prime }\left(0\right)=0,\hfill & u\left(1\right)+{u}^{\prime }\left(1\right)=0,\hfill \end{array}\right\$
(1.1)

where 1 < α ≤ 2, f : [0, +∞) → (0, +∞).

In [16], relying on the Krasnosel'skiĭ's fixed point theorem as well as on the Leggett-Williams fixed point theorem, Kaufmann and Mboumi discussed the existence of positive solutions for the following fractional BVP

$\left\{\begin{array}{c}{D}_{0+}^{\alpha }u\left(t\right)+a\left(t\right)f\left(u\left(t\right)\right)=0,\phantom{\rule{1em}{0ex}}0

In [17], by applying Altman's fixed point theorem and Leray-Schauder' fixed point theorem, Wang obtained the existence and uniqueness of solutions for the following BVP of nonlinear impulsive differential equations of fractional order q

$\left\{\begin{array}{ccc}{}^{c}{D}^{q}u\left(t\right)=f\left(t,u\left(t\right)\right),\hfill & 1

In [18], relying on the contraction mapping principle and the Krasnosel'skiĭ's fixed point theorem, Zhou and Chu discussed the existence of solutions for a nonlinear multi-point BVP of integro-differential equations of fractional order q ∈ (1, 2]

$\left\{\begin{array}{c}{}^{c}{D}_{0+}^{q}u\left(t\right)=f\left(t,u\left(t\right),\left(Ku\right)\left(t\right),\left(Hu\right)\left(t\right)\right),\phantom{\rule{1em}{0ex}}1

On the other hand, integer-order p-Laplacian boundary value problems have been widely studied owing to its importance in theory and application of mathematics and physics, see for example, [2933] and the references therein. Especially, in [29], by using the fixed point index method, Yang and Yan investigated the existence of positive solution for the third-order Sturm-Liouville boundary value problems with p-Laplacian operator

$\left\{\begin{array}{l}\left({\varphi }_{p}\left({u}^{″}\left(t\right){\right)}^{\prime }+f\left(t,u\left(t\right)\right)=0,t\left(0,1\right),\hfill \\ au\left(0\right)-b{u}^{\prime }\left(0\right)=0cu\left(1\right)+{u}^{\prime }\left(1\right)=0,{u}^{″}\left(0\right)=0,\hfill \end{array}$
(1.2)

where φ p (s) = |s|p-2s.

However, there are few articles dealing with the existence of solutions to boundary value problems for fractional differential equation with p-Laplacian operator. In [24], the authors investigated the nonlinear nonlocal problem

$\left\{\begin{array}{c}{D}_{0+}^{\beta }\left({\phi }_{p}\left({D}_{0+}^{\alpha }u\right)\right)\left(t\right)+f\left(t,u\left(t\right)\right)=0,\phantom{\rule{1em}{0ex}}0
(1.3)

where 0 < β ≤ 1, 1 < α ≤ 2, 0 ≤ a ≤ 1, 0 < ξ < 1. By using Krasnosel'skiĭ's fixed point theorem and Leggett-Williams theorem, some sufficient conditions for the existence of positive solutions to the above BVP are obtained.

In [25], by using upper and lower solutions method, under suitable monotone conditions, the authors investigated the existence of positive solutions to the following nonlocal problem

$\left\{\begin{array}{c}{D}_{0+}^{\beta }\left({\phi }_{p}\left({D}_{0+}^{\alpha }u\right)\right)\left(t\right)+f\left(t,u\left(t\right)\right),\phantom{\rule{1em}{0ex}}0
(1.4)

where 1 < α, β ≤ 2, 0 ≤ a, b ≤ 1, 0 < ξ, η < 1.

No contribution exists, as far as we know, concerning the existence of solutions for the fractional differential equation with p-Laplacian operator

$\left\{\begin{array}{cc}{D}_{0+}^{\beta }\left({\phi }_{p}\left({D}_{0+}^{\alpha }u\right)\right)\left(t\right)+f\left(t,u\left(t\right)\right)=0,\hfill & 0
(1.5)

where ${D}_{0+}^{\beta },{D}_{0+}^{\alpha }$ and ${D}_{0+}^{\gamma }$ are the standard Riemann-Liouville derivative with 1 < α ≤ 2, 0 < β ≤ 1, 0 < γ ≤ 1, 0 ≤ α - γ - 1, the constant σ is a positive number, the p-Laplacian operator is defined as φ p (s) = |s|p-2s, p > 1, and function f is assumed to satisfy certain conditions, which will be specified later. To obtain the existence and multiplicity of positive solutions to BVP (1.5), the fixed point theorem on cones will be applied.

It is worth emphasizing that our work presented in this article has the following features which are different from those in [24, 25]. Firstly, BVP (1.5) is an important two point BVP. When γ = 1, the boundary value conditions in (1.5) reduce to u(0) = 0, u(1) + σu'(1) = 0, which are the well-known Sturm-Liouville boundary value conditions u(0) + bu'(0) = 0, u(1) + σu'(1) = 0 (such as BVP (1.1)) with b = 0. It is a well-known fact that the boundary value problems with Sturm-Liouville boundary value conditions for integral order differential equations have important physical and applied background and have been studied in many literatures, while BVPs (1.3) and (1.4) are the nonlocal boundary value problems, which are not able to substitute BVP (1.5). Secondly, when α = 2, β = 1, γ = 1, then BVP (1.5) reduces to BVP (1.2) with b = 0. So, BVP (1.5) is an important generalization of BVP (1.2) from integral order to fractional order. Thirdly, in BVPs (1.3) or (1.4), the boundary value conditions u(1) = au(ξ), ${D}_{0+}^{\alpha }u\left(1\right)=b{D}_{0+}^{\alpha }u\left(\eta \right)$ show the relations between the derivatives of same order ${D}_{0+}^{\mu }u\left(1\right)$ and ${D}_{0+}^{\mu }u\left(\zeta \right)\left(\mu =0,\alpha \right)$. By contrast with that, the condition $u\left(1\right)+\sigma {D}_{0+}^{\gamma }u\left(1\right)=0$ in BVP (1.5) shows that relation between the derivatives of different order u(1) and ${D}_{0+}^{\gamma }u\left(1\right)$ is regarded as the derivative value of zero order of u at t = 1), which brings about more difficulties in deducing the property of green's function than the former. Finally, order α + β satisfies that 2 < α + β ≤ 4 in BVP (1.4), while order α + β satisfies that 1 < α + β ≤ 3 in BVP (1.5). In the case for α, β taking integral numbers, the BVPs (1.5) and (1.4) are the third-order BVP and the fourth-order BVP, respectively. So, BVP (1.5) differs essentially from BVP (1.4). In addition, the conditions imposed in present paper are easily verified.

The organization of this article is as follows. In Section 2, we present some necessary definitions and preliminary results that will be used to prove our main results. In Section 3, we put forward and prove our main results. Finally, we will give two examples to demonstrate our main results.

## 2 Preliminaries

In this section, we introduce some preliminary facts which are used throughout this article.

Let ℕ be the set of positive integers, be the set of real numbers and + be the set of nonnegative real numbers. Let I = [0, 1]. Denote by C(I, ) the Banach space of all continuous functions from I into with the norm

$∥u∥=\text{max}\left\{\left|u\left(t\right)\right|:t\in I\right\}.$

Define the cone P in C(I, ) as P = {uC(I, ): u(t) ≥ 0, tI}. Let q > 1 satisfy the relation $\frac{1}{q}+\frac{1}{p}=1$, where p is given by (1. 5).

Definition 2.1. [6] The Riemann-Liouville fractional integral of order α > 0 of a function y : (a, b] → is given by

${I}_{a+}^{\alpha }y\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}\underset{a}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -1}y\left(s\right)ds,\phantom{\rule{1em}{0ex}}t\in \left(a,b\right].$

Definition 2.2. [6] The Riemann-Liouville fractional derivative of order α > 0 of function y : (a, b] → is given by

${D}_{a+}^{\alpha }y\left(t\right)=\frac{1}{\Gamma \left(n-\alpha \right)}{\left(\frac{d}{dt}\right)}^{n}\underset{a}{\overset{t}{\int }}\frac{y\left(s\right)}{{\left(t-s\right)}^{\alpha -n+1}}ds,\phantom{\rule{1em}{0ex}}t\in \left(a,b\right],$

where n = [α] + 1 and [α] denotes the integer part of α.

Lemma 2.1. [34] Let α > 0. If uC(0, 1) ⋂ L(0, 1) possesses a fractional derivative of order α that belongs to C(0, 1) ⋂ L(0, 1), then

${I}_{0+}^{\alpha }{D}_{0+}^{\alpha }u\left(t\right)=u\left(t\right)+{c}_{1}{t}^{\alpha -1}+{c}_{2}{t}^{\alpha -2}+\cdots +{c}_{n}{t}^{\alpha -n},$

for some c i , i = 1, 2,..., n, where n = [α] + 1.

A function uC(I, ) is called a nonnegative solution of BVP (1.5), if u ≥ 0 on [0, 1] and satisfies (1.5). Moreover, if u(t) > 0, t ∈ (0, 1), then u is said to be a positive solution of BVP (1.5).

For forthcoming analysis, we first consider the following fractional differential equation

$\left\{\begin{array}{c}{D}_{0+}^{\alpha }u\left(t\right)+\varphi \left(t\right)=0,\phantom{\rule{1em}{0ex}}0
(2.1)

where α, γ, σ are given by (1.5) and ϕC(I, ).

By Lemma 2.1, we have

$u\left(t\right)={c}_{1}{t}^{\alpha -1}+{c}_{2}{t}^{\alpha -2}-{I}_{0+}^{\alpha }\varphi \left(t\right),\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].$

From the boundary condition u(0) = 0, we have c2 = 0, and so

$u\left(t\right)={c}_{1}{t}^{\alpha -1}-{I}_{0+}^{\alpha }\varphi \left(t\right),\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].$
(2.2)

Thus,

${D}_{0+}^{\gamma }u\left(t\right)={c}_{1}\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{t}^{\alpha -\gamma -1}-{I}_{0+}^{\alpha -\gamma }\varphi \left(t\right)$

and

$u\left(1\right)={c}_{1}-{I}_{0+}^{\alpha }\varphi \left(1\right),\phantom{\rule{1em}{0ex}}{D}_{0+}^{\gamma }u\left(1\right)={c}_{1}\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}-{I}_{0+}^{\alpha -\gamma }\varphi \left(1\right).$

From the boundary condition $u\left(1\right)+\sigma {D}_{0+}^{\gamma }u\left(1\right)=0$, it follows that

$\left(1+\sigma \frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}\right){c}_{1}-\left({I}_{0+}^{\alpha }\varphi \left(1\right)+\sigma {I}_{0+}^{\alpha -\gamma }\varphi \left(1\right)\right)=0.$

Let $\delta ={\left[1+\sigma \frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}\right]}^{-1}$. Then

${c}_{1}=\delta \left[{I}_{0+}^{\alpha }\varphi \left(1\right)+\sigma {I}_{0+}^{\alpha -\gamma }\varphi \left(1\right)\right].$
(2.3)

Substituting (2.3) into (2.2), we have

$\begin{array}{ll}\hfill u\left(t\right)& =\delta \left[{I}_{0+}^{\alpha }\varphi \left(1\right)+\sigma {I}_{0+}^{\alpha -\gamma }\varphi \left(1\right)\right]{t}^{\alpha -1}-{I}_{0+}^{\alpha }\varphi \left(t\right)\phantom{\rule{2em}{0ex}}\\ =\delta {t}^{\alpha -1}\left[\frac{1}{\Gamma \left(\alpha \right)}\underset{0}{\overset{1}{\int }}{\left(1-s\right)}^{\alpha -1}\varphi \left(s\right)ds+\frac{1}{\Gamma \left(\alpha -\gamma \right)}\sigma \underset{0}{\overset{1}{\int }}{\left(1-s\right)}^{\alpha -\gamma -1}\varphi \left(s\right)ds\right]\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{1}{\Gamma \left(\alpha \right)}\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -1}\varphi \left(s\right)ds\phantom{\rule{2em}{0ex}}\\ =\frac{1}{\Gamma \left(\alpha \right)}\left\{\delta {t}^{\alpha -1}\underset{0}{\overset{1}{\int }}\left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right]\varphi \left(s\right)ds-\underset{0}{\overset{t}{\int }}{\left(t-s\right)}^{\alpha -1}\varphi \left(s\right)ds\right\}\phantom{\rule{2em}{0ex}}\\ =\frac{1}{\Gamma \left(\alpha \right)}\left\{\underset{0}{\overset{t}{\int }}\left[\delta {t}^{\alpha -1}\left({\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right)-{\left(t-s\right)}^{\alpha -1}\right]\varphi \left(s\right)ds\right\\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\delta {t}^{\alpha -1}\underset{t}{\overset{1}{\int }}\left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right]\varphi \left(s\right)ds}\phantom{\rule{2em}{0ex}}\\ =\underset{0}{\overset{1}{\int }}G\left(t,s\right)\varphi \left(s\right)ds,\phantom{\rule{2em}{0ex}}\end{array}$
(2.4)

where

$G\left(t,s\right)=\frac{1}{\Gamma \left(\alpha \right)}\cdot \left\{\begin{array}{cc}{g}_{1}\left(t,s\right),\hfill & 0\le s\le t\le 1,\hfill \\ {g}_{2}\left(t,s\right),\hfill & 0\le t\le s\le 1,\hfill \end{array}\right\$

and

$\begin{array}{ll}\hfill {g}_{1}\left(t,s\right)& =\delta {t}^{\alpha -1}\left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right]-{\left(t-s\right)}^{\alpha -1},\phantom{\rule{1em}{0ex}}0\le s\le t,\phantom{\rule{2em}{0ex}}\\ \hfill {g}_{2}\left(t,s\right)& =\delta {t}^{\alpha -1}\left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right],\phantom{\rule{1em}{0ex}}t\le s\le 1.\phantom{\rule{2em}{0ex}}\end{array}$

So, we obtain the following lemma.

Lemma 2.2. The solution of Equation (2.1) is given by

$u\left(t\right)=\underset{0}{\overset{1}{\int }}G\left(t,s\right)\varphi \left(s\right)ds,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].$

Also, we have the following lemma.

Lemma 2.3. The Green's function G(t, s) has the following properties

1. (i)

G(t, s) is continuous on [0, 1] × [0, 1],

2. (ii)

G(t, s) > 0, s, t ∈ (0, 1).

Proof. (i) Owing to the fact 1 < α ≤ 2, 0 < γ ≤ 1, 0 ≤ α - γ - 1, from the expression of G, it is easy to see that conclusion (i) of Lemma 2.3 is true.

1. (ii)

There are two cases to consider.

2. (1)

If 0 < st < 1, then

$\begin{array}{ll}\hfill \Gamma \left(\alpha \right){g}_{1}\left(t,s\right)& ={t}^{\alpha -1}\left\{\delta \left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right]-{\left(1-\frac{s}{t}\right)}^{\alpha -1}\right\}\phantom{\rule{2em}{0ex}}\\ >{t}^{\alpha -1}\left\{\delta \left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right]-{\left(1-s\right)}^{\alpha -1}\right\}\phantom{\rule{2em}{0ex}}\\ ={t}^{\alpha -1}{\left(1-s\right)}^{\alpha -1}\left\{\delta \left[1+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{-\gamma }\right]-1\right\}\phantom{\rule{2em}{0ex}}\\ \ge {t}^{\alpha -1}{\left(1-s\right)}^{\alpha -1}\left\{\delta \left[1+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}\right]-1\right\}=0.\phantom{\rule{2em}{0ex}}\end{array}$
3. (2)

If 0 < ts < 1, then conclusion (ii) of Lemma 2.3 is obviously true from the expression of G.

We need to introduce some notations for the forthcoming discussion.

Let ${\eta }_{0}={\left(\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}\right)}^{\frac{1}{\alpha -1}}$. Denote $\eta \left(s\right)=\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha }$, s ∈ [0, 1]. Set g(s) = G(s, s), s ∈ [0, 1]. From 0 < γ ≤ 1, σ > 0, 1 < α ≤ 2 and $\delta ={\left[1+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}\right]}^{-1}$, we know that η0 ∈ (0, 1).

Lemma 2.4. The Green's function G has the properties

1. (i)

G(t, s) ≤ G(s, s),s, t ∈ [0, 1].

2. (ii)

G(t, s) ≥ η(s)G(s, s), t ∈ [η0, 1], s ∈ [0, 1].

Proof. (i) There are two cases to consider.

Case 1. 0 ≤ st ≤ 1. In this case, since the following relation

$\begin{array}{ll}\hfill \frac{\partial {g}_{1}\left(t,s\right)}{\partial t}& =\left(\alpha -1\right)\left\{\delta {t}^{\alpha -2}\left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right]-{\left(t-s\right)}^{\alpha -2}\right\}\phantom{\rule{2em}{0ex}}\\ \le \left(\alpha -1\right)\left\{\delta {t}^{\alpha -2}\left[1+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}\right]-{\left(t-s\right)}^{\alpha -2}\right\}\phantom{\rule{2em}{0ex}}\\ =\left(\alpha -1\right)\left\{{t}^{\alpha -2}-{\left(t-s\right)}^{\alpha -2}\right\}<0.\phantom{\rule{2em}{0ex}}\end{array}$

holds for 0 < s < t ≤ 1, we have

$G\left(t,s\right)\le G\left(s,s\right),\phantom{\rule{1em}{0ex}}0\le s\le t\le 1.$

Case 2. 0 ≤ ts ≤ 1. In this case, from the expression of g2(t, s), it is easy to see that

$G\left(t,s\right)\le G\left(s,s\right),\phantom{\rule{1em}{0ex}}0\le t\le s\le 1.$
1. (ii)

We will consider the following two cases.

Case 1. When 0 < sη0, η0t ≤ 1, then from the above argument in (i) of proof, we know that g1(t, s) is decreasing with respect to t on [η0, 1]. Thus

$\underset{t\in \left[{\eta }_{0},1\right]}{\text{min}}G\left(t,s\right)=G\left(1,s\right)={g}_{1}\left(1,s\right)/\Gamma \left(\alpha \right),\phantom{\rule{1em}{0ex}}s\in \left(0,\eta 0\right],$
(2.5)

and so

$\underset{t\in \left[{\eta }_{0},1\right]}{\text{min}}\frac{G\left(t,s\right)}{G\left(s,s\right)}=\frac{{g}_{1}\left(1,s\right)}{g\left(s\right)},\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}s\in \left(0,\eta 0\right].$

Case 2. η0 < s < 1, η0t ≤ 1.

1. (a)

If st, then by similar arguments to (2.5), we also have

$\underset{t\in \left[s,1\right]}{\text{min}}G\left(t,s\right)=G\left(1,s\right)={g}_{1}\left(1,s\right)/\Gamma \left(\alpha \right).$
2. (b)

If η0ts, then the following relation

$\underset{t\in \left[{\eta }_{0},s\right]}{\text{min}}G\left(t,s\right)={g}_{2}\left({\eta }_{0},s\right)/\Gamma \left(\alpha \right)$

holds in view of the expression of g2(t, s).

To summarize,

$\underset{t\in \left[{\eta }_{0},1\right]}{\text{min}}\frac{G\left(t,s\right)}{G\left(s,s\right)}\ge \text{min}\left\{\frac{{g}_{1}\left(1,s\right)}{g\left(s\right)},\frac{{g}_{2}\left({\eta }_{0},s\right)}{g\left(s\right)}\right\},\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{all}}\phantom{\rule{2.77695pt}{0ex}}s\in \left({\eta }_{0},1\right)$
(2.6)

Now, we shall show that

$\underset{t\in \left[{\eta }_{0},1\right]}{\text{min}}\frac{G\left(t,s\right)}{G\left(s,s\right)}\ge \frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha },s\in \left(0,1\right).$
(2.7)

In fact, for s ∈ (0, 1), we have

$\begin{array}{c}{g}_{1}\left(1,s\right)=\delta \left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right]-{\left(1-s\right)}^{\alpha -1}\\ =\delta \left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -1}\right]\\ +\delta \frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}\left[{\left(1-s\right)}^{\alpha -\gamma -1}-{\left(1-s\right)}^{\alpha -1}\right]-{\left(1-s\right)}^{\alpha -1}\\ =\delta \left[1+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}\right]{\left(1-s\right)}^{\alpha -1}\\ +\delta \frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\left[1-{\left(1-s\right)}^{\gamma }\right)\right]-{\left(1-s\right)}^{\alpha -1}\\ =\delta \frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\left[1-{\left(1-s\right)}^{\gamma }\right]\\ >\delta \frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\gamma s,\end{array}$

and so

$\begin{array}{ll}\hfill \frac{{g}_{1}\left(1,s\right)}{g\left(s\right)}& >\frac{\gamma \delta \frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}s}{\delta {s}^{\alpha -1}\left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right]}\phantom{\rule{2em}{0ex}}\\ =\frac{\gamma \frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha }}{{\left(1-s\right)}^{\gamma }+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}}\phantom{\rule{2em}{0ex}}\\ >\frac{\gamma \frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha }}{1+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}}\phantom{\rule{2em}{0ex}}\\ =\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha },\phantom{\rule{1em}{0ex}}s\in \left(0,1\right).\phantom{\rule{2em}{0ex}}\end{array}$
(2.8)

On the other hand, for s ∈ (η0, 1), we have

$\frac{{g}_{2}\left({\eta }_{0},s\right)}{g\left(s\right)}={\eta }_{0}^{\alpha -1}{s}^{1-\alpha }.$
(2.9)

Since ${\eta }_{0}^{\alpha -1}=\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}$, the equality

${\eta }_{0}^{\alpha -1}{s}^{1-\alpha }=\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha }$

holds for s = 1. Thus,

${\eta }_{0}^{\alpha -1}{s}^{1-\alpha }>\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha },\phantom{\rule{1em}{0ex}}s\in \left(0,1\right).$
(2.10)

Since 1 < α ≤ 2, it follows from (2.9) that

$\frac{{g}_{2}\left({\eta }_{0},s\right)}{g\left(s\right)}>\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha },\phantom{\rule{1em}{0ex}}s\in \left({\eta }_{0},1\right).$
(2.11)

Hence, from (2.8) and (2.11), we immediately have

$\text{min}\left\{\frac{{g}_{1}\left(1,s\right)}{g\left(s\right)},\frac{{g}_{2}\left({\eta }_{0},s\right)}{g\left(s\right)}\right\}>\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha },\phantom{\rule{1em}{0ex}}s\in \left({\eta }_{0},1\right).$
(2.12)

Thus, from (2.6) and (2.12 ), it follows that

$\underset{t\in \left[{\eta }_{0},1\right]}{\text{min}}\frac{G\left(t,s\right)}{G\left(s,s\right)}>\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha },\phantom{\rule{1em}{0ex}}s\in \left({\eta }_{0},1\right).$
(2.13)

Also, by (2.8), the following inequality

$\frac{{g}_{1}\left(1,s\right)}{g\left(s\right)}>\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha },\phantom{\rule{1em}{0ex}}s\in \left(0,{\eta }_{0}\right]$

holds, and therefore

$\underset{t\in \left[{\eta }_{0},1\right]}{\text{min}}\frac{G\left(t,s\right)}{G\left(s,s\right)}>\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha },\phantom{\rule{1em}{0ex}}s\in \left(0,{\eta }_{0}\right]$
(2.14)

from the proof in Case 1.

Summing up the above relations (2.13)-(2.14), we have

$\underset{t\in \left[{\eta }_{0},1\right]}{\text{min}}\frac{G\left(t,s\right)}{G\left(s,s\right)}>\frac{\gamma \delta \sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{s}^{2-\alpha },\phantom{\rule{1em}{0ex}}s\in \left(0,1\right),$

and so

$\underset{t\in \left[{\eta }_{0},1\right]}{\text{min}}G\left(t,s\right)\ge \eta \left(s\right)G\left(s,s\right),\phantom{\rule{1em}{0ex}}s\in \left[0,1\right].$

The proof of Lemma 2.4 is complete.

To study BVP (1. 5), we first consider the associated linear BVP

$\left\{\begin{array}{c}{D}_{0+}^{\beta }\left({\phi }_{p}\left({D}_{0+}^{\alpha }u\right)\right)\left(t\right)+h\left(t\right)=0,\phantom{\rule{1em}{0ex}}0
(2.15)

where hP.

Let $w={D}_{0+}^{\alpha }u,v={\phi }_{p}\left(w\right)$. By Lemma 2.1, the solution of initial value problem

$\left\{\begin{array}{cc}{D}_{0+}^{\beta }u\left(t\right)+h\left(t\right)=0,\hfill & t\in \left(0,1\right),\hfill \\ v\left(0\right)=0\hfill \end{array}\right\$

is given by

$v\left(t\right)={C}_{1}{t}^{\beta -1}-{I}_{0+}^{\beta }h\left(t\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right].$

From the relations v(0) = 0, 0 < β ≤ 1, it follows that C1 = 0, and so

$v\left(t\right)=-{I}_{0+}^{\beta }h\left(t\right),\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].$
(2.16)

Noting that ${D}_{0+}^{\alpha }u=w,w={\phi }_{p}^{-1}\left(v\right)$, from (2.16), we know that the solution of (2.15) satisfies

$\left\{\begin{array}{l}{D}_{0+}^{\alpha }u\left(t\right)={\phi }_{p}^{-1}\left(-{I}_{0}^{\beta }h\right)\left(t\right)\right),t\in \left(0,1\right),\hfill \\ u\left(0\right)=0,u\left(1\right)+\sigma {D}_{0+}^{\gamma }u\left(1\right)=0.\hfill \end{array}$
(2.17)

By Lemma 2.2, the solution of Equation (2.17) can be written as

$u\left(t\right)=-\underset{0}{\overset{1}{\int }}G\left(t,s\right){\phi }_{p}^{-1}\left(-{I}_{0}^{\beta }h\right)\left(s\right)ds,\phantom{\rule{1em}{0ex}}t\in I.$
(2.18)

Since h(s) ≥ 0, s ∈ [0, 1], we have ${\phi }_{p}^{-1}\left(-{I}_{0+}^{\beta }h\left(s\right)\right)=-{\left(\left({I}_{0+}^{\beta }h\right)\left(s\right)\right)}^{q-1}$, s ∈ [0, 1], and so

$\begin{array}{ll}\hfill u\left(t\right)& =\underset{0}{\overset{1}{\int }}G\left(t,s\right){\left({I}_{0+}^{\beta }h\left(s\right)\right)}^{q-1}ds\phantom{\rule{2em}{0ex}}\\ =\frac{1}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(t,s\right)ds{\left(\underset{0}{\overset{s}{\int }}h\left(\tau \right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}\phantom{\rule{2em}{0ex}}\end{array}$
(2.19)

from (2.18). Thus, by Lemma 2.3, we have obtained the following lemma.

Lemma 2.5. Let hP. Then the solution of Equation (2.15) in P is given by

$u\left(t\right)=\frac{1}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(t,s\right)ds{\left(\underset{0}{\overset{s}{\int }}h\left(\tau \right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}.$

We also need the following lemmas to obtain our results.

Lemma 2.6. If a, b ≥ 0, γ > 0, then

${\left(a+b\right)}^{\gamma }\le \text{max}\left\{{2}^{\gamma -1},1\right\}\left({a}^{\gamma }+{b}^{\gamma }\right).$

Proof. Obviously, without loss of generality, we can assume that 0 < a < b, γ ≠ 1.

Let ϕ(t) = tγ, t ∈ [0, +∞).

1. (i)

If γ > 1, then ϕ(t) is convex on (0, +∞), and so

$\varphi \left(\frac{1}{2}a+\frac{1}{2}b\right)\le \frac{1}{2}\varphi \left(a\right)+\frac{1}{2}\varphi \left(b\right),$

.

i.e., $\frac{1}{{2}^{\gamma }}{\left(a+b\right)}^{\gamma }\le \frac{1}{2}\left({a}^{\gamma }+{b}^{\gamma }\right)$. Thus

${\left(a+b\right)}^{\gamma }\le {2}^{\gamma -1}\left({a}^{\gamma }+{b}^{\gamma }\right).$
1. (ii)

If 0 < γ < 1, then ϕ(t) is concave on [0, +∞), and so

$\begin{array}{ll}\hfill \varphi \left(a\right)& =\varphi \left(\frac{b}{a+b}\cdot 0+\frac{a}{a+b}\cdot \left(a+b\right)\right)\phantom{\rule{2em}{0ex}}\\ \ge \frac{b}{a+b}\varphi \left(0\right)+\frac{a}{a+b}\cdot \varphi \left(a+b\right)\phantom{\rule{2em}{0ex}}\\ =\frac{a}{a+b}\varphi \left(a+b\right),\phantom{\rule{2em}{0ex}}\end{array}$
$\begin{array}{ll}\hfill \varphi \left(b\right)& =\varphi \left(\frac{b}{a+b}\cdot 0+\frac{a}{a+b}\cdot \left(a+b\right)\right)\phantom{\rule{2em}{0ex}}\\ \ge \frac{a}{a+b}\varphi \left(0\right)+\frac{b}{a+b}\varphi \left(a+b\right)\phantom{\rule{2em}{0ex}}\\ =\frac{b}{a+b}\varphi \left(a+b\right).\phantom{\rule{2em}{0ex}}\end{array}$

Thus, ϕ(a) + ϕ(b) ≥ ϕ(a + b), namely,

${\left(a+b\right)}^{\gamma }\le {a}^{\gamma }+{b}^{\gamma }.$

By (i), (ii) above, we know that the conclusion of Lemma 2.6 is true.

Lemma 2.7. Let c > 0, γ > 0. For any x, y ∈ [0, c], we have that

1. (i)

If γ > 1, then |xγ- yγ| ≤ γcγ-1|x - y|,

2. (ii)

If 0 < γ ≤ 1, then |xγ- yγ| ≤ |x - y|γ.

Proof. Obviously, without loss of generality, we can assume that 0 < y < x since the variables x and y are symmetrical in the above inequality.

1. (i)

If γ > 1, then we set ϕ(t) = tγ, t ∈ [0, c]. by virtue of mean value theorem, there exists a ξ ∈ (0, c) such that

$\begin{array}{ll}\hfill {x}^{\gamma }-{y}^{\gamma }& =\gamma {\xi }^{\gamma -1}\left(x-y\right)\phantom{\rule{2em}{0ex}}\\ \le \gamma {c}^{\gamma -1}\left(x-y\right),\phantom{\rule{2em}{0ex}}\end{array}$

i.e.,

$\left|{x}^{\gamma }-{y}^{\gamma }\right|\le \gamma {c}^{\gamma -1}\left|x-y\right|.$
1. (ii)

If 0 < γ < 1, then by Lemma 2.6, it is easy to see that

$\begin{array}{ll}\hfill {x}^{\gamma }-{y}^{\gamma }& ={\left(x-y-y\right)}^{\gamma }-{y}^{\gamma }\phantom{\rule{2em}{0ex}}\\ \le {\left(x-y\right)}^{\gamma }+{y}^{\gamma }-{y}^{\gamma }={\left(x-y\right)}^{\gamma },\phantom{\rule{2em}{0ex}}\end{array}$

and so

$\left|{x}^{\gamma }-{y}^{\gamma }\right|\le {\left|x-y\right|}^{\gamma }.$

Now we introduce some notations, which will be used in the sequel.

Let $D={\int }_{0}^{1}G\left(s,s\right){s}^{\beta \left(q-1\right)}ds,Q={\int }_{0}^{1}\eta \left(s\right)G\left(s,s\right){s}^{\beta \left(q-1\right)}ds$,

$l=\Gamma \left(\beta +1\right){\left(D\cdot \text{max}\left\{{2}^{q-1},1\right\}\right)}^{\frac{1}{1-q}},\mu =\frac{\Gamma \left(\beta +1\right)}{{Q}^{\frac{1}{q-1}}}.$

By simple calculation, we know that

$D=\frac{\delta }{\Gamma \left(\alpha \right)}B\left(\alpha ,\alpha +\beta \left(q-1\right)\right)+\frac{\sigma \delta }{\Gamma \left(\alpha -\gamma \right)}B\left(\alpha -\gamma ,\alpha +\beta \left(q-1\right)\right),$
(2.20)
$Q=\frac{\gamma {\delta }^{2}\sigma }{\Gamma \left(\alpha -\gamma \right)}\left[B\left(\alpha ,2+\beta \left(q-1\right)\right)+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}B\left(\alpha -\gamma ,2+\beta \left(q-1\right)\right)\right].$
(2.21)

(H1) fC(I × +, +).

(H2) $\overline{\underset{x\to +\infty }{\text{lim}}}\underset{t\in I}{\text{max}}\frac{f\left(t,x\right)}{{x}^{p-1}}\mu .$.

(H3) There exists a r0 > 0 such that f(t, x) is nonincreasing relative to x on [0, r0] for any fixed tI.

By Lemma 2.5, it is easy to know that the following lemma is true.

Lemma 2.8. If (H1) holds, then BVP (1.5) has a nonnegative solution if and only if the integral equation

$u\left(t\right)=\frac{1}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(t,s\right){\left(\underset{0}{\overset{s}{\int }}f\left(\tau ,u\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}ds,\phantom{\rule{1em}{0ex}}t\in I$
(2.22)

has a solution in P. Let c be a positive number, P be a cone and P c = {yP : ∥y∥ ≤ c}. Let α be a nonnegative continuous concave function on P and

$P\left(\alpha ,a,b\right)=\left\{u\in P|a\le \alpha \left(u\right),∥u∥\le b\right\}.$

We will use the following lemma to obtain the multiplicity results of positive solutions.

Lemma 2.9. [35] Let $A:\overline{{P}_{c}}\to \overline{{P}_{c}}$ be completely continuous and α be a nonnegative continuous concave function on P such that α(y) ≤ ∥y∥ for all $y\in \overline{{P}_{c}}$. Suppose that there exist a, b and d with 0 < a < b < dc such that

(C1) $\left\{y\in P\left(\alpha ,b,d\right)\right\}|\alpha \left(y\right)>b\right\}\ne \overline{)0}$ and α(Ay) > b, for all yP(α, b, d);

(C2) ∥Ay∥ < a, for ∥y∥ ≤ a;

(C3) α(Ay) > b, for yP(α, b, c) with ∥Ay∥ > d.

Then A has at least three fixed points y1, y2, y3 satisfying

$∥{y}_{1}∥a\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{with}}\phantom{\rule{2.77695pt}{0ex}}\alpha \left({y}_{3}\right)

## 3 Main results

In this section, our objective is to establish existence and multiplicity of positive solution to the BVP (1.5). To this end, we first define the operator on P as

$Au=\frac{1}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}{G\left(t,s\right)\left(\underset{0}{\overset{s}{\int }}f\left(\tau ,u\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}ds,\phantom{\rule{1em}{0ex}}u\in P.$
(3.1)

The properties of the operator A are given in the following lemma.

Lemma 3.1. Let (H1) hold. Then A : PP is completely continuous.

Proof. First, under assumption (H1), it is obvious that APP from Lemma 2.3. Next, we shall show that operator A is completely continuous on P. Let $E={\int }_{0}^{1}G\left(s,s\right)ds$. The following proof will be divided into two steps.

Step 1. We shall show that the operator A is compact on P.

Let B be an arbitrary bounded set in P. Then exists an M > 0 such that ∥u∥ ≤ M for all uB. According to the continuity of f, we have $L\triangleq \underset{\left(t,x\right)\in I×\left[0,M\right]}{\mathrm{max}f\left(t,x\right)}<+\infty$. Thus, by Lemmas 2.3 and 2.4, it follows that

$\begin{array}{ll}\hfill 0\le \left(\mathsf{\text{A}}u\right)\left(t\right)& \le \frac{{L}^{q-1}}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(s,s\right){\left(\underset{0}{\overset{s}{\int }}{\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}ds\phantom{\rule{2em}{0ex}}\\ <\frac{{L}^{q-1}}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(s,s\right)ds\phantom{\rule{2em}{0ex}}\\ =\frac{{L}^{q-1}}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}E,\phantom{\rule{1em}{0ex}}t\in I.\phantom{\rule{2em}{0ex}}\end{array}$

Thus,

$∥\mathsf{\text{A}}u∥\le \frac{{L}^{q-1}}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}E.$

That is, the set AB is uniformly bounded.

On the other hand, the uniform continuity of G(t, s) on I × I implies that for arbitrary ε > 0, there exists a δ > 0 such that whenever t1, t2I with |t1 - t2| < δ, the following inequality

$\left|G\left({t}_{1},s\right)-G\left({t}_{2},s\right)\right|<\epsilon \frac{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}{{L}^{q-1}}\left(\beta \left(q-1\right)+1\right)$

holds for all sI. Therefore,

$\begin{array}{ll}\hfill \left|\mathsf{\text{A}}u\left({t}_{1}\right)-\mathsf{\text{A}}u\left({t}_{2}\right)\right|& \le \frac{1}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}\left|G\left({t}_{1},s\right)-G\left({t}_{2},s\right)\right|{\left(\underset{0}{\overset{s}{\int }}f\left(\tau ,u\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}ds\phantom{\rule{2em}{0ex}}\\ \le \frac{{L}^{q-1}}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}{s}^{\left(q-1\right)\beta }ds\frac{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}{{L}^{q-1}}\left(\beta \left(q-1\right)+1\right)\epsilon =\epsilon .\phantom{\rule{2em}{0ex}}\end{array}$

Thus, AB is equicontinuous. Consequently, the operator is compact on P by Arzelà-Ascoli theorem.

Step 2. The operator A is continuous.

Let {u n } be an arbitrary sequence in P with u n u0P. Then exists an L > 0 such that

$0\le f\left(\tau ,{u}_{n}\left(\tau \right)\right)\le L,\phantom{\rule{1em}{0ex}}\tau \in \left[0,1\right],\phantom{\rule{1em}{0ex}}n\ge 0.$

Thus,

$\underset{0}{\overset{s}{\int }}f\left(\tau ,{u}_{n}\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \le L\underset{0}{\overset{s}{\int }}{\left(s-\tau \right)}^{\beta -1}d\tau \le \frac{L}{\beta }\triangleq c,\phantom{\rule{1em}{0ex}}s\in \left[0,1\right].$

On the other hand, the uniform continuity of f combined with the fact that ∥u n - u0∥ → 0 yields that there exists a N ≥ 1 such that the following estimate

$|f\left(\tau ,{u}_{n}\left(\tau \right)-f\left(t,{u}_{0}\left(\tau \right)\right)|<\epsilon$

holds for nN.

1. (1)

If 1 < q ≤ 2, then from Lemma 2.7 (ii), we have

$\begin{array}{l}\left|{\left(\underset{0}{\overset{s}{\int }}f\left(\tau ,{u}_{n}\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}-{\left(\underset{0}{\overset{s}{\int }}f\left(\tau ,{u}_{0}\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}\right|\phantom{\rule{2em}{0ex}}\\ \le {\left(\underset{0}{\overset{s}{\int }}\left|f\left(\tau ,{u}_{n}\left(\tau \right)\right)-f\left(\tau ,{u}_{0}\left(\tau \right)\right)\right|{\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}\phantom{\rule{2em}{0ex}}\\ <{\epsilon }^{q-1}\frac{1}{{\beta }^{q-1}}{s}^{\beta \left(q-1\right)},\phantom{\rule{1em}{0ex}}s\in \left[0,1\right].\phantom{\rule{2em}{0ex}}\end{array}$

Hence, by Lemmas 2.3 and 2.4, from (3.1), we obtain

$\begin{array}{ll}\hfill \left|\mathsf{\text{A}}{u}_{n}\left(t\right)-\mathsf{\text{A}}{u}_{0}\left(t\right)\right|& <\frac{{\epsilon }^{q-1}}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(s,s\right)ds\phantom{\rule{2em}{0ex}}\\ =\frac{E}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}{\epsilon }^{q-1}.\phantom{\rule{2em}{0ex}}\end{array}$

Thus,

$∥\mathsf{\text{A}}{u}_{n}-\mathsf{\text{A}}{u}_{0}∥\le \frac{E}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}{\epsilon }^{q-1}.$
(3.2)
1. (2)

If q > 2, then from Lemma 2.7 (i), we have

$\begin{array}{c}\left|{\left(\underset{0}{\overset{s}{\int }}f\left(\tau ,{u}_{n}\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}-{\left(\underset{0}{\overset{s}{\int }}f\left(\tau ,{u}_{0}\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}\right|\\ \le \left(q-1\right){c}^{q-2}\underset{0}{\overset{s}{\int }}\left|f\left(\tau ,{u}_{n}\left(\tau \right)\right)-f\left(\tau ,{u}_{0}\left(\tau \right)\right)\right|{\left(s-\tau \right)}^{\beta -1}d\tau \\ <\frac{q-1}{\beta }{c}^{q-2}{s}^{\beta }\epsilon ,\phantom{\rule{1em}{0ex}}s\in \left[0,1\right].\end{array}$

Thus, we have

$\begin{array}{ll}\hfill \left|\mathsf{\text{A}}{u}_{n}\left(t\right)-\mathsf{\text{A}}{u}_{0}\left(t\right)\right|& <\frac{\left(q-1\right){c}^{q-2}\epsilon }{\beta {\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(s,s\right)ds\phantom{\rule{2em}{0ex}}\\ =\frac{\left(q-1\right){c}^{q-2}E}{\beta {\left(\Gamma \left(\beta \right)\right)}^{q-1}}\epsilon ,\phantom{\rule{2em}{0ex}}\end{array}$

and so

$∥\mathsf{\text{A}}{u}_{n}-\mathsf{\text{A}}{u}_{0}∥\le \frac{\left(q-1\right){c}^{q-2}E}{\beta {\left(\Gamma \left(\beta \right)\right)}^{q-1}}\epsilon .$
(3.3)

From (3.2)-(3.3), it follows that ∥Au n - Au0∥ → 0(n → ∞).

Summing up the above analysis, we obtain that the operator A is completely continuous on P.

We are now in a position to state and prove the first theorem in this article.

Theorem 3.1. Let (H1), (H2), and (H3) hold. Then BVP (1.5) has at least one positive solution.

Proof. By Lemma 2.8, it is easy to know that BVP (1.5) has a nonnegative solution if and only if the operator A has a fixed point in P. Also, we know that A : PP is completely continuous by Lemma 3.1.

The following proof is divided into two steps.

Step 1. From (H2), we can choose a ε0 ∈ (0, l) such that

$\overline{\underset{x\to +\infty }{\text{lim}}}\underset{t\in I}{\text{max}}\frac{f\left(t,x\right)}{{x}^{p-1}}

Therefore, there exists a R0 > 0 such that the inequality

$f\left(t,x\right)<\left(l-{\epsilon }_{0}\right){x}^{p-1},\phantom{\rule{1em}{0ex}}t\in I$
(3.4)

holds for xR0.

Let $M=\underset{\left(t,x\right)\in I×\left[0,{R}_{0}\right]}{\text{max}}f\left(t,x\right)$. It follows from (3.4) that

$f\left(t,x\right)\le \left(l-{\epsilon }_{0}\right){x}^{p-1}+M,\phantom{\rule{1em}{0ex}}\forall x\in {ℝ}_{+},\phantom{\rule{1em}{0ex}}t\in I.$
(3.5)

From the fact that (l - ε0)q-1< lq-1, we can choose a k > 0 such that (l - ε0)q-1< lq-1- k.

Set

${D}_{1}=\frac{\text{max}\left\{{2}^{q-2},1\right\}D}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}},E={D}_{1}k,G={D}_{1}{M}^{q-1}.$
(3.6)

where D is as (2.20). Take $R>\frac{G}{E}$. Set Ω R = {uP : ∥u∥ < R}. We shall show that the relation

$\mathsf{\text{A}}u\ne \mu u,\phantom{\rule{1em}{0ex}}\forall u\in \partial {\Omega }_{R}\phantom{\rule{1em}{0ex}}\mu \ge 1$
(3.7)

holds.

In fact, if not, then there exists a u0 Ω R and a μ0 ≥ 1 with

${\mu }_{0}{u}_{0}=\mathsf{\text{A}}{u}_{0}.$

By (3.5), we have

$\begin{array}{ll}\hfill f\left(t,{u}_{0}\left(t\right)\right)& \le \left(l-{\epsilon }_{0}\right){u}_{0}^{p-1}\left(t\right)+M\phantom{\rule{2em}{0ex}}\\ \le \left(l-{\epsilon }_{0}\right){∥{u}_{0}∥}^{p-1}+M\phantom{\rule{2em}{0ex}}\\ =\left(l-{\epsilon }_{0}\right){R}^{p-1}+M,\phantom{\rule{1em}{0ex}}t\in I.\phantom{\rule{2em}{0ex}}\end{array}$

Therefore, in view of Lemmas 2.3, 2.4, from (3.1), it follows that

$\begin{array}{ll}\hfill Au\left(t\right)& \le \frac{{\left(\left(l-{\epsilon }_{0}\right){R}^{p-1}+M\right)}^{q-1}}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(s,s\right){s}^{\beta \left(q-1\right)}ds\phantom{\rule{2em}{0ex}}\\ =\frac{D}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}{\left(\left(l-{\epsilon }_{0}\right){R}^{p-1}+M\right)}^{q-1},\phantom{\rule{1em}{0ex}}t\in I.\phantom{\rule{2em}{0ex}}\end{array}$
(3.8)

Also, keeping in mind that (p - 1)(q - 1) = 1, by Lemma 2.6, we have

$\begin{array}{ll}\hfill {\left(\left(l-{\epsilon }_{0}\right){R}^{p-1}+M\right)}^{q-1}& \le \text{max}\left\{{2}^{q-2},1\right\}\left({\left(l-{\epsilon }_{0}\right)}^{q-1}R+{M}^{q-1}\right)\phantom{\rule{2em}{0ex}}\\ <\text{max}\left\{{2}^{q-2},1\right\}\left(\left({l}^{q-1}-k\right)R+{M}^{q-1}\right).\phantom{\rule{2em}{0ex}}\end{array}$
(3.9)

Hence, from (3.6), (3.8), and (3.9), it follows that

${u}_{0}\le \mu {u}_{0}=\mathsf{\text{A}}{u}_{0}\le \left({D}_{1}{l}^{q-1}-E\right)R+G,\phantom{\rule{1em}{0ex}}t\in I.$
(3.10)

By definition of l, we have D1lq-1= 1. From (3. 10), it follows that R = ∥u0∥ ≤ (1 - E)R + G, and so $R\le \frac{G}{E}$, which contradicts the choice of R. Hence, the condition (3.7) holds. By virtue of the fixed point index theorem, we have

$i\left(A,{\Omega }_{R},P\right)=1.$
(3.11)

Step 2. By (H2), we can choose a ε0 > 0 such that

$\underset{x\to 0+}{\underset{¯}{\mathrm{lim}}}\underset{t\in I}{\mathrm{min}}\frac{f\left(t,x\right)}{{x}^{p-1}}>\mu +{\epsilon }_{0}.$

Hence, there exists a r1 ∈ (0, r0) such that

$f\left(t,x\right)>\left(\mu +{\epsilon }_{0}\right){x}^{p-1},\phantom{\rule{1em}{0ex}}t\in I,\phantom{\rule{1em}{0ex}}x\in \left[0,{r}_{1}\right],$
(3.12)

where r0 is given by (H3).

Take 0 < r < min {R, r1}, and set Ω r = {uP : ∥u∥ < r}. Now, we show that

1. (i)

$\underset{u\in \partial {\Omega }_{r}}{\text{inf}}∥\mathsf{\text{A}}u∥>0$,

2. (ii)

Auμu, ∀u Ω r , μ ∈ [0, 1].

We first prove that (i) holds. In fact, for any u Ω r , we have 0 ≤ u(t) ≤ r. By (H3), the function f(t, x) is nonincreasing relative to x on [0, r] for any tI, and so

$f\left(t,u\left(t\right)\right)\ge f\left(t,r\right)\ge \left(\mu +{\epsilon }_{0}\right){r}^{p-1},\phantom{\rule{1em}{0ex}}t\in \left[0,1\right]$
(3.13)

from (3.12).

Thus, in view of Lemma 2.4 combined with (3.1) and (3.13), we have

$\begin{array}{ll}\hfill \mathsf{\text{A}}u\left(t\right)& \ge \frac{1}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(t,s\right){\left(\underset{0}{\overset{s}{\int }}\left(\mu ,+{\epsilon }_{0}\right){r}^{p-1}{\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}ds\phantom{\rule{2em}{0ex}}\\ =\frac{{\left(\mu +{\epsilon }_{0}\right)}^{q-1}r}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(t,s\right){s}^{\beta \left(q-1\right)}ds\phantom{\rule{2em}{0ex}}\\ \ge \frac{{\left(\mu +{\epsilon }_{0}\right)}^{q-1}r}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}\eta \left(s\right)G\left(s,s\right){s}^{\beta \left(q-1\right)}ds\phantom{\rule{2em}{0ex}}\\ =\frac{{\left(\mu +{\epsilon }_{0}\right)}^{q-1}Q}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}r,\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{all}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[{\eta }_{0},1\right],\phantom{\rule{2em}{0ex}}\end{array}$

where Q is as (2.21). Consequently,

$∥\mathsf{\text{A}}u∥\ge \frac{{\left(\mu +{\epsilon }_{0}\right)}^{q-1}Q}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}r\doteq {c}_{0}>0.$
(3.14)

Thus $\underset{u\in \partial {\Omega }_{r}}{\text{inf}}∥\mathsf{\text{A}}u∥\ge {c}_{0}>0$.

1. (ii)

Suppose on the contrary that there exists a u0 Ω r and μ0 ∈ [0, 1] such that

${\mu }_{0}{u}_{0}=\mathsf{\text{A}}{u}_{0}.$
(3.15)

Then, by similar arguments to (3.14), we have

$∥\mathsf{\text{A}}{u}_{0}∥\ge Br,$
(3.16)

where $B=\frac{{\left(\mu +{\epsilon }_{0}\right)}^{q-1}Q}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}$.

By (3.15)-(3.16), we obtain

$r=∥{u}_{0}∥\ge {\mu }_{0}∥{u}_{0}∥=∥A{u}_{0}∥\ge Br.$

The hypothesis $\mu =\frac{\Gamma \left(\beta +1\right)}{{Q}^{\frac{1}{q-1}}}$ implies that B > 1, and so r > r from above inequality, which is a contradiction. That means that (ii) holds.

Hence, applying fixed point index theorem, we have

$i\left(A,{\Omega }_{R},P\right)=0.$
(3.17)

By (3.11) and (3.17), we have

$i\left(A,{\Omega }_{R}\{\Omega }_{r,}P\right)=1,$

and so, there exists ${u}_{*}\in {\Omega }_{R}\{\stackrel{̄}{\Omega }}_{r}$ with Au* = u*, ∥u*∥ > r. Hence, u* is a nonnegative solution of BVP (1.5) satisfying ∥u*∥ > r. Now, we show that u*(t) > 0, t ∈ (0, 1).

In fact, since ∥u*∥ > r, u*P, G(t, s) > 0, t, s ∈ (0, 1), from (3.1), we have

$\underset{0}{\overset{s}{\int }}f\left(\tau ,{u}_{*}\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \not\equiv 0,\phantom{\rule{1em}{0ex}}s\in \left(0,1\right),$

and so

${u}_{*}\left(t\right)=A{u}_{*}\left(t\right)>0,\phantom{\rule{1em}{0ex}}t\in \left(0,1\right)$
(3.18)

from the fact that G(t, s) > 0 and ${\int }_{0}^{s}f\left(\tau ,u\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \ge 0$, s ∈ [0, 1]. That is, u* is a positive solution of BVP (1.5).

The proof is complete.

Now, we state another theorem in this article. First, let me introduce some notations which will be used in the sequel.

Let ${M}_{1}=\frac{\Gamma \left(\beta +1\right)}{{D}^{\frac{1}{q-1}}},{M}_{2}=\frac{\Gamma \left(\beta +1\right)}{{B}_{2}^{\frac{1}{q-1}}}$, where D is as (2.20).

Let

${B}_{2}\frac{\gamma {\delta }^{2}\sigma {\left(1-{\eta }_{0}\right)}^{\alpha +\beta \left(q-1\right)+1}}{\Gamma \left(\alpha -\gamma \right)}\left[B\left(\alpha ,\beta \left(q-1\right)+2\right)+{\left(1-{\eta }_{0}\right)}^{-\gamma }\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}B\left(\alpha -\gamma ,\beta \left(q-1\right)+2\right)\right].$
(3.19)

Set P r = {uP : ∥u∥ < r}, for r > 0. Let $\omega \left(u\right)=\underset{t\in \left[{\eta }_{0},1\right]}{\text{min}}u\left(t\right)$, for uP. Obviously, ω is a nonnegative continuous concave functional on P.

Theorem 3.2. Let (H1) hold. Assume that there exist constants a, b, c, l1, l2 with 0 < a < b < c and l1 ∈ (0, M1), l2 ∈ (M2, ∞) such that

(D1) f(t, x) ≤ l1cp-1, x ∈ [0,c], tI; f(t, x)≤ l1ap-1, x ∈ [0, a], tI,

(D2) f(t, x) ≥ l1bp-1, x ∈ [b,c], t ∈ [η0, 1].

Then BVP (1.5) has at least one nonnegative solution u1 and two positive solutions u2, u3 with

$∥{u}_{1}∥

Proof. By Lemmas 2.3 and 2.4, for $u\in {\stackrel{̄}{P}}_{c}$, from (3.1) and condition (D1), it follows that

$\begin{array}{ll}\hfill \mathsf{\text{A}}u\left(t\right)& \le \frac{{\left({l}_{1}\right)}^{q-1}c}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(s,s\right){s}^{\beta \left(q-1\right)}ds\phantom{\rule{2em}{0ex}}\\ =\frac{{\left({l}_{1}\right)}^{q-1}c}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}D,\phantom{\rule{1em}{0ex}}t\in I,\phantom{\rule{2em}{0ex}}\end{array}$

and so

$∥\mathsf{\text{A}}u∥\le \frac{{\left({l}_{1}\right)}^{q-1}D}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}c

from the hypothesis l1 < M1.

Thus, we obtain $A:{\stackrel{̄}{P}}_{c}\to {P}_{c}$. Similarly, we can also obtain $A:{\stackrel{̄}{P}}_{a}\to {P}_{a}$ by condition (D1). Take ${u}_{0}=\frac{b+c}{2}$. Then ω(u0) > b, and so $\left\{u\in P\left(\omega ,b,c\right)|\omega \left(u\right)>b\right\}\ne \varnothing$.

For any uP(ω, b, c), we have that u(t) ≥ b, t ∈ [η0, 1] and ∥u∥ ≤ c. Consequently, by Lemma 2.3, 2.4 and the formula (3.1), for any t ∈ [η0, 1], it follows from condition (D2) that

$\begin{array}{ll}\hfill \mathsf{\text{A}}u\left(t\right)& =\frac{1}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}G\left(t,s\right){\left(\underset{0}{\overset{s}{\int }}f\left(\tau ,u\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}ds\phantom{\rule{2em}{0ex}}\\ \ge \frac{1}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{0}{\overset{1}{\int }}\eta \left(s\right)G\left(s,s\right){\left(\underset{0}{\overset{s}{\int }}f\left(\tau ,u\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}ds\phantom{\rule{2em}{0ex}}\\ \ge \frac{1}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{{\eta }_{0}}{\overset{1}{\int }}\eta \left(s\right)G\left(s,s\right){\left(\underset{0}{\overset{s}{\int }}f\left(\tau ,u\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}ds\phantom{\rule{2em}{0ex}}\\ \ge \frac{1}{{\left(\Gamma \left(\beta \right)\right)}^{q-1}}\underset{{\eta }_{0}}{\overset{1}{\int }}\eta \left(s\right)G\left(s,s\right){\left(\underset{{\eta }_{0}}{\overset{s}{\int }}f\left(\tau ,u\left(\tau \right)\right){\left(s-\tau \right)}^{\beta -1}d\tau \right)}^{q-1}ds\phantom{\rule{2em}{0ex}}\\ \ge \frac{{{l}_{2}}^{q-1}b}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}\underset{{\eta }_{0}}{\overset{1}{\int }}\eta \left(s\right)G\left(s,s\right){\left(s-{\eta }_{0}\right)}^{\beta \left(q-1\right)}ds.\phantom{\rule{2em}{0ex}}\end{array}$
(3.20)

Also, by changing the variable $\theta =\frac{s-{\eta }_{0}}{1-{\eta }_{0}}$, we have

$\begin{array}{l}\phantom{\rule{1em}{0ex}}\underset{{\eta }_{0}}{\overset{1}{\int }}\eta \left(s\right)G\left(s,s\right){\left(s-{\eta }_{0}\right)}^{\beta \left(q-1\right)}ds\phantom{\rule{2em}{0ex}}\\ =\frac{\gamma {\delta }^{2}\sigma }{\Gamma \left(\alpha -\gamma \right)}\underset{{\eta }_{0}}{\overset{1}{\int }}s\left[{\left(1-s\right)}^{\alpha -1}+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-s\right)}^{\alpha -\gamma -1}\right]{\left(s-{\eta }_{0}\right)}^{\beta \left(q-1\right)}ds\phantom{\rule{2em}{0ex}}\\ >\frac{\gamma {\delta }^{2}\sigma {\left(1-{\eta }_{0}\right)}^{\alpha +\beta \left(q-1\right)+1}}{\Gamma \left(\alpha -\gamma \right)}\underset{0}{\overset{1}{\int }}{\theta }^{1+\beta \left(q-1\right)}{\left(1-\theta \right)}^{\alpha -1}\left[1+\frac{\sigma \Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{\left(1-{\eta }_{0}\right)}^{-\gamma }{\left(1-\theta \right)}^{-\gamma }\right]d\theta \phantom{\rule{2em}{0ex}}\\ ={B}_{2,}\phantom{\rule{2em}{0ex}}\end{array}$
(3.21)

where B2 is given by (3.19).

Substituting (3.21) into (3.20), we obtain

$\mathsf{\text{A}}u\left(t\right)\ge \frac{{{l}_{2}}^{q-1}{B}_{2}}{{\left(\Gamma \left(\beta +1\right)\right)}^{q-1}}b,\phantom{\rule{1em}{0ex}}t\in \left[{\eta }_{0},1\right],$

and so ω(Au) > b from the hypothesis l2 > M2.

Summing up the above analysis, we know that all the conditions of Lemma 2.9 with c = d are satisfied, and so BVP (1.5) has at least three nonnegative solutions u1, u2, u3 with

$∥{u}_{1}∥

By similar argument to (3.18), we can deduce that u2 and u3 are two positive solutions.

The proof is complete.

Example 3.1. Consider the following BVP

$\left\{\begin{array}{c}{D}_{0+}^{\beta }\left({\phi }_{p}\left({D}_{0+}^{\alpha }u\right)\right)\left(t\right)+\left({t}^{2}+{\text{cos}}^{2}u\right)=0,\phantom{\rule{1em}{0ex}}0
(3.22)

where 1 < α < 2, 0 < β < 1, 0 < γ < 1, 0 ≤ α - γ - 1, σ > 0 and the p-Laplacian operator is defined as φ p (s) = |s|p-2s, p > 1.

It is easy to verify that all assumptions of Theorem 3.1 are satisfied. Hence, by the conclusion of Theorem 3.1, BVP (3.22) has at least one positive solution on [0, 1].

Example 3.2. Consider the following BVP

$\left\{\begin{array}{c}{D}_{0+}^{\frac{1}{2}}\left({\phi }_{\frac{3}{2}}\left({D}_{0+}^{\frac{3}{2}}u\right)\right)\left(t\right)+f\left(t,u\right)=0,\phantom{\rule{1em}{0ex}}0
(3.23)

where $\alpha =\frac{2}{3},\beta =\gamma =\frac{1}{2},\sigma =\frac{2}{\sqrt{\pi }},p=\frac{3}{2}$ relative to Theorem 3.2. With the aid of computation we have that ${M}_{1}=\sqrt{\pi }\sqrt[4]{\pi }\sqrt{\frac{20}{32+5\pi }}=1.527...,{M}_{2}=\frac{64}{15}\sqrt{\pi }\sqrt[4]{\pi }{\left(\frac{7}{35+4\sqrt{15}}\right)}^{\frac{1}{2}}=3.74...,{\eta }_{0}=\frac{1}{16}$. Take l1 = 1.5, l2 = 4. Then l1 ∈ (0, M1), l2 ∈ (M2, ∞). Again choosing $a=\frac{\sqrt[3]{2.25}}{4}$, b = 1, c = 196, and setting

$f\left(t,x\right)=\left\{\begin{array}{c}\left(\text{sin}\frac{\pi }{90}\left(32t+13\right)\right)\left({x}^{\frac{1}{2}}+7\right),\phantom{\rule{1em}{0ex}}x\in \left[1,+\infty \right),\hfill \\ 8\left(\text{sin}\frac{\pi }{90}\left(32t+13\right)\right){x}^{2},\phantom{\rule{1em}{0ex}}x\in \left[0,1\right),\hfill \end{array}\right\$

for t ∈ [0, 1], then we see that f satisfies the following relations:

$\begin{array}{ll}\hfill f\left(t,x\right)& \le 21=1.5×19{6}^{\frac{1}{2}}={l}_{1}{c}^{p-1},\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],\phantom{\rule{1em}{0ex}}x\in \left[0,196\right]=\left[0,c\right],\phantom{\rule{2em}{0ex}}\\ \hfill f\left(t,x\right)& \ge 4={l}_{2}{b}^{p-1},\phantom{\rule{1em}{0ex}}t\in \left[\frac{1}{16},1\right]=\left[{\eta }_{0},1\right],\phantom{\rule{1em}{0ex}}x\in \left[1,196\right]=\left[b,c\right],\phantom{\rule{2em}{0ex}}\\ \hfill f\left(t,x\right)& \le \frac{3\sqrt[3]{1.5}}{4}=1.5×\frac{\sqrt[3]{1.5}}{2}={l}_{1}{a}^{p-1},\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],\phantom{\rule{1em}{0ex}}t\in \left[0,\frac{\sqrt[3]{2.25}}{4}\right]=\left[0,a\right].\phantom{\rule{2em}{0ex}}\end{array}$

So, all the assumptions of Theorem 3.2 are satisfied. By Theorem 3.2, we arrive at BVP (3.23) has at least one nonnegative