Positive solutions of fractional p-Laplacian equations with integral boundary value and two parameters

Abstract

We consider a class of Caputo fractional p-Laplacian differential equations with integral boundary conditions which involve two parameters. By using the Avery–Peterson fixed point theorem, we obtain the existence of positive solutions for the boundary value problem. As an application, we present an example to illustrate our main result.

1 Introduction

In this paper, we investigate the following integral boundary value problem (short for BVP) of Caputo fractional differential equations with p-Laplacian operator and parameters:

$$ \left \{ \textstyle\begin{array}{l} D_{0+}^{\beta}\varphi_{p} ( D_{0+}^{\alpha}x(t) ) + f (t,x(t),D_{0+} ^{\beta}x(t) )= 0,\quad t \in(0,1),\\ ( \varphi_{p} ( D_{0+}^{\alpha}x(0) ) )^{(i)} =\varphi_{p} ( D_{0+}^{\alpha}x(1) )=0,\quad i=1,2, \ldots, m-1 , \\ x(0) + x^{{\prime}}(0) = \int_{0}^{1}g_{0}(s)x(s)\,ds + a, \\ x(1) + x^{{\prime}}(1) = \int_{0}^{1}g_{1}(s)x(s)\,ds + b, \\ x^{(j)}(0) = 0,\quad j= 2, 3, \ldots, n-1, \end{array}\displaystyle \right . $$

(1)

where $1< n-1<\alpha<n$, $1< m-1<\beta< m$, $\alpha-\beta>1$, $D_{{0^{+}}}^{\alpha}$ and $D_{{0^{+}}}^{\beta}$ are the Caputo fractional derivatives. $\varphi_{p}$ is the p-Laplacian operator, $\varphi _{p}(s)=\vert s\vert^{p-2}s$, $p>1$, $\varphi_{p}^{-1}=\varphi_{q}$, $1/p+1/q=1$. $g_{0},g_{1} \in C([0,1], [0,+\infty))$, $f \in C([0,1] \times[0,+\infty) \times[0,+\infty), [0,+\infty))$ are given functions. $a,b>0$ are disturbance parameters.

As we all know, fractional differential equation theory is becoming more and more perfect because of its extensive application, and many significant achievements have been made; see [1–12]. As one of many applications, turbulence problem can be well characterized by the p-Laplacian operator; see [13]. Fractional p-Laplacian equations are becoming more and more important, they can be used to describe a class of diffusion phenomena, which have been widely used in the fields of fluid mechanics, material memory, biology, plasma physics, finance and chemistry. Many important results related to the boundary value problems of fractional differential equations with p-Laplacian operator have been obtained; see [14–24]. But in practical problems, disturbance is objective. As a boundary value problem with disturbance parameter can describe real problems better, many scholars turn their attention to it.

In [6], Jia et al. consider the fractional-order differential equation integral boundary value problem with disturbance parameters

$$\left \{ \textstyle\begin{array}{l} - {}^{C}{D^{\delta}}u ( t ) = f ( {t,u ( t )} ),\quad t \in J,\\ {m_{1}}u ( 0 ) - {n_{1}}u' ( 0 ) = 0,\\ {m_{2}}u ( 1 ) - {n_{2}}u' ( 1 ) = \int_{0}^{1} {g ( s )u ( s )\,ds} + a, \end{array}\displaystyle \right . $$

where $J=[0,1]$, $1 < \delta\leq2$, $f \in C( [0,1]\times[0,+\infty) , [0,+\infty) )$, $m_{i} \geq0$, $n_{i} \geq0$, $m^{2}_{i}+n^{2}_{i}>0$, ${i=1,2}$, $g \in C( [0,1] , [0,+\infty))$, disturbance parameter $a >0$, and $^{C}{D^{\delta}}$ is the Caputo fractional derivative of order δ. By using an upper and lower solution method, the fixed point index theorem and the Schauder fixed point theorem, sufficient conditions are obtained for the problem to have at least one positive solution, two positive solutions and no solution.

In [25], Wang et al. consider a class of fractional differential equations with integral boundary conditions which involve two disturbance parameters. By using the Guo–Krasnoselskii fixed point theorem, new results on the existence and nonexistence of positive solutions for the boundary value problem are obtained. The problem is given by

$$\left \{ \textstyle\begin{array}{l} D_{0{+}}^{\alpha}x(t)=f (t,x(t) ), \quad t \in(0,1), \\ x(0)=x'(0)=0,\\ x(1)=\int_{0}^{1}g_{1}(s)x(s)\, {d}s+a, \\ x'(1)=\int_{0}^{1}g_{2}(s)x(s)\, {d}s-b, \end{array}\displaystyle \right . $$

where $D_{0+}^{\alpha}$ is the standard Riemann–Liouville fractional derivative with $3<\alpha\leq4$, ${f:[0,1]\times[0,+\infty) \rightarrow[0,+\infty)}$ is a continuous function, $g_{1},g_{2}\in L^{1}[0,1]$ and $a,b\geq0$.

In [26] Hao et al. consider the existence of positive solutions of higher order fractional integral boundary value problem with a parameter

$$\left \{ \textstyle\begin{array}{l} -{D^{\eta-2}_{0+} } (u''( t ))+ \lambda f ( {t,u ( t )} ) =0,\quad t \in J,\\ u''(0)=u'''(0)= \cdots=u^{n-2}(0)=0,\\ ^{C}{D^{k-2}_{0+} } (u''( t )) \vert_{t=1}=0,\\ \alpha u(0) - \beta u'(0)= \int_{0}^{1}u(s)\,d A(s),\\ \gamma u(1) + \delta u'(1)= \int_{0}^{1}u(s)\,d B(s), \end{array}\displaystyle \right . $$

where $D^{\eta-2}_{0+}$, $D^{k-2}_{0+}$ are the standard Riemann–Liouville fractional derivative, $n-1 < \eta\leq n$, $\eta \geq4$, $2 \leq k \leq n-2$, $\alpha, \beta, \gamma, \delta>0$. $\int_{0}^{1}u(s)\,d A(s)$ and $\int_{0}^{1}u(s)\,d B(s)$ denote the Riemann–Stieltjes integrals of u with respect to A and B. $A(t)$, $B(t)$ are nondecreasing on $[0, 1]$, $f : [0, 1] \times [0, +\infty) \rightarrow[0, +\infty)$ is continuous, $\lambda> 0$ is a parameter. By using the Guo–Krasnoselskii fixed point theorem on cones, under different conditions of nonlinearity, existence and nonexistence, results for positive solutions are derived in terms of different parameter intervals.

The purpose of this paper is to establish conditions ensuring the existence of three positive solutions of BVP (1) and give an estimate of these solutions by using the Avery–Peterson fixed point theorem. Our supposed problem is different from the problems studied before and mentioned above. Our result is new and our work extends the application of the theorem.

In this paper, a positive solution $x=x(t)$ of BVP (1) means a solution of (1) satisfying $x(t)>0$, $t\in[0,1]$.

Throughout this paper, we always assume that the following condition is satisfied:

(L₀):

$0< a< b<2a<+\infty$, $0\leq g_{0}(t) \leq g_{1}(t) \leq 2g_{0}(t)$, $0\leq\int_{0}^{1}g_{0}(s)\,ds$, $\int_{0}^{1}g_{1}(s)\,ds < 1$.

2 Preliminaries and lemmas

The basic theory of fractional-order differential equation and boundary value problem can be obtained from many places in the literature, which will not be repeated here; see [1–9]. Here we present some necessary basic results that will be used.

Lemma 2.1

(see [2])

The Caputo fractional derivative of order $n-1<\alpha<n$ for $t^{\beta}$ is given by

$$ D_{0 + }^{\alpha}{t^{\beta}} = \left \{ { \textstyle\begin{array}{l@{\quad}l} {\frac{{\varGamma(\beta + 1)}}{{\varGamma(\beta - \alpha + 1)}}{t^{\beta - \alpha}},}&{\beta \in N\textit{ and }\beta \ge n\textit{ or }\beta \notin N\textit{ and }\beta > n - 1,}\\ {0,}&{\beta \in\{ 0,1, \ldots,n - 1\} .} \end{array}\displaystyle } \right . $$

Lemma 2.2

(see [19])

Let $h\in C[0,1]$and $1< m-1<\beta<m$. Then the BVP

$$ \left \{ \textstyle\begin{array}{l} D_{{0^{+} }}^{\beta}u(t)=h(t),\quad 0< t< 1, \\ u(1) = u^{(i)}(0) = 0, \quad i=1,2,\ldots,m-1, \end{array}\displaystyle \right . $$

has an unique solution

$$ u(t)=- \int_{0}^{1}H(t,s)h(s)\,ds, $$

where

$$ H(t,s)=\frac{1}{\varGamma(\beta)}\left \{ \textstyle\begin{array}{l@{\quad}l} (1-s)^{\beta-1}-(t-s)^{\beta-1}, & 0\leq s\leq t\leq1, \\ (1-s)^{\beta-1},&0\leq t\leq s\leq1. \end{array}\displaystyle \right . $$

(2)

Denote

$$ \begin{gathered} M_{1}= \int_{0}^{1}g_{0}(s)\,ds,\qquad M_{2}= \int_{0}^{1}sg_{0}(s)\,ds,\\ N_{1}= \int_{0}^{1}g_{1}(s)\,ds,\qquad N_{2}= \int_{0}^{1}sg_{1}(s)\,ds,\\ \delta^{-1}=1+M_{2}+N_{1}-M_{2}N_{1}+M_{1}(N_{2}-2)-N_{2},\\ \omega(t) =\delta \bigl( b(M_{2}-1+t-M_{1}t)+a(2-N_{2}-t+N_{1}t) \bigr).\end{gathered} $$

(3)

From (L₀), we know, for $t \in(0,1)$,

$$ 1>N_{1}>N_{2}>M_{2}>0,\qquad 1>N_{1}>M_{1}>M_{2}>0,\qquad 2M_{1}>N_{1},2M_{2}>N_{2}. $$

(4)

Thus

$$\begin{aligned} \delta^{-1} =& 1+M_{2}+N_{1}-M_{2}N_{1}+M_{1}(N_{2}-2)-N_{2} \\ =& 1+M_{2}+N_{1}-M_{2}N_{1}+M_{1}N_{2}-2M_{1}-N_{2} \\ =& (1+M_{1}N_{2}-2M_{1})+M_{2}+N_{1}-M_{2}N_{1}-N_{2} \\ =& (1-M_{1})+(M_{1}N_{2}-N_{2})+(M_{2}-M_{2}N_{1})+(N_{1}-M_{1}) \\ =& (1-M_{1})+N_{2}(M_{1}-1)+M_{2}(1-N_{1})+(N_{1}-M_{1}) \\ =& (1-M_{1}) (1-N_{2})+M_{2}(1-N_{1})+(N_{1}-M_{1}) \\ >& 0. \end{aligned}$$

Thus, the following lemma holds.

Lemma 2.3

Let (L₀) hold, $y\in C[0,1]$and $1< n-1<\alpha<n$, $1< m-1<\beta <m$, then the following boundary value problem:

$$ \left \{ \textstyle\begin{array}{l} D_{{0^{+} }}^{\alpha}x(t)=y(t),\quad0< t< 1, \\ x^{(j)}(0) = 0, \quad j= 2, 3, \ldots, n-1,\\ x(0) + x^{{\prime}}(0) = \int_{0}^{1}g_{0}(s)x(s)\,ds + a, \\ x(1) + x^{{\prime}}(1) = \int_{0}^{1}g_{1}(s)x(s)\,ds + b, \end{array}\displaystyle \right . $$

(5)

has an unique solution

$$ x(t)= \int_{0}^{1}G(t,s)y(s)\,ds+\omega(t) $$

(6)

and

$$ D_{0+}^{\beta}x(t)=\frac{1}{\varGamma(\alpha-\beta)} \int_{0}^{t}(t-s)^{\alpha-\beta-1}y(s)\,ds, $$

(7)

where

$$\begin{aligned}& G(t,s) = G_{1}(t,s)+G_{2}(t,s), \end{aligned}$$

(8)

$$\begin{aligned}& G_{1}(t,s) = \frac{1}{\varGamma(\alpha)}\left \{ \textstyle\begin{array}{l@{\quad}l} (t-s)^{\alpha-1}+(1-t)(1-s)^{\alpha-2}(\alpha -s), & 0\leq s\leq t\leq1, \\ (1-t)(1-s)^{\alpha-2}(\alpha-s),&0\leq t\leq s\leq1, \end{array}\displaystyle \right . \end{aligned}$$

(9)

$$\begin{aligned}& G_{2}(t,s) = \delta \int_{0}^{1} \bigl((M_{2}-1+t-M_{1}t)g_{1}( \tau)+(2-N_{2}-t+N_{1}t)g_{0}(\tau) \bigr) G_{1}(\tau,s)\,d\tau. \end{aligned}$$

(10)

Proof

Consider BVP (5), we have

$$ x(t)=\frac{1}{\varGamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}y(s) \,ds+C_{0}+C_{1}t+C_{2}t^{2}+ \cdots+C_{n-1}t^{n-1}. $$

In view of $x^{(j)}(0)=0$ ($j=2, 3,\ldots,n-1$), we know $C_{2}=C_{3}=\cdots=C_{n-1}=0$, and

$$ \begin{gathered} x(0)=C_{0},\qquad x^{{\prime}}(0)=C_{1}, \qquad x(1)= \frac{1}{\varGamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-1}y(s) \,ds+C_{0}+C_{1}, \\ x^{{\prime}}(1)=\frac{1}{\varGamma(\alpha-1)} \int_{0}^{1}(1-s)^{\alpha-2}y(s) \,ds+C_{1},\end{gathered} $$

so that

$$ x(t)=\frac{1}{\varGamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}y(s) \,ds+C_{0}+C_{1}t . $$

(11)

From the boundary condition of BVP (5), by methods similar to Lemma 2.4 in [19], through traditional analytical calculation and integration techniques, we have

$$\begin{aligned} x(t) =& \int_{0}^{1}G_{1}(t,s)y(s)\,ds+ \int_{0}^{1}G_{2}(t,s)y(s)\,ds+\omega(t) \\ =& \int_{0}^{1}G(t,s)y(s)\,ds+\omega(t), \end{aligned}$$

where $\omega(t)$, $G_{1}(t,s)$ and $G_{2}(t,s)$ are given by (3), (9) and (10).

On the other hand, in view of (11), because $1< m-1<\beta<\alpha -1<n-1$, by Lemma 2.1, we have

$$\begin{aligned} D_{0+}^{\beta}x(t) =& D_{0+}^{\beta} \biggl(\frac{1}{\varGamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}y(s) \,ds+C_{0}+C_{1}t \biggr) \\ =& D_{0+}^{\beta} \bigl(I_{0+}^{\alpha}y(t)+C_{0}+C_{1}t \bigr) \\ =& D_{0+}^{\beta}I_{0+}^{\alpha}y(t)+D_{0+}^{\beta }(C_{0})+D_{0+}^{\beta}(C_{1}t) \\ =& D_{0+}^{\beta}I_{0+}^{\alpha}y(t) \\ =& I_{0+}^{\alpha-\beta}y(t) \\ =& \frac{1}{\varGamma(\alpha-\beta)} \int_{0}^{t}(t-s)^{\alpha-\beta-1}y(s)\,ds. \end{aligned}$$

(12)

 □

Lemma 2.4

The BVP (1) is equivalent to the following integral equation:

$$ x(t)= \int_{0}^{1}G(t,s)\varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x( \tau),D_{0+}^{\beta}x(\tau) \bigr)\,d\tau \biggr)\,ds+\omega(t) $$

(13)

and

$$ D_{0+}^{\beta}x(t)=\frac{1}{\varGamma(\alpha-\beta)} \int_{0}^{t}(t-s)^{\alpha-\beta-1} \varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x( \tau),D_{0+}^{\beta}x(\tau) \bigr)\,d\tau \biggr)\,ds, $$

(14)

where $H(t,s)$, $\omega(t)$and $G(t,s)$are given by (2), (3) and (8).

Proof

From Lemma 2.2 and Lemma 2.3, let $y(t)=\varphi_{q}(u(t))$, $h(t)=-f (t,x(t),D_{0+}^{\beta}x(t) )$, we have

$$\begin{aligned} y(t) =& \varphi_{q} \bigl(u(t) \bigr)= \varphi_{q} \biggl( \int_{0}^{1}H(t,s)f \bigl(s,x(s),D_{0+}^{\beta}x(s) \bigr)\,ds \biggr) \\ =& \varphi_{q} \biggl( \int_{0}^{1}H(t,s)f \bigl(s,x(s),D_{0+}^{\beta}x(s) \bigr)\,ds \biggr). \end{aligned}$$

Immediately we obtain

$$\begin{aligned} x(t)&= \int_{0}^{1}G(t,s)y(s)\,ds+\omega(t) \\ &= \int_{0}^{1}G(t,s)\varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau ),D_{0+}^{\beta}x(\tau) \bigr)\,d\tau \biggr)\,ds+\omega(t). \end{aligned} $$

From (12), (14) holds.

On the other hand, if $x(t)$ satisfies (13), we can easily prove that $x(t)$ satisfies BVP (1). □

Lemma 2.5

Assume (L₀) hold, then the function $H(t,s)$defined by (2), the function $G(t,s)$defined by (8), and then the function $\omega(t)$defined by (3) satisfies

• (1) $H(t,s)\geq0$is continuous for all $t,s\in[0,1]$;

• (2) $H(t,s)\leq H(s,s)$for all $t,s\in[0,1]$;

• (3) $\int_{0}^{1}H(t,s)\,ds=\frac{1-t^{\beta}}{\varGamma(\beta +1)} \leq\frac{1}{\varGamma(\beta+1)}$for all $t\in[0,1]$;

• (4) $G(t,s)\geq0$is continuous for all $t,s\in[0,1]$;

• (5) $\omega(t)>0$for all $t\in[0,1]$.

Proof

(1) and (2) are proved in [19], we omit the proofs.

(3) For $t \in[0,1]$, by a simple integral operation, we can obtain

$$\begin{aligned} \int_{0}^{1}H(t,s)\,ds&= \int_{0}^{t} \bigl((1-s)^{\beta-1}-(t-s)^{\beta-1} \bigr)\,ds + \int_{t}^{1} (1-s)^{\beta-1}\,ds\\ &= \frac{1-t^{\beta}}{\varGamma(\beta +1)} \leq\frac{1}{\varGamma(\beta+1)}.\end{aligned} $$

(4) From (9), we know $G_{1}(t,s)\geq0$, $t,s\in[0,1]$, and $G_{1}(t,s)>0$, $t,s\in(0,1)$. Combined with (L₀), for $t\in[0,1]$, we have

$$\begin{aligned} \frac{\partial G_{2}(t,s)}{\partial t} =& \delta \int_{0}^{1} \bigl((1-M_{1})g_{1}( \tau)+(-1+N_{1})g_{0}(\tau) \bigr)G_{1}(\tau,s)\,d \tau \\ =& \delta \int_{0}^{1} \bigl(g_{1}( \tau)-M_{1}g_{1}(\tau)-g_{0}(\tau)+N_{1}g_{0}( \tau) \bigr)G_{1}(\tau,s)\,d\tau \\ =& \delta \int_{0}^{1} \bigl(g_{1}( \tau)-g_{0}(\tau)-M_{1}g_{1}(\tau)+N_{1}g_{0}( \tau) \bigr)G_{1}(\tau,s)\,d\tau \\ >& \delta \int_{0}^{1} \bigl(g_{1}( \tau)-g_{0}(\tau)-M_{1}g_{1}(\tau)+M_{1}g_{0}( \tau) \bigr)G_{1}(\tau,s)\,d\tau \\ =& \delta \int_{0}^{1} \bigl(g_{1}( \tau)-g_{0}(\tau)-M_{1} \bigl(g_{1}( \tau)-g_{0}(\tau) \bigr) \bigr)G_{1}(\tau,s)\,d\tau \\ =& \delta \int_{0}^{1} \bigl(g_{1}( \tau)-g_{0}(\tau) \bigr) (1-M_{1}) G_{1}(\tau,s)\,d \tau \\ >&0, \end{aligned}$$

so that $G_{2}(t,s)$ monotonically increase with respect to t.

As a consequence, from (10), we get

$$\begin{aligned} G_{2}(t,s) \geq& G_{2}(0,s)=\delta \int_{0}^{1} \bigl((M_{2}-1)g_{1}( \tau)+(2-N_{2})g_{0}(\tau) \bigr)G_{1}(\tau,s)\,d \tau \\ \geq&\delta \int_{0}^{1} \biggl((M_{2}-1)g_{1}( \tau)+\frac{1}{2}(2-N_{2})g_{1}(\tau) \biggr)G_{1}(\tau,s)\,d\tau \\ =&\delta \int_{0}^{1} \biggl(M_{2}- \frac{1}{2}N_{2} \biggr)g_{1}(\tau)G_{1}( \tau,s)\,d\tau \\ \geq&0. \end{aligned}$$

Hence, $G(t,s)\geq0$.

(5) From (3), for $t\in[0,1]$, we know

$$\begin{aligned} \omega'(t) =& b(1-M_{1})+a(N_{1}-1) \\ >& a(1-M_{1})+a(N_{1}-1) \\ >& a(1-N_{1})+a(N_{1}-1) \\ =& 0, \end{aligned}$$

so that

$$\begin{aligned} \omega(t) \geq& \omega(0)=b(M_{2}-1)+a(2-N_{2}) \\ >& b(M_{2}-1)+a(2-2M_{2}) \\ =& (1-M_{2}) (2a-b) \\ >& 0,\quad t\in[0,1]. \end{aligned}$$

 □

Lemma 2.6

Let $\eta\in(0,\frac{1}{2})$, then

$$\begin{gathered} \max_{t\in[0,1] }G(t,s) \leq G_{1}(0,s)+G_{2}(1,s) , \\ \min_{t\in[0,\eta]}G(t,s) \geq \rho \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr),\end{gathered} $$

where

$$\begin{aligned} \rho=\frac{M_{2}-\frac{1}{2}N_{2}}{1-M_{1}+M_{2}+N_{1}-N_{2}}. \end{aligned}$$

(15)

Proof

Step 1: We prove

$$\begin{aligned} \min_{t\in[0,\eta]}G_{1}(t,s)\geq (1-\eta)G_{1}(0,s)> \frac{1}{2} \max_{t\in[0,1] }G_{1}(t,s). \end{aligned}$$

(16)

For $0\leq s < t\leq1$ and $t\in[0,\eta]$,

$$\varGamma(\alpha)\frac{\partial G_{1}(t,s)}{\partial t}=(\alpha -1) (t-s)^{\alpha-2}-(\alpha-s) (1-s)^{\alpha-2}\leq(\alpha -s) (t-s)^{\alpha-2}-(\alpha-s) (1-s)^{\alpha-2}< 0, $$

so that

$$ G_{1}(t,s)\leq G_{1}(s,s)=(\alpha-s) (1-s)^{\alpha-1}< ( \alpha-s) (1-s)^{\alpha-2}=G_{1}(0,s) $$

and

$$\begin{aligned} \frac{G_{1}(t,s)}{G_{1}(0,s)} =& \frac{(t-s)^{\alpha-1}+(\alpha -s)(1-t)(1-s)^{\alpha-2}}{(\alpha-s)(1-s)^{\alpha-2}} \\ =&\frac{(t-s)^{\alpha-1}}{(\alpha-s)(1-s)^{\alpha-2}}+1-t \\ \geq& 1-t \\ \geq& 1-\eta> \frac{1}{2}. \end{aligned}$$

For $s\geq t$ and $t \in[0,\eta]$,

$$ \varGamma(\alpha)\frac{\partial G_{1}(t,s)}{\partial t}=-(\alpha-s) (1-s)^{\alpha-2}< 0 $$

so that

$$ G_{1}(t,s)\leq G_{1}(0,s)=(\alpha-s) (1-s)^{\alpha-2} $$

and

$$\begin{aligned} \frac{G_{1}(t,s)}{G_{1}(0,s)} =& \frac{(\alpha-s)(1-t)(1-s)^{\alpha -2}}{(\alpha-s)(1-s)^{\alpha-2}}=1-t \geq1-\eta > \frac{1}{2}. \end{aligned}$$

Therefore, (16) holds.

Step 2: We prove

$$\begin{aligned} \min_{t\in[0,\eta]}G_{2}(t,s)\geq \rho G_{2}(1,s)=\rho\max_{t\in[0,1] }G_{2}(t,s). \end{aligned}$$

(17)

From Lemma 2.5, we know that $G_{2}(t,s)$ is a monotone increasing function with respect to $t \in[0,1]$, so that

$$\min_{t\in[0,\eta]}G_{2}(t,s)=G_{2}(0,s),\qquad \max _{t\in[0,1] }G_{2}(t,s)=G_{2}(1,s). $$

By (L₀) and (4), we have

$$\begin{aligned} \frac{G_{2}(0,s)}{G_{2}(1,s)} =& \frac{\delta\int_{0}^{1} ((M_{2}-1)g_{1}(\tau)+(2-N_{2})g_{0}(\tau) ) G_{1}(\tau,s)\,d\tau }{\delta\int_{0}^{1} ((M_{2}-1+1-M_{1})g_{1}(\tau )+(2-N_{2}-1+N_{1})g_{0}(\tau) ) G_{1}(\tau,s)\,d\tau} \\ \geq& \frac{ \int_{0}^{1} ((M_{2}-1)g_{1}(\tau)+\frac {1}{2}(2-N_{2})g_{1}(\tau) ) G_{1}(\tau,s)\,d\tau}{ \int_{0}^{1} ((M_{2}-M_{1})g_{1}(\tau)+(1-N_{2}+N_{1})g_{1}(\tau) ) G_{1}(\tau,s)\, d\tau} \\ =& \frac{ \int_{0}^{1}(M_{2}-1+1-\frac{1}{2}N_{2})g_{1}(\tau)G_{1}(\tau ,s)\,d\tau}{ \int_{0}^{1}(M_{2}-M_{1}+1-N_{2}+N_{1})g_{1}(\tau) G_{1}(\tau,s)\,d\tau} \\ =& \frac{ (M_{2}-\frac{1}{2}N_{2}) \int_{0}^{1}g_{1}(\tau)G_{1}(\tau ,s)\,d\tau}{ (1-M_{1}+M_{2}+N_{1}-N_{2})\int_{0}^{1}g_{1}(\tau) G_{1}(\tau,s)\,d\tau} \\ =& \frac{ M_{2}-\frac{1}{2}N_{2}}{ 1-M_{1}+M_{2}+N_{1}-N_{2}} \\ =&\rho. \end{aligned}$$

Obviously, $\rho>0$ and

$$\rho- \frac{1}{2}=\frac{ M_{2}-\frac{1}{2}N_{2}}{ 1-M_{1}+M_{2}+N_{1}-N_{2}}- \frac{1}{2}= \frac{ M_{2}- 1+M_{1}-N_{1}}{ 2 (1-M_{1}+M_{2}+N_{1}-N_{2} ) }< 0, $$

so that $0<\rho<\frac{1}{2}$ and (17) hold.

Finally, from (16) and (17), we can easily show that the following results hold:

$$\begin{aligned} \max_{t\in[0,1] }G(t,s) =&\max_{t\in[0,1] } \bigl( G_{1}(t,s)+G_{2}(t,s) \bigr) \\ \leq& \max_{t\in[0,1] }G_{1}(t,s)+\max _{t\in[0,1] }G_{2}(t,s) \\ = & G_{1}(0,s)+G_{2}(1,s) \end{aligned}$$

and

$$\begin{aligned} \min_{t\in[0,\eta]}G(t,s) \geq&\min _{t\in[0,\eta]}G_{1}(t,s)+\min_{t\in[0,\eta]}G_{2}(t,s) \\ \geq& \frac{1}{2}G_{1}(0,s)+\rho G_{2}(1,s) \\ >&\rho G_{1}(0,s)+\rho G_{2}(1,s) \\ =&\rho \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr) \\ \geq&\rho\max_{t\in[0,1] }G(t,s). \end{aligned}$$

 □

Lemma 2.7

Assume (L₀) hold, then the function $\omega(t)$satisfies the following properties:

• (1) $\omega(t)\leq\omega(1)=\max_{t\in[0,1]}\omega(t)$;

• (2) $\min_{t\in[0,\eta]}\omega(t)\geq\rho\max_{t\in [0,1]}\omega(t)$, whereρis given by (15).

Proof

From Lemma 2.5 and (3), we have

$$ \begin{gathered}\min_{t\in[0,\eta]}\omega(t)=\omega(0)= \delta \bigl(b(M_{2}-1)+a(2-N_{2}) \bigr), \\ \max_{t\in[0,1]}\omega(t)=\omega(1)=\delta \bigl(b(M_{2}-M_{1})+a(1-N_{2}+N_{1}) \bigr), \end{gathered} $$

and

$$\begin{aligned} \frac{\omega(0)}{\omega(1)} =& \frac{\delta (b(M_{2}-1)+a(2-N_{2}))}{\delta(b(M_{2}-M_{1})+a(1-N_{2}+N_{1}))} \\ \geq& \frac{b(M_{2}-1)+\frac {1}{2}b(2-N_{2})}{b(M_{2}-M_{1})+b(2-N_{2}-1+N_{1})} \\ =& \frac{ M_{2}-\frac{1}{2}N_{2}}{ 1-M_{1}+M_{2}+N_{1}-N_{2}} \\ =& \rho. \end{aligned}$$

Hence,

$$ \omega(t)\leq\omega(1)=\max_{t\in[0,1]}\omega(t),\qquad \min _{t\in[0,\eta]}\omega(t)\geq\rho\max_{t\in[0,1]}\omega(t). $$

 □

To finish this section, we present the well-known Avery–Peterson fixed point theorem as follows.

Let γ and θ be nonnegative continuous convex functionals on P, φ be a nonnegative continuous concave functional on P, and ψ be a nonnegative continuous functional on P. For $A,B,C,D>0$, we define the following convex set:

$$\begin{gathered} P(\gamma; D)= \bigl\{ x\in P: \gamma(x)< D \bigr\} , \\ P(\gamma, \varphi; B, D) = \bigl\{ x\in P: B\leq\varphi(x),\gamma (x)\leq D \bigr\} , \\ P(\gamma, \theta, \varphi; B, C, D) = \bigl\{ x\in P: B\leq\varphi (x),\theta(x) \leq C, \gamma(x)\leq D \bigr\} ,\end{gathered} $$

and a closed set

$$ P(\gamma, \psi; A,D) = \bigl\{ x\in P: A\leq\psi(x),\gamma(x)\leq D \bigr\} . $$

Lemma 2.8

(see [27])

LetPbe a cone in a real Banach spaceE. Letγandθbe nonnegative continuous convex functionals onP, φbe a nonnegative continuous concave functional onP, andψbe a nonnegative continuous functional onPsatisfying $\psi(\lambda x)\leq\lambda\psi(x)$for $0\leq\lambda\leq1$, such that, for some positive numbersMandD, $\varphi(x)\leq\psi(x)$, $\Vert x \Vert\leq M\gamma(x)$for all $x\in\overline{P(\gamma;D)}$. Suppose

$$T:\overline{P(\gamma;D)}\rightarrow\overline{P(\gamma;D)} $$

is completely continuous and there exist positive numbersA, B, andCwith $A < B$such that

$(H1)$:

$\{x\in P(\gamma,\theta,\varphi;B,C,D):\varphi(x)>B\} \neq{\O}$, and $\varphi(x)>B$for $x\in P(\gamma,\theta,\varphi;B,C,D)$;

$(H2)$:

$\varphi(Tx)>B$for $x\in P(\gamma,\varphi;B,D)$with $\theta(Tx)>A$;

$(H3)$:

$0\notin P(\gamma,\psi;A,D)$and $\psi(Tx)< A$for $x\in P(\gamma,\psi;A,D)$with $\psi(x)=A$.

Then T has at least three fixed point $x_{1},x_{2},x_{3}\in\overline {P(\gamma;D)}$ such that

$$ \gamma(x_{i})\leq D,\quad i=1,2,3; \qquad\varphi(x_{1})>B, \qquad A< \varphi(x_{2}),\qquad \psi(x_{2})< B; \qquad \psi(x_{3})< A. $$

3 Main results

In this section, we prove the existence of positive solution of BVP (1) by applying the following Avery–Peterson fixed point theorem.

We consider the Banach space $E=\{x\in C[0,1]:D_{0+}^{\beta}x\in C[0,1]\}$ with the norm

$$ \Vert x \Vert =\max \Bigl\{ \max_{t\in[0,1]} \bigl\vert x(t) \bigr\vert ,\max_{t\in[0,1]} \bigl\vert D_{0+}^{\beta}x(t) \bigr\vert \Bigr\} . $$

Let

$$ P= \Bigl\{ x\in E: x(t)\geq0, D_{0+}^{\beta}x(t) \geq0, \min _{t\in[0,\eta]}x(t) \geq\rho\max_{t\in[0,1]} x(t) \Bigr\} , $$

then P is a cone in E.

Define the operator $T:P \rightarrow E$ by

$$ Tx(t):= \int_{0}^{1}G(t,s)\varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x( \tau),D_{0+}^{\beta}x(\tau) \bigr)\,d\tau \biggr)\,ds+\omega(t). $$

Lemma 3.1

Assume (L₀) hold, then $T:P \rightarrow P$is a completely continuous operator.

Proof

For $x \in P$, it is easy to see that T is continuous operator and $Tx(t)\geq0$. By (14), we have

$$D_{0+}^{\beta}Tx(t)=\frac{1}{\varGamma(\alpha-\beta)} \int _{0}^{t}(t-s)^{\alpha-\beta-1} \varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x( \tau),D_{0+}^{\beta}x(\tau ) \bigr)\,d\tau\biggr)\,ds \geq0. $$

From Lemma 2.5 and Lemma 2.6 and Lemma 2.7, similar to Lemma 3.1 in [19], we can easily prove that T is a completely continuous operator. □

Define continuous nonnegative convex functionals as

$$ \gamma(x)= \Vert x \Vert ,\qquad \theta(x)=\psi(x)=\max_{t \in[0,1]} \bigl\vert x(t) \bigr\vert . $$

Define continuous nonnegative concave functionals as

$$ \varphi(x)=\min_{t \in[0,\eta]} \bigl\vert x(t) \bigr\vert . $$

Thus

$$ \rho\theta(x) \leq\varphi(x)\leq\theta(x)=\psi(x),\qquad \Vert x \Vert \leq M \gamma(x), $$

where $M=1$.

Let

$$\begin{gathered} J_{1}= \int_{0}^{1} \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr)\varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds , \\ J_{2}=\frac{1}{\varGamma(\alpha-\beta)} \int_{0}^{1}(1-s)^{\alpha-\beta-1} \varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds ,\end{gathered} $$

and

$$ J_{3}= \int_{0}^{1} \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr)\varphi_{q} \biggl( \int_{0}^{\eta}H(s,\tau)\,d\tau \biggr)\,ds. $$

Theorem 3.1

Suppose (L₀) hold, and there exist constants $A, B, D \geq\omega (1)$with $A < B <\rho D \min\{\frac{J_{3}}{J_{1}},\frac {J_{3}}{J_{2}}\}$and $C=\frac{B}{\rho}$, such that

$(L_{1})$:

$f(t,x,y) \leq\min \{\varphi_{p}(\frac{D-\omega (1)}{J_{1}}), \varphi_{p}(\frac{D}{J_{2}}) \}$, $(t,x,y)\in[0,1] \times [0,D] \times[0,D]$;

$(L_{2})$:

$f(t,x,y) > \varphi_{p}(\frac{B-\rho\omega(1)}{\rho J_{3}})$, $(t,x,y)\in[0,\eta]\times[B,\frac{B}{\rho}]\times[0,D]$;

$(L_{3})$:

$f(t,x,y) < \varphi_{p}(\frac{A-\omega(1)}{J_{1}})$, $(t,x,y)\in[0,1]\times[0,A]\times[0,D]$.

Then BVP (1) has at least three positive solutions $x_{1}$, $x_{2}$, $x_{3}$, satisfying

$$\begin{aligned}& \Vert x_{i} \Vert \leq D\quad (i=1,2,3), \end{aligned}$$

(18)

$$\begin{aligned}& \min_{t\in[0,\eta]} \bigl\vert x_{1}(t) \bigr\vert > B,\qquad A< \min_{t\in[0,\eta]} \bigl\vert x_{2}(t) \bigr\vert , \qquad\max_{t\in[0,1]} \bigl\vert x_{2}(t) \bigr\vert < B, \qquad\max_{t\in[0,1]} \bigl\vert x_{3}(t) \bigr\vert < A. \end{aligned}$$

(19)

Proof

Obviously, the function x is a positive solution of BVP (1) if and only if x is a fixed point of the operator T in P.

For $x \in\overline{P(\gamma;D)}$, we get

$$ \max_{t \in[0,1]} \bigl\vert x(t) \bigr\vert \leq D,\qquad \max _{t \in[0,1]} \bigl\vert D_{0+}^{\beta}x(t) \bigr\vert \leq D, $$

this implies

$$ 0 \leq x(t),\qquad D_{0+}^{\beta}x(t) \leq D. $$

From $(L_{1})$, we get

$$\begin{aligned} \max_{t \in[0,1]} \bigl\vert Tx(t) \bigr\vert \leq& \int_{0}^{1} \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr)\varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x( \tau),D_{0+}^{\beta}x(\tau) \bigr)\,d\tau \biggr)\,ds+\omega(1) \\ \leq& \int_{0}^{1} \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr)\varphi_{q} \biggl(\varphi_{p} \biggl( \frac{D-\omega(1)}{J_{1}} \biggr) \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds+\omega(1) \\ =& \frac{D-\omega(1)}{J_{1}} \int_{0}^{1} \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr)\varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds+\omega(1) \\ =& D \end{aligned}$$

and

$$\begin{aligned}& \max_{t \in[0,1]} \bigl\vert D_{0+}^{\beta}Tx(t) \bigr\vert \\& \quad= \max_{t \in[0,1]} \biggl\vert \frac{1}{\varGamma(\alpha-\beta)} \int_{0}^{t}(t-s)^{\alpha-\beta-1} \varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x( \tau),D_{0+}^{\beta}x(\tau) \bigr)\,d\tau \biggr)\,ds \biggr\vert \\& \quad\leq \frac{1}{\varGamma(\alpha-\beta)} \int_{0}^{1}(1-s)^{\alpha-\beta-1} \varphi_{q} \biggl(\varphi_{p} \biggl(\frac{D}{J_{2}} \biggr) \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds \\& \quad= \frac{D}{J_{2}} \frac{1}{\varGamma(\alpha-\beta)} \int_{0}^{1}(1-s)^{\alpha-\beta-1} \varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds \\& \quad= D, \end{aligned}$$

so that

$$ \gamma(Tx)= \Vert Tx \Vert = \max \Bigl\{ \max_{t \in[0,1]} \bigl\vert Tx(t) \bigr\vert , \max_{t \in[0,1]} \bigl\vert D_{0+}^{\beta}Tx(t) \bigr\vert \Bigr\} \leq D. $$

Therefore $T:\overline{P(\gamma;D)} \rightarrow\overline{P(\gamma;D)}$.

From $\frac{B}{\rho} \in P(\gamma,\theta,\varphi;B,C,D)$ and $\varphi (\frac{B}{\rho}) > B$, we have

$$ \bigl\{ x \in P(\gamma,\theta,\varphi;B,C,D):\varphi(x) > B \bigr\} \neq {\O}. $$

For $x \in P(\gamma,\theta,\varphi;B,C,D)$, we know that $B \leq x(t) \leq C=\frac{B}{\rho}$ for $t\in[0,\eta]$ and $0 \leq{ D_{0+}^{\beta}x(t) \leq D}$.

By $(L_{2})$,

$$\begin{aligned} \varphi(Tx) =& \min_{t \in[0,\eta]} \bigl\vert Tx(t) \bigr\vert \\ \geq& \int_{0}^{1} \rho \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr)\varphi_{q} \biggl( \int_{0}^{\eta}H(s,\tau) \varphi_{p} \biggl( \frac{B-\rho\omega(1)}{\rho J_{3}} \biggr)\,d\tau \biggr)\,ds+\rho\omega (1) \\ =& \rho\frac{B-\rho\omega(1)}{\rho J_{3}} \int_{0}^{1} \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr)\varphi_{q} \biggl( \int_{0}^{\eta}H(s,\tau)\,d\tau \biggr)\,ds+\rho \omega(1) \\ =& B. \end{aligned}$$

So $\varphi(Tx)>B$ for all $x \in P(\gamma,\theta,\varphi;B,C,D)$. Hence, the condition $(H1)$ of Lemma 2.8 is satisfied.

For all $x \in P(\gamma,\varphi;B,D)$ with $\theta(Tx)>C=\frac{B}{\rho }$, we have

$$ \varphi(Tx) \geq\rho\theta(Tx) > \rho C= \rho\frac{B}{\rho} =B. $$

Thus, the condition $(H2)$ of Lemma 2.8 holds.

Because of $\psi(0) = 0 < A$, then $0 \notin P(\gamma,\psi;A,D)$. For $x \in P(\gamma,\psi;A,D)$ with $\psi(x)=A$, we know $\gamma(x) \leq D$. It means that $\max_{t \in[0,1]}x(t)=A$ and $0 \leq D_{0+}^{\beta }x(t) \leq D$.

From $(L_{3})$, we can obtain

$$\begin{aligned} \psi(Tx) =& \max_{t \in[0,1]} \bigl\vert Tx(t) \bigr\vert \\ \leq& \max_{t \in[0,1]} \int_{0}^{1}G(t,s)\varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x( \tau),D_{0+}^{\beta}x(\tau) \bigr)\,d\tau \biggr)\,ds+\omega(1) \\ < & \int_{0}^{1} \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr)\varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)\varphi_{p} \biggl( \frac{A-\omega(1)}{J_{1}} \biggr)\,d\tau \biggr)\,ds+\omega(1) \\ =& \frac{A-\omega(1)}{J_{1}} \int_{0}^{1} \bigl(G_{1}(0,s)+G_{2}(1,s) \bigr)\varphi_{q} \biggl( \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds+\omega(1) \\ =& A. \end{aligned}$$

Therefore, the condition $(H3)$ of Lemma 2.8 holds.

To sum up, the conditions of Lemma 2.8 are all verified and we notice that $x_{i}(t)\geq{\omega(0)>0}$. Hence, BVP (1) has at least three positive solutions $x_{1}$, $x_{2}$, $x_{3}$ satisfying (18) and (19). □

4 Example

Consider the following boundary value problem:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} D_{0+}^{\frac{7}{3}}\varphi_{\frac{3}{2}} ( D_{0+}^{\frac{11}{3}}x(t) ) =f (t,x(t),D_{0+} ^{\frac{7}{3}}x(t) ),\quad t \in(0,1),\\ ( \varphi_{\frac{3}{2}} ( D_{0+}^{\frac{11}{3}}x(0) ) )^{\prime} =\varphi_{\frac{3}{2}} ( D_{0+}^{\frac{11}{3}}x(1) )=0, \\ x(0)+x^{\prime}(0)= \int_{0}^{1}sx(s)\,ds+\frac{5}{3}, \\ x(1)+x^{\prime}(1) = \int_{0}^{1}(s^{2}+s)x(s)\,ds+\frac{7}{3}, \\ x^{\prime\prime}(0)=x^{(3)} = 0 , \end{array}\displaystyle \right . $$

(20)

where $\alpha=\frac{11}{3}$, $\beta=\frac{7}{3}$, $p=\frac{3}{2}$, $g_{0}(t)=t$, $g_{1}(t)=t^{2}+t$, and

$$\begin{aligned} f(t,x,y) =& \left \{ \textstyle\begin{array}{l@{\quad}l} \tan(\frac{t}{100})+2^{x/10} x^{2}+\cos(y),& 0\leq x\leq65,\\ \tan(\frac{t}{100})+\cos(y)+160 \sqrt{5},&65< x\leq10{,}000. \end{array}\displaystyle \right . \end{aligned}$$

Choose $A=3$, $B=65$, $D=25{,}000$, $\eta=\frac{1}{4}$. By simple computation, we have

$$ \begin{gathered} \rho= 0.0384615,\qquad M=1,\qquad\omega(1)=2.83721, \\ M_{1} = \frac{1}{2},\qquad M_{2} = \frac{1}{3},\qquad N_{1} = \frac{5}{6},\qquad N_{2} =\frac{7}{12},\qquad\delta= \frac{72}{43}, \\ J_{1} =0.147109,\qquad J_{2} = 0.0722385,\qquad J_{3}=0.0310138. \end{gathered} $$

We can check that the nonlinear term $f(t,x,y)$ satisfies

$(L_{1})$:

$\max f(t,x,y)\approx383.413 \leq\min \{\varphi _{p}(\frac{D-\omega(1)}{J_{1}}), \varphi_{p}(\frac{D}{J_{2}}) \}\approx 412.216$, $(t,x,y)\in[0,1] \times [0,25{,}000] \times[0,25{,}000]$;

$(L_{2})$:

$\min f(t,x,y) \approx381.413 > \varphi_{p}(\frac {B-\rho\omega(1)}{\rho J_{3}})\approx228.283$, $(t,x,y)\in[0,\frac {1}{3}]\times[65,\frac{B}{\rho}]\times [0,25{,}000]$;

$(L_{3})$:

$\max f(t,x,y) \approx1.02108 < \varphi_{p}(\frac {A-\omega(1)}{J_{1}})\approx1.05195$, $(t,x,y)\in[0,1]\times[0,3]\times [0,25{,}000]$.

Thus, from Theorem 3.1, we know that BVP (20) has at least three positive solutions $x_{1}$, $x_{2}$, $x_{3}$, satisfying

$$\begin{gathered} \Vert x_{i} \Vert \leq25{,}000 \quad(i=1,2,3), \\ \min_{t\in[0,\eta]} \vert x_{1} \vert > 65,\qquad 3< \min _{t\in[0,\eta]} \vert x_{2} \vert ,\qquad \max _{t\in[0,1]} \vert x_{2} \vert < 65,\qquad \max _{t\in[0,1]} \vert x_{3} \vert < 3.\end{gathered} $$