Existence and uniqueness of solutions for a mixed p-Laplace boundary value problem involving fractional derivatives

Abstract

In this article, the existence and uniqueness of solutions for a multi-point fractional boundary value problem involving two different left and right fractional derivatives with p-Laplace operator is studied. A novel approach is used to acquire the desired results, and the core of the method is Banach contraction mapping principle. Finally, an example is given to verify the results.

1 Introduction

In recent years, fractional-order calculus theory has been widely used in mathematics, science, engineering, etc. As a result, the studies of such equation have gained considerable popularity, see [1–8]. Also, fractional-order mixed differential or integral equation involving different fractional derivatives, such as conformable fractional, Riemann–Liouville, and Caputo, has got a lot of interest, and even fractional-order differential or integral equations with p-Laplace operator have been extensively discussed by more and more researchers [9–14].

In [11], a mixed fractional p-Laplace boundary value problem was studied by Liu et al.

$$ \textstyle\begin{cases} D_{0+}^{\alpha }( \varphi _{p} ({}^{c} D_{0+}^{\beta } u(t)))= f(t, u(t),{}^{c} D_{0+}^{\beta } u(t)), \\ {}^{c} D_{0+}^{\beta } u(0)= u'(0)= 0, \\ u(1)= r_{1}u(\eta ), \quad \quad {}^{c} D_{0+}^{\beta } u(1)= r_{2}{}^{c} D_{0+}^{\beta } u(\xi ), \end{cases} $$

where $\varphi _{p}(t)= \vert t \vert ^{p-2}\cdot t$, $p>1$, $1< \alpha , \beta \leq 2$, $r_{1}, r_{2} \geq 0$, $D_{0+}^{\alpha }$ is Riemann–Liouville fractional derivative, ${}^{c} D_{0+}^{\beta }$ is Caputo fractional derivative, and $f:[0,1] \times [0, +\infty ) \times (-\infty , 0]\rightarrow [0, + \infty )$ is continuous. The minimum upper solution and the maximum lower solution of the above boundary value problem were given by applying the lower and upper solutions method.

In [2], Bai investigated the uniqueness and existence of solutions of the following fractional-order differential equation:

$$ \textstyle\begin{cases} ( \varphi _{p} (D_{0+}^{\alpha } u(t)))' + f(t, u(t)) = 0, \\ u(0)= D_{0+}^{\alpha } u(0) = 0, \quad \quad {}^{c} D_{0+}^{\beta } u(1)= {}^{c} D_{0+}^{\beta } u(1)=0, \end{cases} $$

where $D_{0+}^{\alpha }$ is Riemann–Liouville fractional derivative, ${}^{c} D_{0+}^{\beta }$ is Caputo fractional derivative, $0< \beta \leq 1$, $2 < \alpha \leq 2+ \beta$, $p> 1$, and $f:[a,b] \times \mathbb{R} \rightarrow \mathbb{R}$ is continuous. The uniqueness result of a solution and the existence of specific solutions to problem were showed by applying Guo–Krasnoselskii’s fixed point theorem.

In [4], Dang et al. proposed a fresh approach to gain the existence and uniqueness of solutions of a fourth-order two-point nonlinear differential equation

$$\begin{aligned}& u^{(4)}(x)= f \bigl(x, u(x), u''(x) \bigr), \\& u(0)= u(1)= u''(0)= u''(1)= 0, \end{aligned}$$

where $f:[0,1] \times \mathbb{R}^{2} \rightarrow \mathbb{R}$ is continuous. In [3], Bai et al. considered a class of fourth-order nonlinear differential equation with p-Laplace operator, and the boundary conditions change from two points to multiple points compared to the above problem. In both papers, the main results were given by applying the Banach contraction mapping principle.

As far as we know, nobody used the method which was put forward by Dang [6] and Bai [3] to prove the existence and uniqueness of solutions of a nonlinear multi-point fractional-order p-Laplace boundary value problem, which has at least two different kinds of fractional derivatives. Inspired by the above-mentioned articles, the following mixed boundary value problem is studied in this work:

$$ \textstyle\begin{cases} {}^{c} D_{1-}^{\gamma }( \varphi _{p} (D_{0+}^{\delta } y(x)))= g(x, y(x),D_{0+}^{\delta } y(x)), \\ y(0)= 0, \quad \quad y(1)= r_{1}y(\mu ), \\ D_{0+}^{\delta } y(1)= 0, \quad\quad \varphi _{p}( D_{0+}^{\delta } y(0))= r_{2} \varphi _{p} (D_{0+}^{\delta } y(\eta )), \end{cases} $$

(1.1)

where $1< \gamma , \delta \leq 2$, $0<\mu ,\eta < 1$, $0\leq r_{1} < \frac{1}{\mu ^{\delta - 1}}$, $0\leq r_{2}< \frac{1}{1- \eta }$, $\varphi _{p}(x)= \vert x \vert ^{p-2}\cdot x$, $\frac{1}{p}+ \frac{1}{q}= 1$, $p, q> 1$, $D_{0+}^{\delta }$ is left Riemann–Liouville fractional derivative, ${}^{c} D_{1-}^{\gamma }$ is right Caputo fractional derivative, and function $g\in C([0,1] \times \mathbb{R}^{2}, \mathbb{R})$.

The rest of this work is organized as follows. In Sect. 2, some related definitions and necessary lemmas of fractional calculus theory are presented, which will be applied in the main results of this article. In Sect. 3, the existence and uniqueness of solutions of the mixed fractional-order p-Laplace differential equation are proved by applying the method which was put forward by Dang and Bai. In Sect. 4, a particular example is constructed to verify the main conclusions of the paper.

2 Preliminaries

This section introduces some related definitions and necessary lemmas of fractional calculus theory.

Definition 2.1

For given $\gamma > 0$ and the function $y:(0,\infty )\rightarrow \mathbb{R}$, define the left and right Riemann–Liouville fractional integrals respectively as follows:

$$ \begin{aligned} & I_{0+}^{\gamma } y(x)= \frac{1}{\Gamma (\gamma )} \int _{0}^{x}(x- s)^{ \gamma - 1}y(s)\,ds, \\ & I_{1-}^{\gamma } y(x)=\frac{1}{\Gamma (\gamma )} \int _{x}^{1}(s- x)^{ \gamma - 1}y(s)\,ds. \end{aligned} $$

Definition 2.2

For given $\gamma > 0$, $\gamma \in (n, n+ 1)$ and the function $y:(0,\infty )\rightarrow \mathbb{R}$, define the left Riemann–Liouville fractional derivative and the right Caputo fractional derivative respectively as follows:

$$ \begin{aligned} & D_{0+}^{\gamma } y(x)= \frac{d^{n}}{dx^{n}} \frac{1}{\Gamma (n- \gamma )} \int _{0}^{x}(x- s)^{n- \gamma - 1}y(s)\,ds, \\ & {}^{c} D_{1-}^{\gamma } y(x)=(-1)^{n} \frac{1}{\Gamma (n- \gamma )} \int _{x}^{1}(s- x)^{n- \gamma - 1}y^{(n)}(s) \,ds. \end{aligned} $$

Let $E= C[0,1]$, whose norm $\Vert \cdot \Vert$ is the maximum norm. Given $\phi \in C[0,1]$ and the constants $r_{1},r_{2}\in \mathbb{R}$, discuss the following fractional-order mixed p-Laplace boundary value problem:

$$ \textstyle\begin{cases} {}^{c} D_{1-}^{\gamma }( \varphi _{p} (D_{0+}^{\delta } y(x)))= \phi (x), \\ y(0)= 0, \quad \quad y(1)= r_{1}y(\mu ), \\ D_{0+}^{\delta } y(1)= 0, \quad \quad \varphi _{p}( D_{0+}^{\delta } y(0))= r_{2} \varphi _{p}( D_{0+}^{\delta } y(\eta )), \end{cases} $$

(2.1)

where $1< \gamma , \delta \leq 2$, $0<\mu ,\eta < 1$, $0\leq r_{1}< \frac{1}{\mu ^{\delta - 1}}$, $0\leq r_{2} < \frac{1}{1- \eta }$.

Lemma 2.1

The unique solution of the fractional-order mixed p-Laplace boundary value problem (2.1) is equivalent to

$$\begin{aligned}& y(x)= \int _{0}^{1} G_{1}(x, \tau ) \varphi _{q} \biggl( \int _{0}^{1} G_{2}(\tau , s) \phi (s) \,ds \biggr) \,d\tau , \end{aligned}$$

(2.2)

where

$$ G_{1}(x, \tau ) = \frac{1}{\Gamma (\delta )} \textstyle\begin{cases} \Lambda _{1} [(1 - \tau )^{\delta - 1} - r_{1} (\mu - \tau )^{\delta - 1}] - (x - \tau )^{\delta - 1},& 0 < \tau \leq \min \{x,\mu \}; \\ \Lambda _{1} [(1 - \tau )^{\delta - 1} - r_{1} (\mu - \tau )^{\delta - 1}], & x \leq \tau \leq \mu ; \\ \Lambda _{1} (1 - \tau )^{\delta - 1} - (x - \tau )^{ \delta - 1}, & \mu \leq \tau \leq x; \\ \Lambda _{1} (1 - \tau )^{\delta - 1},& \max \{x,\mu \} \leq \tau < 1, \end{cases} $$

and

$$ G_{2}(\tau , s) = \frac{1}{\Gamma (\gamma )} \textstyle\begin{cases} \Lambda _{2}s^{\gamma - 1}, & 0 < s \leq \min \{\tau ,\eta \}; \\ \Lambda _{2}s^{\gamma - 1}- (s-\tau )^{\gamma - 1}, & \tau \leq s \leq \eta ; \\ \Lambda _{2}[s^{\gamma - 1}- r_{2}(s-\eta )^{\gamma - 1}], & \eta \leq s \leq \tau ; \\ \Lambda _{2}[s^{\gamma - 1}- r_{2}(s-\eta )^{\gamma - 1}]- (s-\tau )^{ \gamma - 1},& \max \{\tau ,\eta \} \leq s < 1, \end{cases} $$

with

$$\begin{aligned}& \Lambda _{1}=\frac{x^{\delta - 1}}{1-r_{1}\mu ^{\delta - 1}}, \qquad \Lambda _{2}= \frac{1-x}{1-r_{2}(1-\eta )}. \end{aligned}$$

Moreover, $G_{1}(x, \tau )>0$, $G_{2}(\tau , s)>0$ for $x,\tau ,s\in (0,1)$.

Proof

Let $-\varphi _{p} (D_{0+}^{\delta } y(x))= k(x)$, then the mixed boundary value problem (2.1) changes to

$$ \textstyle\begin{cases} {}^{c} D_{1-}^{\gamma } k(x)= -\phi (x), \\ k(1)=0, \quad \quad k(0)= r_{2}k(\eta ), \end{cases} $$

(2.3)

and

$$ \textstyle\begin{cases} D_{0+}^{\delta } y(x)= -\varphi _{q}(k(x)), \\ y(0)=0, \quad \quad y(1)= r_{1}y(\mu ). \end{cases} $$

(2.4)

Reduce the equation ${}^{c} D_{1-}^{\gamma } k(x)= -\phi (x)$ as an equivalent equation

$$\begin{aligned}& k(x)= -\frac{1}{\Gamma (\gamma )} \int _{x}^{1} (s-x) ^{\gamma -1} \phi (s)\,ds+ c_{1}+ c_{2}(1-x). \end{aligned}$$

Using the condition $k(1)=0$ yields $c_{1}= 0$. Since $k(0)= r_{2}k(\eta )$, then

$$\begin{aligned}& k(0)= -\frac{1}{\Gamma (\gamma )} \int _{0}^{1} s ^{\gamma -1}\phi (s)\,ds+ c_{2} \end{aligned}$$

and

$$\begin{aligned}& k(\eta )= -\frac{1}{\Gamma (\gamma )} \int _{\eta }^{1} (s-\eta ) ^{ \gamma -1}\phi (s) \,ds+c_{2}(1-\eta ). \end{aligned}$$

By calculation, we can get

$$\begin{aligned}& c_{2}= \frac{1}{\Gamma (\gamma )[1-r_{2}(1-\eta )]} \biggl( \int _{0}^{1} s^{\gamma -1}\phi (s)\,ds- \int _{\eta }^{1} r_{2} (s-\eta ) ^{\gamma -1} \phi (s)\,ds \biggr). \end{aligned}$$

So,

$$ \begin{aligned} k(x) &= -\frac{1}{\Gamma (\gamma )} \int _{x}^{1} (s-x) ^{ \gamma -1}\phi (s)\,ds \\ & \quad {} + \frac{1-x}{\Gamma (\gamma )[1 - r_{2}(1 - \eta )]} \biggl( \int _{0}^{1} s^{\gamma -1}\phi (s)\,ds - \int _{\eta }^{1} r_{2} (s-\eta ) ^{\gamma -1}\phi (s)\,ds \biggr) \\ &= \int _{0}^{1} G_{2}(x,s)\phi (s)\,ds. \end{aligned} $$

Similarly, the solution of boundary value problem (2.4) is given by

$$\begin{aligned}& y(x)= \int _{0}^{1} G_{1}(x,s)\varphi _{q} \bigl(k(s) \bigr)\,ds. \end{aligned}$$

Consequently, boundary value problem (2.1) is equivalent to (2.2).

From the monotonicity, for $x,\tau ,s\in (0,1)$, $G_{1}(x, \tau ),G_{2}(\tau , s)>0$ are verified easily. The proof is completed. □

Lemma 2.2

For given $\phi (s) \in C[0,1]$, set

$$\begin{aligned}& w(\tau )= \varphi _{q} \biggl( \int _{0}^{1} G_{2}(\tau , s) \phi (s) \,ds \biggr), \quad\quad y(x)= \int _{0}^{1} G_{1}(x, \tau ) w(\tau ) \,d \tau , \\& M_{1}= \max_{0\leq x\leq 1} \int _{0}^{1} G_{1}(x, \tau ) \,d \tau , \quad\quad M_{2}= \max_{0\leq \tau \leq 1} \int _{0}^{1} G_{2}( \tau , s) \,ds. \end{aligned}$$

Then

$$ \begin{aligned} \Vert w \Vert \leq M_{2}^{q-1} \Vert \phi \Vert ^{q-1}, \qquad \Vert y \Vert \leq M_{1}M_{2}^{q-1} \Vert \phi \Vert ^{q-1}. \end{aligned} $$

Proof

Since $\frac{1}{p}+ \frac{1}{q}= 1$ and $\varphi _{p}$ is increasing, then

$$ \begin{aligned} w(\tau ) &= \varphi _{q} \biggl( \int _{0}^{1} G_{2}( \tau , s) \phi (s) \,ds \biggr) \\ &\leq \varphi _{q} \biggl( \int _{0}^{1} G_{2}(\tau , s) \,ds \Vert \phi \Vert \biggr) \\ &\leq M_{2}^{q-1} \Vert \phi \Vert ^{q-1}. \end{aligned} $$

So, $\Vert w \Vert \leq M_{2}^{q-1} \Vert \phi \Vert ^{q-1}$. Similarly, $\Vert y \Vert \leq M_{1}M_{2}^{q-1} \Vert \phi \Vert ^{q-1}$. The proof is completed. □

Lemma 2.3

([7, 13])

The following relations of p-Laplace operator hold:

• (i) There are $1< p\leq 2$, $\vert k_{1} \vert , \vert k_{2} \vert \geq n> 0$, and $k_{1}k_{2}> 0$ such that

  $$\begin{aligned}& \bigl\vert \varphi _{p}(k_{2})- \varphi _{p}(k_{1}) \bigr\vert \leq (p- 1)n^{p-2} \vert k_{2}-k_{1} \vert ; \end{aligned}$$

• (ii) There are $p> 2$ and $\vert k_{1} \vert , \vert k_{2} \vert \leq N$ such that

  $$\begin{aligned}& \bigl\vert \varphi _{p}(k_{2})- \varphi _{p}(k_{1}) \bigr\vert \leq (p- 1)N^{p-2} \vert k_{2}-k_{1} \vert . \end{aligned}$$

3 Main results

This section studies the existence and uniqueness of solutions for mixed fractional-order p-Laplace boundary value problem (1.1) by applying the Banach contraction mapping principle.

Given number $M> 0$, denote

$$\begin{aligned}& D_{M}= \bigl\{ (x,y,w)| 0\leq x\leq 1, \vert y \vert \leq M_{1}M_{2}^{q-1}M^{q-1}, \vert w \vert \leq M_{2}^{q-1}M^{q-1} \bigr\} , \end{aligned}$$

and by a closed ball $B[O,M]$ in the space of continuous functions $C[0,1]$.

Theorem 3.1

Assume that $1< p\leq 2$, and there exist some numbers $M, Q_{1}, Q_{2}>0$ such that the following conditions hold:

• (H1) $\vert g(x, y, w) \vert \leq M$ for $(x, y, w)\in D_{M}$;

• (H2) $\vert g(x,y_{2},w_{2}) - g(x,y_{1},w_{1}) \vert \leq Q_{1} \vert y_{2}-y_{1} \vert + Q_{2} \vert w_{2}-w_{1} \vert$ for $(x,y_{i},w_{i})\in D_{M}$, $i=1,2$;

• (H3) $L_{1}:= (q-1) M^{q-2} M_{2}^{q-1} (Q_{1} M_{1} + Q_{2})<1$.

Then the mixed boundary value problem (1.1) has a unique solution satisfying the estimation

$$\begin{aligned}& \bigl\vert y(x) \bigr\vert \leq M_{1}M_{2}^{q-1}M^{q-1}, \quad\quad \bigl\vert D_{0+}^{\delta }y(x) \bigr\vert \leq M_{2}^{q-1}M^{q-1} \quad \textit{for } x\in [0,1]. \end{aligned}$$

Proof

First of all, define an operator $A:C[0,1]\rightarrow C[0,1]$ by

$$\begin{aligned}& \begin{aligned} (A\phi ) (x)={}& g \biggl( x, \int _{0}^{1} G_{1}(x, \tau ) \varphi _{q} \biggl( \int _{0}^{1} G_{2}(\tau ,s) \phi (s) \,ds \biggr) \,d\tau , \\ &{}\varphi _{q} \biggl( \int _{0}^{1} G_{2}(x,s) \phi (s) \,ds \biggr) \biggr). \end{aligned} \end{aligned}$$

From the continuity of $G_{1}(x,\tau )$, $G_{2}(\tau ,s)$, and $g(x,y,w)$, it is not hard to see that the operator A is a continuous operator. According to Lemma 2.1, it is easy to get the following conclusion that if the mixed boundary value problem (1.1) has a solution $y(x)$, then $\phi (x)= {}^{c} D_{1-}^{\gamma }( \varphi _{p} (D_{0+}^{\delta } y(x)))$ is the fixed point of the operator A. On the contrary, if $\phi (x)$ is a fixed point of the operator A, then

$$\begin{aligned}& y(x)= \int _{0}^{1} G_{1}(x, \tau ) \varphi _{q} \biggl( \int _{0}^{1} G_{2}(\tau , s) \phi (s) \,ds \biggr) \,d\tau \end{aligned}$$

also is the solution of the mixed boundary value problem (1.1).

Next, what we need to prove is that the operator A maps $B[O,M]$ into itself. Given $\phi (x)\in B[O,M]$, by Lemma 2.2, there is

$$\begin{aligned}& \bigl\vert y(x) \bigr\vert \leq M_{1}M_{2}^{q-1}M^{q-1}, \qquad \bigl\vert w(x) \bigr\vert \leq M_{2}^{q-1}M^{q-1}. \end{aligned}$$

Consequently, for any $x\in [0,1]$, there is $(x,y(x),w(x))\in D_{M}$. So, from $(H1)$, we can conclude that

$$\begin{aligned}& \bigl\vert (A\phi ) (x) \bigr\vert = \bigl\vert g \bigl(x,y(x),w(x) \bigr) \bigr\vert \leq M, \end{aligned}$$

therefore $(A\phi )(x)\in B[O, M]$. Namely, the operator A is an operator that maps $B[O,M]$ into itself.

Finally, the operator $A:B[O,M]\rightarrow B[O,M]$ is proven to be a contraction mapping. Obviously, $B[O,M]$ is proven to be a complete distance space on account of that $B[O,M]$ is a subspace of $C([0,1], \Vert \cdot \Vert )$. From $(H2)$, Lemma 2.2, and (ii) of Lemma 2.3, there is $\vert \int _{0}^{1} G_{2}(\tau ,s) \phi (s) \,ds \vert \leq M_{2}M := N$ for each $\phi _{1}(x), \phi _{2}(x)\in B[O,M]$ and $1< p\leq 2$, that is, $q \geq 2$, we gain

$$\begin{aligned} & \bigl\vert (A\phi _{2}) (x)- (A\phi _{1}) (x) \bigr\vert \\ &\quad = \bigl\vert g \bigl(x,y_{2}(x),w_{2}(x) \bigr)- g \bigl(x,y_{1}(x),w_{1}(x) \bigr) \bigr\vert \\ & \quad \leq Q_{1} \bigl\vert y_{2}(x)-y_{1}(x) \bigr\vert + Q_{2} \bigl\vert w_{2}(x)-w_{1}(x) \bigr\vert \\ & \quad \leq Q_{1} \biggl\vert \int _{0}^{1} G_{1}(x, \tau ) \biggl[ \varphi _{q} \biggl( \int _{0}^{1} G_{2}(\tau , s) \phi _{2}(s) \,ds \biggr) - \varphi _{q} \biggl( \int _{0}^{1} G_{2}(\tau , s) \phi _{1}(s) \,ds \biggr) \biggr] \,d\tau \biggr\vert \\ & \quad \quad {} + Q_{2} \biggl\vert \varphi _{q} \biggl( \int _{0}^{1} G_{2}(\tau , s) \phi _{2}(s) \,ds \biggr) - \varphi _{q} \biggl( \int _{0}^{1} G_{2}( \tau , s) \phi _{1}(s) \,ds \biggr) \biggr\vert \\ & \quad \leq Q_{1}(q-1) (M_{2}M)^{q-2} \biggl[ \int _{0}^{1} G_{1}(x, \tau ) \biggl\vert \int _{0}^{1} G_{2}(\tau ,s) \bigl(\phi _{2}(s)- \phi _{1}(s) \bigr) \,ds \biggr\vert \,d\tau \biggr] \\ & \quad \quad {} + Q_{2}(q-1) (M_{2}M)^{q-2} \biggl\vert \int _{0}^{1} G_{2}(\tau ,s) \bigl(\phi _{2}(s)- \phi _{1}(s) \bigr) \,ds \biggr\vert \\ &\quad \leq Q_{1}(q-1) (M_{2}M)^{q-2} \biggl[ \int _{0}^{1} G_{1}(x, \tau ) \int _{0}^{1} G_{2}(\tau ,s) \,ds \,d\tau \Vert \phi _{2}- \phi _{1} \Vert \biggr] \\ & \quad \quad {} + Q_{2}(q-1) (M_{2}M)^{q-2} \biggl[ \int _{0}^{1} G_{2}(\tau ,s) \,ds \Vert \phi _{2}- \phi _{1} \Vert \biggr] \\ & \quad \leq Q_{1}(q-1) (M_{2}M)^{q-2} M_{1} M_{2} \Vert \phi _{2}- \phi _{1} \Vert \\ & \quad \quad {} + Q_{2}(q-1) (M_{2}M)^{q-2} M_{2} \Vert \phi _{2}- \phi _{1} \Vert \\ & \quad = (q-1) M^{q-2} M_{2}^{q-1} (Q_{1} M_{1} + Q_{2}) \Vert \phi _{2}- \phi _{1} \Vert \\ & \quad = L_{1} \Vert \phi _{2}- \phi _{1} \Vert . \end{aligned}$$

Hence,

$$\begin{aligned}& \bigl\Vert (A\phi _{2})- (A\phi _{1}) \bigr\Vert \leq L_{1} \Vert \phi _{2}- \phi _{1} \Vert , \end{aligned}$$

where $L_{1}<1$ is given by $(H3)$. Therefore, the operator $A:B[O,M]\rightarrow B[O,M]$ is a contraction mapping. That is to say, there is a unique solution y(x) of the mixed boundary value problem (1.1) such that

$$\begin{aligned}& \bigl\vert y(x) \bigr\vert \leq M_{1}M_{2}^{q-1}M^{q-1}, \quad\quad \bigl\vert D_{0+}^{\delta }y(x) \bigr\vert \leq M_{2}^{q-1}M^{q-1} \quad \text{for } x\in [0,1]. \end{aligned}$$

The proof is completed. □

4 An example

Example 4.1

Consider the following mixed fractional-order p-Laplace differential equation:

$$\begin{aligned}& {}^{c} D_{1-}^{\frac{3}{2}} \bigl( \varphi _{p} \bigl(D_{0+}^{\frac{3}{2}} y(x) \bigr) \bigr)= -2y^{2}(x)D_{0+}^{\frac{3}{2}}y(x) + 3y(x) - 2D_{0+}^{\frac{3}{2}}y(x) + 4\sin (\pi x), \\& y(0)= y(1)= D_{0+}^{\frac{3}{2}} y(0)= D_{0+}^{\frac{3}{2}} y(1)=0, \end{aligned}$$

where $g(x,y,w)= -2y^{2}w + 3y - 2w + 4\sin (\pi x)$.

We choose $p=2$, that is, $q=2$. By a simple computation, we obtain that $M_{1}=\frac{1}{8}$, $M_{2}=\frac{1}{3}$. Then a suitable number $M > 0$ is chosen to satisfy all the conditions of Theorem 3.1, and

$$\begin{aligned}& \vert y \vert \leq \frac{M}{24}, \qquad \vert w \vert \leq \frac{M}{3}. \end{aligned}$$

Obviously, for each $(x,y,w)\in D_{M}$, there is

$$ \begin{aligned} \bigl\vert g(x,y,w) \bigr\vert &\leq \bigl\vert -2y^{2}w + 3y - 2w + 4\sin (\pi x) \bigr\vert \\ &\leq 2 \biggl(\frac{M}{24} \biggr)^{2} \biggl( \frac{M}{3} \biggr) + 3 \biggl(\frac{M}{24} \biggr) + 2 \biggl( \frac{M}{3} \biggr) + 4\sin ( \pi x) \\ &\leq M. \end{aligned} $$

We can easily get that $0 < M < 13.4164$. Therefore, choose $M = 1$, condition $(H1)$ holds.

Meanwhile, for $(x,y,w)\in D_{M}$,

$$\begin{aligned}& \vert g_{y} \vert = \vert -4yw + 3 \vert \leq 4 \frac{M}{24} \frac{M}{3} + 3 \leq 3.5, \\& \vert g_{w} \vert = \bigl\vert -2y^{2} - 2 \bigr\vert \leq 2 \biggl(\frac{M}{24} \biggr)^{2} + 2 \leq 2.5. \end{aligned}$$

So we choose $Q_{1}=3.5$, $Q_{2}=2.5$, condition $(H2)$ holds.

Moreover,

$$\begin{aligned}& L_{1}= (q-1) M^{q-2} M_{2}^{q-1} (Q_{1} M_{1} + Q_{2})= 0.979 < 1, \end{aligned}$$

condition $(H3)$ holds.

So, the above boundary value problem has a unique solution satisfying

$$\begin{aligned}& \bigl\vert y(x) \bigr\vert \leq \frac{1}{24}, \qquad \bigl\vert D_{0+}^{\frac{3}{2}}y(x) \bigr\vert \leq \frac{1}{3}. \end{aligned}$$

5 Conclusion

In this article, we have discussed a novel approach which has been put forward by Dang and Bai to prove the existence and uniqueness of solution for a nonlinear multi-point fractional-order p-Laplace differential equation, which has at least two different kinds of left and right fractional derivatives. As far as we know, almost nobody explored the work in this area. The advantage of this method is that it is very easy to verify and can be applied to many conditions. Of course, if the boundary condition is changed, the assumptions can be weakened appropriately. Last but not least, the approach mentioned above can be applied to some other boundary value problems, such as conformable fractional order. Furthermore, the boundary conditions can also become integral boundary conditions, and so on.