Extremal solutions for p-Laplacian fractional differential systems involving the Riemann-Liouville integral boundary conditions

Abstract

In this paper, we investigate the existence of extremal solutions for fractional differential systems involving the p-Laplacian operator and Riemann-Liouville integral boundary conditions. We derive our results based on the monotone iterative technique, combined with the method of upper and lower solutions. An example is added to illustrate the main result.

1 Introduction

In this paper, we study the existence of extremal solutions of the following fractional differential systems involving the p-Laplacian operator and Riemann-Liouville integral boundary conditions:

$$ \textstyle\begin{cases} -D^{\alpha}(\phi_{p}(-D^{\beta}x(t)))= f(t,x(t),D^{\beta}x(t)),\quad 0< t< 1,\\ D^{\beta}x(0)=0,\qquad(\phi_{p}(-D^{\beta}x(0)))'=0,\\ D^{\gamma}(\phi_{p}(-D^{\beta}x(1)))= I^{\nu}(\phi_{p}(-D^{\beta}x(\eta)))=\int_{0}^{\eta}(\eta-s)^{\nu-1}\phi _{p}(-D^{\beta}x(s))\,ds,\\ x(0)=0,\qquad D^{\beta-1}x(1)=I^{\omega}g(\xi,x(\xi))+k=\frac{1}{\Gamma (\omega)} \int_{0}^{\xi}(\xi-s)^{\omega-1}g(s,x(s))\,ds+k, \end{cases} $$

(1.1)

where $D^{\alpha}$, $D^{\beta}$, and $D^{\gamma}$ are the standard Riemann-Liouville fractional derivatives, $I^{\nu}$ and $I^{\omega}$ are the Riemann-Liouville fractional integrals, and $0<\gamma<1<\beta <2<\alpha<3$, $\nu,\omega>0$, $0<\eta,\xi<1$, $k\in\mathbb{R}$, $f\in C([0,1]\times\mathbb{R}\times \mathbb{R},\mathbb{R})$, $g\in C([0,1]\times\mathbb{R},\mathbb{R})$. The p-Laplacian operator is defined as $\phi_{p}(t)= \vert t \vert ^{p-2}t$, $p>1$, and $(\phi_{p})^{-1}=\phi_{q}$, $\frac{1}{p}+\frac{1}{q}=1$.

The study of boundary value problems in the setting of fractional calculus has received a great attention in the last decade, and a variety of results concerning the of existence of solutions, based on various analytic techniques, can be found in the literature [1–10]. In particular, much effort has been made toward the study of the existence of solutions for fractional differential equations involving p-Laplacian operators; see [11–14]. Using the monotone iteration method, Ding [15] investigated a fractional boundary value problem with p-Laplacian operator

$$ \textstyle\begin{cases} D^{\beta}(\phi_{p}(D^{\alpha}u(t)))= f(t,u(t),D^{\alpha}u(t)),\quad t\in(0,1],\\ t^{\frac{1-\beta}{p-1}}D^{\alpha}u(t)\vert_{t=0}=0,\qquad g(\widetilde{u}(0),\widetilde{u}(1))=0, \end{cases} $$

(1.2)

where $0<\alpha,\beta\leq1$, $1<\alpha+\beta\leq2$, and $D^{\alpha}$ is the standard Riemann-Liouville fractional derivative, and established the existence and uniqueness of extremal solutions for the BVP (1.2) under the condition that the nonlinear functions f and g are continuous and satisfy certain growth conditions. Zhang [16] considered the following nonlinear fractional integral boundary value problem:

$$ \textstyle\begin{cases} {}^{C}\!D^{\alpha}u(t)=f(t,u(t),u(\theta(t)),\quad n< \alpha\leq n+1, n\geq 2, t\in[0,1],\\ u'(0)=u''(0)=u'''(0)=\cdots=u^{n}(0)=0,\\ u(0)=\int_{0}^{1}g(s,u(s))\,ds+\lambda, \end{cases} $$

(1.3)

where $\lambda\geq0$, ${}^{C}\!D^{\alpha}$ is the Caputo fractional derivative, and f and g are continuous functions. The authors constructed two well-defined monotone iterative sequences of upper and lower solutions and proved that they converge uniformly to the actual solution of problem (1.3). A numerical iterative scheme is also introduced to obtain an accurate approximate solution for the problem.

In this paper, we consider a kind of fractional differential equations involving p-Laplacian operators and nonlocal boundary conditions based on the Riemann-Liouville integral. To the best of our knowledge, little work has been conducted to deal with this kind of problem, and no work has been done concerning the maximal and minimal solutions of (1.1) by using the method of upper and lower solutions and the monotone iteration technique.

The rest of this paper is organized as follows. In Section 2, we give some useful preliminaries and lemmas. In Section 3, the main result and proof are given, and an example is presented to illustrate the main results.

2 Preliminaries

In this section, we deduce some preliminary results that will be used in the next section to attain existence results for the nonlinear system (1.1)

Lemma 2.1

Assume that $h(t)\in C[0,1]$, $l\in\mathbb {R}$. Then the fractional value boundary problem

$$ \textstyle\begin{cases} -D^{\alpha}v(t)=h(t),\quad 0< t< 1,\\ v(0)=0,\qquad v'(0)=0,\qquad D^{\gamma}v(1)=l, \end{cases} $$

(2.1)

has a unique solution

$$v(t)= \int_{0}^{1}G(t,s)h(s)\,ds+\frac{\Gamma(\alpha-\gamma)lt^{\alpha -1}}{\Gamma(\alpha)}, $$

where

$$G(t,s)= \textstyle\begin{cases} \frac{t^{\alpha-1}(1-s)^{\alpha-\gamma-1}-(t-s)^{\alpha-1}}{\Gamma (\alpha)},&0\leq s\leq t\leq1,\\ \frac{t^{\alpha-1}(1-s)^{\alpha-\gamma-1}}{\Gamma(\alpha)}, &0\leq t\leq s\leq1. \end{cases} $$

Proof

We can transform the equation $-D^{\alpha}v(t)=h(t)$ to the equivalent integral equation

$$v(t)=-I^{\alpha}h(t)+ C_{1}t^{\alpha-1}+C_{2}t^{\alpha-2}+C_{3}t^{\alpha-3}. $$

Noting that $v(0)=0$ and $v'(0)=0$, we have $C_{2}=C_{3}=0$. Consequently, we have the form

$$ v(t)=-I^{\alpha}h(t)+ C_{1}t^{\alpha-1} $$

(2.2)

and

$$\begin{aligned} D^{\gamma}v(t) =&-D^{\gamma}I^{\alpha}h(t)+ C_{1}D^{\gamma}t^{\alpha-1} \\ =&-I^{\alpha-\gamma} h(t)+C_{1}D^{\gamma}t^{\alpha-1} \\ =&-\frac{1}{\Gamma(\alpha-\gamma)} \int_{0}^{t}(t-s)^{\alpha-\gamma-1}h(s)\,ds +C_{1}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\gamma)}t^{\alpha-1}. \end{aligned}$$

So,

$$ D^{\gamma}v(1)=-\frac{1}{\Gamma(\alpha-\gamma)} \int_{0}^{1}(1-s)^{\alpha-\gamma-1}h(s)\,ds +C_{1}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\gamma)}. $$

(2.3)

On the other hand, $D^{\gamma}v(1)=l$, and combining with (2.3), we obtain

$$C_{1}=\frac{1}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-\gamma-1}h(s)\,ds + \frac{\Gamma(\alpha-\gamma)}{\Gamma(\alpha)}l. $$

Therefore, the unique solution of problem (2.1) is

$$\begin{aligned} v(t) =&-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}h(s)\,ds+ \frac{t^{\alpha-1}}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-\gamma-1}h(s)\,ds + \frac{\Gamma(\alpha-\gamma)lt^{\alpha-1}}{\Gamma(\alpha)} \\ =& \int_{0}^{1}G(t,s)h(s)\,ds+\frac{\Gamma(\alpha-\gamma)lt^{\alpha-1}}{\Gamma (\alpha)}. \end{aligned}$$

The proof is completed. □

Lemma 2.2

([17])

Assume that $z(t)\in C[0,1]$, $\lambda ,k \in\mathbb{R}$, and $\Gamma(\beta+\omega)\neq\lambda\xi^{\beta +\omega-1}$. Then the fractional boundary value problem

$$ \textstyle\begin{cases} -D^{\beta}x(t)= z(t), \quad0< t< 1,\\ x(0)=0, \qquad D^{\beta-1}x(1)=\lambda I^{\omega}x(\xi)+k, \end{cases} $$

(2.4)

has a unique solution

$$x(t)= \int_{0}^{1}H(t,s)z(s)\,ds+\frac{k\Gamma(\beta+\omega)t^{\beta-1}}{\Gamma (\beta)[\Gamma(\beta+\omega)- \lambda\xi^{\beta+\omega-1}]}, $$

where

$$H(t,s)=\frac{1}{\Delta} \textstyle\begin{cases} [\Gamma(\beta+\omega)-\lambda(\xi-s)^{\beta+\omega-1}]t^{\beta-1}\\ \quad {}- [\Gamma(\beta+\omega)-\lambda\xi^{\beta+\omega-1}](t-s)^{\beta-1}, &s\leq t,s\leq\xi,\\ \Gamma(\beta+\omega)t^{\beta-1}-\lambda(\xi-s)^{\beta+\omega-1}t^{\beta-1}, &t\leq s\leq\xi,\\ \Gamma(\beta+\omega)[t^{\beta-1}-(t-s)^{\beta-1}]+\lambda\xi^{\beta +\omega-1} (t-s)^{\beta-1}, &\xi\leq s\leq t,\\ \Gamma(\beta+\omega)t^{\beta-1}, &s\geq t,s\geq\xi, \end{cases} $$

and $\Delta=\Gamma(\beta)[\Gamma(\beta+\omega)-\lambda\eta^{\beta+\omega-1}]$.

Lemma 2.3

([17])

Suppose that $\lambda\geq0$ and $\Gamma(\beta+\omega)>\lambda\xi^{\beta+\omega-1}$. Then the functions $G(t,s)$ and $H(t,s)$ are continuous, and $0\leq G(t,s)\leq\frac{t^{\alpha-1}(1-s)^{\alpha-\gamma-1}}{\Gamma(\alpha)}$ and $0\leq H(t,s)\leq\frac{\Gamma(\beta+\omega)}{\Gamma(\beta)[\Gamma(\beta +\omega)- \lambda\xi^{\beta+\omega-1}]}(1+t^{\beta-1})$ for $(t,s)\in [0,1]\times[0,1]$.

Lemma 2.4

If $v(t)\in C[0,1]$ satisfies

$$ \textstyle\begin{cases} -D^{\alpha}v(t)\geq0,\quad0< t< 1,\\ v(0)=0, \qquad v'(0)=0,\qquad D^{\gamma}v(1)\geq0, \end{cases} $$

(2.5)

then $v(t)\geq0$ for $t \in[0,1]$.

Proof

By Lemma 2.1 we know that (2.1) has a unique solution

$$v(t)= \int_{0}^{1}G(t,s)h(s)\,ds+\frac{\Gamma(\alpha-\gamma)lt^{\alpha -1}}{\Gamma(\alpha)}. $$

In view of Lemma 2.3, we have $G(t,s)\geq0$ for $t, s\in[0,1]$. Let $h(t)\geq0$ and $l\geq0$. Then we obtain (2.5) and $v(t)\geq0$ for $t \in[0,1]$. □

Lemma 2.5

Let $\lambda\geq0$ and $\Gamma(\beta+\omega )>\lambda\xi^{\beta+\omega-1}$. If $x(t)\in C[0,1]$ satisfies

$$ \textstyle\begin{cases} -D^{\beta} x(t)\geq0,\quad0< t< 1,\\ x(0)=0, \qquad D^{\beta-1}x(1)\geq\lambda I^{\omega}x(\xi), \end{cases} $$

(2.6)

then $x(t)\geq0$ for $t\in[0,1]$.

Proof

We can easily prove Lemma 2.5 similarly to Lemma 2.4. □

Lemma 2.6

Assume that $h(t)\in C[0,1]$ and $\Gamma (\beta+\omega)\neq\lambda\xi^{\beta+\omega-1}$. Then the following boundary value problem

$$ \textstyle\begin{cases} -D^{\alpha}(\phi_{p}(-D^{\beta}x(t)))= h(t),\quad 0< t< 1,\\ D^{\beta}x(0)=0, \qquad(\phi_{p}(-D^{\beta}x(0)))'=0,\qquad D^{\gamma }(\phi_{p}(-D^{\beta}x(1)))=l,\\ x(0)=0, \qquad D^{\beta-1}x(1)=\lambda I^{\omega}x(\xi)+k, \end{cases} $$

(2.7)

has a unique solution

$$\begin{aligned} x(t)&= \int_{0}^{1}H(t,s)\phi_{q} \biggl( \int_{0}^{1}G(s,\tau)h(\tau)\,d\tau+\frac {\Gamma(\alpha-\gamma)ls^{\alpha-1}}{ \Gamma(\alpha)} \biggr)\,ds\\ &\quad {}+\frac{k\Gamma(\beta+\omega)t^{\beta-1}}{\Gamma (\beta)[\Gamma(\beta+\omega)- \lambda\xi^{\beta+\omega-1}]}. \end{aligned} $$

Proof

Let $\phi_{p}(-D^{\beta}x(t))=v(t)$ and consider the boundary value problem

$$ \textstyle\begin{cases} -D^{\alpha}v(t)=h(t),\quad0< t< 1,\\ v(0)=0,\qquad v'(0)=0,\qquad D^{\gamma}v(1)=l. \end{cases} $$

(2.8)

By Lemma 2.1 we obtain

$$v(t)= \int_{0}^{1}G(t,s)h(s)\,ds+\frac{\Gamma(\alpha-\gamma)lt^{\alpha -1}}{\Gamma(\alpha)}. $$

Noting that $\phi_{p}(-D^{\beta}x(t))=v(t)$ and $-D^{\beta}x(t)=\phi _{q}(v(t))$, we get that the boundary problem (2.7) is equivalent to the following problem:

$$ \textstyle\begin{cases} -D^{\beta}x(t)= \phi_{q}(v(t)),\quad0< t< 1,\\ x(0)=0, \qquad D^{\beta-1}x(1)=\lambda I^{\omega}x(\xi)+k. \end{cases} $$

(2.9)

By Lemma 2.2 the solution of (2.9) can be written as

$$ x(t)= \int_{0}^{1}H(t,s)\phi_{q}\bigl(v(s) \bigr)\,ds+\frac{k\Gamma(\beta+\omega)t^{\beta -1}}{\Gamma(\beta)[\Gamma(\beta+\omega)- \lambda\xi^{\beta+\omega -1}]}. $$

(2.10)

Combining with (2.8) and (2.9), we assert that the boundary problem (2.7) has a unique solution

$$\begin{aligned} x(t)&= \int_{0}^{1}H(t,s)\phi_{q} \biggl( \int_{0}^{1}G(s,\tau)h(\tau)\,d\tau+\frac {\Gamma(\alpha-\gamma)ls^{\alpha-1}}{ \Gamma(\alpha)} \biggr)\,ds\\ &\quad {}+\frac{k\Gamma(\beta+\omega)t^{\beta-1}}{\Gamma (\beta)[\Gamma(\beta+\omega)- \lambda\xi^{\beta+\omega-1}]}. \end{aligned} $$

 □

3 Main results

Let $E=\{x :x \in C[0,1], D^{\beta}x(t)\in C[0,1]\}$ be endowed with the norm $\Vert x \Vert _{\beta}= \Vert x \Vert + \Vert D^{\beta}x \Vert$, where $\Vert x \Vert =\max_{0\leq t\leq1} \vert x(t) \vert$ and $\Vert D^{\beta}x \Vert =\max_{0\leq t\leq1} \vert D^{\beta}x(t) \vert$. Then $(E, \Vert \cdot \Vert _{\beta})$ is a Banach space.

Definition 3.1

We say that $x(t)\in E$ is a lower solution of problem (1.1), if

$$\textstyle\begin{cases} -D^{\alpha}(\phi_{p}(-D^{\beta}x(t)))\leq f(t,x(t),D^{\beta}x(t)), \quad0< t< 1,\\ D^{\beta}x(0)=0,\qquad(\phi_{p}(-D^{\beta}x(0)))'=0, \qquad D^{\gamma }(\phi_{p}(-D^{\beta}x(1)))\leq I^{\nu}(\phi_{p}(-D^{\beta}x(\eta))),\\ x(0)=0,\qquad D^{\beta-1}x(1)\leq I^{\omega}g(\xi,x(\xi))+k, \end{cases} $$

and it is an upper solution of (1.1) if the above inequalities are reversed.

We further need the following assumptions.

$(H_{1})$ :

$x_{0}, y_{0}\in E$ are lower and upper solutions of problem (1.1), respectively, and $x_{0}(t)\leq y_{0}(t)$ and $D^{\beta}y_{0}(t)\leq D^{\beta}x_{0}(t)$ for $t\in[0,1]$.

$(H_{2})$ :

The function $f\in C([0,1]\times\mathbb{R}\times\mathbb {R},\mathbb{R})$ satisfies

$$f\bigl(t,y,D^{\beta}y\bigr)\geq f\bigl(t,x,D^{\beta}x\bigr) $$

if $x_{0}(t)\leq x(t)\leq y(t) \leq y_{0}(t)$ and $D^{\beta}y_{0}(t)\leq D^{\beta}y(t)\leq D^{\beta}x(t) \leq D^{\beta}x_{0}(t)$ for $t\in[0,1]$.

$(H_{3})$ :

There exists a constant $\lambda\geq0$ such that $\Gamma (\beta+\omega)> \lambda\xi^{\beta+\omega-1}$ and

$$g(t,y)-g(t,x)\geq\lambda(y-x) $$

if $x_{0}(t)\leq x(t)\leq y(t) \leq y_{0}(t)$ for $t\in[0,1]$.

Theorem 3.1

Suppose that $(H_{1})$-$(H_{3})$ hold. Then boundary value problem (1.1) has an extremal solution $x^{*}, y^{*}\in [x_{0},y_{0}]$. Moreover,

$$x_{0}(t)\leq x^{*}(t)\leq y^{*}(t)\leq y_{0}(t) $$

and

$$D^{\beta} y_{0}(t)\leq D^{\beta}y^{*}(t)\leq D^{\beta} x^{*}(t)\leq D^{\beta} x_{0}(t)\quad \textit{for } t \in[0,1]. $$

Proof

For $n=0,1,2\ldots$ , we define

$$ \textstyle\begin{cases} -D^{\alpha}(\phi_{p}(-D^{\beta}x_{n+1}(t)))= f(t,x_{n}(t),D^{\beta }x_{n}(t)),\quad0< t< 1,\\ D^{\beta}x_{n+1}(0)=0,\qquad(\phi_{p}(-D^{\beta}x_{n+1}(0)))'=0,\\ D^{\gamma}(\phi_{p}(-D^{\beta}x_{n+1}(1)))=I^{\nu}(\phi_{p}(-D^{\beta }x_{n}(\eta))),\\ x_{n+1}(0)=0,\qquad D^{\beta-1}x_{n+1}(1)=I^{\omega}\{g(\xi,x_{n}(\xi ))+\lambda[x_{n+1}(\xi)-x_{n}(\xi)]\}+k, \end{cases} $$

(3.1)

and

$$ \textstyle\begin{cases} -D^{\alpha}(\phi_{p}(-D^{\beta}y_{n+1}(t)))= f(t,y_{n}(t),D^{\beta }y_{n}(t)), \quad0< t< 1,\\ D^{\beta}y_{n+1}(0)=0,\qquad(\phi_{p}(-D^{\beta}y_{n+1}(0)))'=0,\\ D^{\gamma}(\phi_{p}(-D^{\beta}y_{n+1}(1)))=I^{\nu}(\phi_{p}(-D^{\beta }y_{n}(\eta))),\\ y_{n+1}(0)=0,\qquad D^{\beta-1}y_{n+1}(1)=I^{\omega}\{g(\xi,y_{n}(\xi ))+\lambda[y_{n+1}(\xi)-y_{n}(\xi)]\}+k. \end{cases} $$

(3.2)

By Lemma 2.6 we know that (3.1) and (3.2) have unique solutions

$$\begin{aligned} x_{n+1}(t) =& \int_{0}^{1}H(t,s)\phi_{q} \biggl( \int_{0}^{1}G(s,\tau) f\bigl(\tau ,x_{n}( \tau),D^{\beta}x_{n}(\tau)\bigr)\,d\tau\\ &{}+\frac{\Gamma(\alpha-\gamma) s^{\alpha-1}I^{\nu}(\phi_{p}(-D^{\beta}x_{n}(\eta)))}{\Gamma(\alpha)} \biggr)\,ds \\ &{} +\frac{\Gamma(\beta+\omega)t^{\beta-1}\{I^{\omega}[g(\xi,x_{n}(\xi ))-\lambda x_{n}(\xi)]+k\}}{\Gamma(\beta)[\Gamma(\beta+\omega)- \lambda \xi^{\beta+\omega-1}]} \end{aligned}$$

and

$$\begin{aligned} y_{n+1}(t) =& \int_{0}^{1}H(t,s)\phi_{q} \biggl( \int_{0}^{1}G(s,\tau) f\bigl(\tau ,y_{n}( \tau),D^{\beta}y_{n}(\tau)\bigr)\,d\tau\\ &{}+\frac{\Gamma(\alpha-\gamma) s^{\alpha-1}I^{\nu}(\phi_{p}(-D^{\beta}y_{n}(\eta)))}{\Gamma(\alpha)} \biggr)\,ds \\ &{}+\frac{\Gamma(\beta+\omega)t^{\beta-1}\{I^{\omega}[g(\xi,y_{n}(\xi ))-\lambda y_{n}(\xi)]+k\}}{\Gamma(\beta)[\Gamma(\beta+\omega)- \lambda \xi^{\beta+\omega-1}]}. \end{aligned}$$

First, we show that $x_{0}(t)\leq x_{1}(t)\leq y_{1}(t)\leq y_{0}(t)$, and $D^{\beta} y_{0}\leq D^{\beta}y_{1}(t)\leq D^{\beta} x_{1}(t)\leq D^{\beta} x_{0}(t)$, $t\in[0,1]$. Let $\varepsilon(t)=\phi_{p}(-D^{\beta}x_{1}(t))-\phi_{p}(-D^{\beta }x_{0}(t))$. From (3.1) and $(H_{1})$ we obtain

$$\textstyle\begin{cases} -D^{\alpha}\varepsilon(t)=-D^{\alpha}(\phi_{p}(-D^{\beta }x_{1}(t)))+D^{\alpha}(\phi_{p}(-D^{\beta}x_{0}(t)))\\ \hphantom{-D^{\alpha}\varepsilon(t)}\geq f(t,x_{0}(t),D^{\beta}x_{0}(t))-f(t,x_{0}(t),D^{\beta}x_{0}(t))= 0,\\ \varepsilon(0)=0,\qquad\varepsilon'(0)=0, \\ D^{\gamma}\varepsilon(1)=D^{\gamma}(\phi_{p}(-D^{\beta }x_{1}(t)))-D^{\gamma}(\phi_{p}(-D^{\beta}x_{0}(t)))\\ \hphantom{D^{\gamma}\varepsilon(1)}\geq I^{\omega}(\phi_{p}(-D^{\beta}x_{0}(\eta)))-I^{\omega}(\phi _{p}(-D^{\beta}x_{0}(\eta)))=0. \end{cases} $$

In view of Lemma 2.4, we have $\phi_{p}(-D^{\beta}x_{1}(t))\geq\phi _{p}(-D^{\beta}x_{0}(t))$, $t\in[0,1]$, since $\phi_{p}(x)$ is nondecreasing, and thus

$$ D^{\beta}x_{1}(t)\leq D^{\beta }x_{0}(t). $$

(3.3)

Let $v(t)=x_{1}(t)-x_{0}(t)$, it follows from (3.1), (3.3) and $(H_{3})$ that

$$\textstyle\begin{cases} -D^{\beta}v(t)=-D^{\beta}x_{1}(t)+D^{\beta}x_{0}(t)\geq0,\quad t\in[0,1],\\ v(0)=0, \\ D^{\beta-1}v(1)\geq I^{\omega}\{g(\xi,x_{0}(\xi))+\lambda[x_{1}(\xi )-x_{0}(\xi)]\}-I^{\omega}g(\xi,x_{0}(\xi)) \geq\lambda I^{\omega} v(\xi). \end{cases} $$

According to Lemma 2.5, we have $x_{1}(t)\geq x_{0}(t)$ for $t\in[0,1]$.

Using similar reasoning, we can show that $y_{0}(t)\geq y_{1}(t)$ and $D^{\beta}y_{0}(t)\leq D^{\beta}y_{1}(t)$. Now, we let $w(t)=\phi_{p}(-D^{\beta}y_{1}(t))-\phi_{p}(-D^{\beta }x_{1}(t))$. From $(H_{1})$ and $(H_{2})$ we have

$$\textstyle\begin{cases} -D^{\alpha}w(t)=-D^{\alpha}\phi_{p}(-D^{\beta}y_{1}(t))+D^{\alpha}\phi _{p}(-D^{\beta}x_{1}(t))\\ \hphantom{-D^{\alpha}w(t)}=f(t,y_{0}(t),D^{\beta}y_{0}(t)) -f(t,x_{0}(t),D^{\beta}x_{0}(t))\geq0,\\ w(0)=0, \qquad w'(0)=0, \\ D^{\gamma}w(1)=D^{\gamma}(\phi_{p}(-D^{\beta}y_{1}(t)))-D^{\gamma}(\phi _{p}(-D^{\beta}x_{1}(t)))\\ \hphantom{D^{\gamma}w(1)}=I^{\nu}(\phi_{p}(-D^{\beta}y_{0}(\eta)))-I^{\nu}(\phi_{p}(-D^{\beta}x_{0}(\eta )))\geq0. \end{cases} $$

In view of Lemma 2.4, we have $w(t)\geq0$ for $t\in[0,1]$. Thus we have $\phi_{p}(-D^{\beta}y_{1}(t))\geq\phi_{p}(-D^{\beta}x_{1}(t))$, that is, $D^{\beta}y_{1}(t)\leq D^{\beta}x_{1}(t)$, since $\phi_{p}(x)$ is nondecreasing. Therefore $D^{\beta} y_{0}(t)\leq D^{\beta}y_{1}(t)\leq D^{\beta} x_{1}(t)\leq D^{\beta} x_{0}(t)$ for $t\in[0,1]$. Let $\delta(t)=y_{1}(t)-x_{1}(t)$. From $(H_{3})$ we get

$$-D^{\beta}\delta(t)=-D^{\beta}y_{1}(t)+D^{\beta}x_{1}(t) \geq0. $$

Also, $\delta(0)=0$, and

$$\begin{aligned} D^{\beta-1}\delta(1) =&I^{\omega}\bigl\{ g\bigl(\xi,y_{0}( \xi)\bigr)+\lambda\bigl[y_{1}(\xi )-y_{0}(\xi)\bigr]\bigr\} - I^{\omega}\bigl\{ g\bigl(\xi,x_{0}(\xi)\bigr)+\lambda \bigl[x_{1}(\xi)-x_{0}(\xi)\bigr]\bigr\} \\ =&I^{\omega}\bigl\{ g\bigl(\xi,y_{0}(\xi)\bigr)-g\bigl( \xi,x_{0}(\xi)\bigr)+\lambda\bigl[y_{1}(\xi) -y_{0}(\xi)\bigr]-\lambda\bigl[x_{1}(\xi)-x_{0}( \xi)\bigr]\bigr\} \\ \geq&I^{\omega}\bigl\{ \lambda\bigl[y_{0}(\xi)-x_{0}( \xi)\bigr]+\lambda\bigl[y_{1}(\xi) -y_{0}(\xi)\bigr]-\lambda \bigl[x_{1}(\xi)-x_{0}(\xi)\bigr]\bigr\} \\ =&\lambda I^{\omega}\delta(\xi). \end{aligned}$$

Moreover, we get $y_{1}(t)\geq x_{1}(t)$ from Lemma 2.5. Hence, we have the relation $x_{0}(t)\leq x_{1}(t)\leq y_{1}(t)\leq y_{0}(t)$ for $t\in[0,1]$.

In the following, we show that $x_{1}(t)$ and $y_{1}(t)$ are lower and upper solutions of problem (1.1), respectively. From (3.1)-(3.2) and $(H_{1})$-$(H_{3})$ we get

$$-D^{\alpha}\bigl(\phi_{p}\bigl(-D^{\beta}x_{1}(t) \bigr)\bigr)=f\bigl(t,x_{0}(t),D^{\beta }x_{0}(t) \bigr)\leq f\bigl(t,x_{1}(t),D^{\beta}x_{1}(t)\bigr), $$

Also,

$$\begin{aligned}& D^{\beta}x_{1}(0)=0, \qquad \bigl(\phi_{p} \bigl(-D^{\beta}x_{0}(0)\bigr)\bigr)'=0, \\& D^{\gamma}\bigl(\phi_{p}\bigl(-D^{\beta}x_{1}(1) \bigr)\bigr)=I^{\nu}\bigl(\phi_{p}\bigl(-D^{\beta }x_{0}( \eta)\bigr)\bigr)\leq I^{\nu}\bigl(\phi_{p} \bigl(-D^{\beta}x_{1}(\eta)\bigr)\bigr), \end{aligned}$$

and $x_{1}(0)=0$,

$$\begin{aligned} D^{\beta-1}x_{1}(1) =&I^{\omega}\bigl\{ g\bigl( \xi,x_{0}(\xi)\bigr)-g\bigl(\xi,x_{1}(\xi)\bigr)+g\bigl(\xi ,x_{1}(\xi)\bigr) +\lambda\bigl[x_{1}(\xi)-x_{0}( \xi)\bigr]\bigr\} +k \\ \leq&I^{\omega}\bigl\{ \lambda\bigl[x_{0}(\xi)-x_{1}( \xi)\bigr]+g\bigl(\xi,x_{1}(\xi)\bigr)+\lambda \bigl[x_{1}( \xi)-x_{0}(\xi)\bigr]\bigr\} +k \\ =&I^{\beta}g\bigl(\xi,x_{1}(\xi)\bigr)+k. \end{aligned}$$

This proves that $x_{1}(t)$ is a lower solution of problem (1.1). Similarly, we can obtain that $y_{1}(t)$ is an upper solution of (1.1).

Using mathematical induction, we see that

$$x_{0}(t)\leq x_{1}(t)\leq\cdots\leq x_{n}(t) \leq\cdots\leq y_{n}(t)\leq\cdots \leq y_{1}(t)\leq y_{0}(t) $$

and

$$D^{\beta}y_{0}(t)\leq D^{\beta}y_{1}(t)\leq \cdots\leq D^{\beta}y_{n}(t) \leq \cdots\leq D^{\beta}x_{n}(t)\leq\cdots\leq D^{\beta}x_{1}(t)\leq D^{\beta}x_{0}(t) $$

for $t\in[0,1]$ and $n=1,2,3,\ldots$ .

Since the sequence $\{x_{n}(t)\}$ is nondecreasing and bounded from above, the sequence $\{y_{n}(t)\}$ is nonincreasing and bounded from below. By standard argument we know that the sequences $\{x_{n}(t)\}$ and $\{ y_{n}(t)\}$ uniformly converge to their limit functions $x^{*}(t)$ and $y^{*}(t)$, respectively, that is,

$$\lim_{n\rightarrow\infty}x_{n}(t)=x^{*}(t), \qquad \lim _{n\rightarrow\infty }y_{n}(t)=x^{*}(t), \quad\forall t\in[0,1], $$

and

$$\lim_{n\rightarrow\infty}D^{\beta}x_{n}(t)=D^{\alpha}x^{*}(t), \qquad\lim_{n\rightarrow\infty}D^{\beta}y_{n}(t)=D^{\alpha}x^{*}(t), \quad\forall t\in[0,1]. $$

Moreover, from (3.1) and (3.2) we can obtain that $x^{*}(t)$ and $y^{*}(t)$ are solutions of problem (1.1).

Finally, we prove that $x^{*}(t)$ and $y^{*}(t)$ are the minimal and maximal solutions of problem (1.1), respectively. Let $x(t)\in[x_{0},y_{0}]$ be any solution of problem (1.1). We suppose that, for some n, $x_{n}(t)\leq x(t)\leq y_{n}(t)$ and $D^{\beta}y_{n}(t)\leq D^{\beta}x(t)\leq D^{\beta}x_{n}(t)$ for $t\in[0,1]$. Let $p(t)=(\phi_{p}(-D^{\beta}x(t)))-\phi_{p}(-D^{\beta}x_{n+1}(t))$ and $q(t)=(\phi_{p}(-D^{\beta}y_{n+1}(t)))-\phi_{p}(-D^{\beta}x(t))$. Then by assumption $(H_{2})$ we see that

$$\textstyle\begin{cases} -D^{\alpha}p(t)=f(t,x(t),D^{\beta}x(t))-f(t,x_{n}(t),D^{\beta}x_{n}(t))\geq 0,\\ p(0)=0, \qquad p'(0)=0, \\ D^{\gamma}p(1)=I^{\nu}(\phi_{p}(-D^{\beta}x(\eta)))-I^{\nu}(\phi _{p}(-D^{\beta}x_{n}(\eta)))\geq0, \end{cases} $$

and

$$\textstyle\begin{cases} -D^{\alpha}q(t)\geq0,\\ q(0)=0, \qquad q'(0)=0, \\ D^{\gamma}q(1)\geq0. \end{cases} $$

Using Lemma 2.4, we have

$$ D^{\beta}y_{n+1}(t)\leq D^{\beta}x(t)\leq D^{\beta}x_{n+1}(t). $$

(3.4)

Let $m(t)=x(t)-x_{n+1}(t)$ and $n(t)=y_{n+1}(t)-x(t)$. By assumption $(H_{3})$ and (3.4) we get

$$\textstyle\begin{cases} -D^{\beta}m(t)=-D^{\beta}x(t)+D^{\beta}x_{n+1}(t)\geq0, \quad t\in [0,1],\\ m(0)=0, \\ D^{\beta-1}m(1) =I^{\omega}\{g(\xi,x(\xi))-g(\xi,x_{n}(\xi))-\lambda[x_{n+1}(\xi )-x_{n}(\xi)]\} \geq\lambda I^{\omega}m(\xi), \end{cases} $$

and

$$\textstyle\begin{cases} -D^{\beta}n(t)\geq0,\quad t\in[0,1],\\ n(0)=0, \\ D^{\beta-1}n(1) \geq\lambda I^{\omega}n(\xi). \end{cases} $$

These and Lemma 2.5 imply that $x_{n+1}(t)\leq x(t)\leq y_{n+1}(t)$, $t\in [0,1]$, so by induction $x^{*}(t)\leq x(t)\leq y^{*}(t)$ and $D^{\beta}y^{*}(t)\leq D^{\beta}x(t)\leq D^{\beta}x^{*}(t)$, $t\in[0,1]$, as $n\to\infty$. The proof is complete. □

Example

Consider the following fractional boundary value problem:

$$ \textstyle\begin{cases} -D^{\frac{5}{2}}(\phi_{4}(-D^{\frac{3}{2}}x(t)))= \frac{1}{10}tx-\frac {1}{(t+3)^{2}}D^{\frac{3}{2}}x(t)+\frac{1}{7}t^{\frac{1}{2}}, \quad 0< t< 1,\\ D^{\frac{3}{2}}x(0)=0,\qquad(\phi_{4}(-D^{\frac{3}{2}}x(0)))'=0,\\ D^{\frac{1}{4}}(\phi_{p}(-D^{\frac{3}{2}}x(1)))=I^{\frac{5}{2}} (\phi _{4}(-D^{\frac{3}{2}}x(\frac{1}{2})) ),\\ x(0)=0,\qquad D^{\frac{1}{2}}x(1)=I^{\frac{3}{2}}g(\frac{1}{4},x(\frac {1}{4}))+0.1=\frac{1}{\Gamma(\frac{3}{2})} \int_{0}^{\frac{1}{4}}(\frac{1}{4}-s)^{\frac{1}{2}}(s+1)x(s)\,ds+0.1,\hspace{-20pt} \end{cases} $$

(3.5)

where $\alpha=\frac{5}{2}$, $\beta=\frac{3}{2}$, $\gamma=\frac{1}{4}$, $\nu =\frac{5}{2}$, $\omega=\frac{3}{2}$, $\eta=\frac{1}{2}$, $\xi=\frac{1}{4}$, $k=0.1$, $p=4$, and

$$\textstyle\begin{cases} f(t,x,D^{\beta}x)=\frac{1}{10}tx-\frac{1}{(t+3)^{2}}D^{\frac {3}{2}}x(t),\\ g(t,x)=(t+1)x. \end{cases} $$

Take $x_{0}(t)=0$ and $y_{0}(t)=t^{\frac{1}{2}}-\frac{\sqrt{\Pi }}{4}t^{2}+\frac{2}{15\sqrt{\Pi}}t^{\frac{5}{2}}$. Then $-1\leq-t^{\frac{1}{2}}+\frac{1}{4}t=D^{\beta}y_{0}(t)\leq D^{\beta}x_{0}(t)=0$. It is not difficult to verify that $x_{0}$, $y_{0}$ are lower and upper solutions of problem (3.5), respectively. Moreover,

$$g(t,y)-g(t,x)=(t+1) (y-x)\geq(y-x), $$

where $x_{0}(t)\leq x\leq y\leq y_{0}(t)$.

For $\lambda=1$, we have

$$\Gamma(\beta+\omega)=\Gamma\biggl(\frac{3}{2}+\frac{3}{2} \biggr)=2> \lambda\xi ^{\beta+\omega-1}=1\cdot\biggl(\frac{1}{4} \biggr)^{2}. $$

All conditions $(H_{1})$, $(H_{2})$, and $(H_{3})$ are satisfied. Therefore by Theorem 3.1 the boundary value problem (3.5) has extremal solutions in $[x_{0}(t),y_{0}(t)]$.