Some existence results on boundary value problems for fractional p-Laplacian equation at resonance

Abstract

Two boundary value problems of the fractional p-Laplacian equation at resonance are considered in this paper. By using the continuation theorem due to Ge, we obtain some existence results for such boundary value problems.

1 Introduction

Consider the following fractional p-Laplacian equation:

$$\begin{aligned} D_{0^{+}}^{\beta}\phi_{p} \bigl(D_{0^{+}}^{\alpha}x(t)\bigr)=f\bigl(t,x(t),D_{0^{+}}^{\alpha}x(t)\bigr), \quad t\in[0,1], \end{aligned}$$

(1.1)

with the boundary value conditions either

$$\begin{aligned} x(0)=x(1), \qquad D_{0^{+}}^{\alpha}x(0)=0 , \end{aligned}$$

(1.2)

or

$$\begin{aligned} x(0)=x(1), \qquad D_{0^{+}}^{\alpha}x(1)=0, \end{aligned}$$

(1.3)

where $0<\alpha$, $\beta\leq1$, $\phi_{p}(s)=\vert s\vert ^{p-2}s$ ($p>1$), $D_{0^{+}}^{\alpha}$ is a Caputo fractional derivative, and $f:[0,1]\times \mathbb{R}^{2}\rightarrow\mathbb{R}$ is a continuous function.

In the last two decades, the theory of fractional calculus has gained popularity due to its wide applications in various fields of engineering and the sciences [1–8]. Moreover, the p-Laplacian equations often exist in non-Newtonian fluid theory, nonlinear elastic mechanics, and so on.

Recently, many important results on the p-Laplacian equations or the fractional differential equations have been given. We refer the reader to [9–31]. However, as far as we know, there is little work about boundary value problems (BVPs for short) for the fractional differential equations with p-Laplacian operator at resonance.

Note that BVP (1.1)-(1.2) (or BVP (1.1)-(1.3)) happens to be at resonance because its associated homogeneous BVP

$$\begin{aligned} \textstyle\begin{cases} D_{0^{+}}^{\beta}\phi_{p}(D_{0^{+}}^{\alpha}x(t))=0, \quad t\in[0,1],\\ x(0)=x(1),\qquad D_{0^{+}}^{\alpha}x(0)=0 \quad (\mbox{or }x(0)=x(1), D_{0^{+}}^{\alpha}x(1)=0), \end{cases}\displaystyle \end{aligned}$$

has a solution $x(t)=c$, $\forall c\in\mathbb{R}$.

The rest of this paper is organized as follows. Section 2 contains some definitions, lemmas and notations. In Section 3, some related lemmas are stated and proved which are useful in the proof of our main results. In Section 4 and Section 5, in view of the continuation theorem due to Ge, we establish two theorems about the existence of solutions for BVP (1.1)-(1.2) (Theorem 4.1) and BVP (1.1)-(1.3) (Theorem 5.1).

2 Preliminaries

We give here some definitions and lemmas about the fractional calculus.

Definition 2.1

[32]

The Riemann-Liouville fractional integral operator of order $\alpha>0$ of a function $x:(0,+\infty )\rightarrow\mathbb{R}$ is given by

$$\begin{aligned} I_{0^{+}}^{\alpha}x(t)=\frac{1}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha-1}x(s)\,ds, \end{aligned}$$

provided that the right side integral is pointwise defined on $(0,+\infty)$.

Definition 2.2

[32]

The Caputo fractional derivative of order $\alpha>0$ of a continuous function $x:(0,+\infty)\rightarrow \mathbb{R}$ is given by

$$\begin{aligned} D_{0^{+}}^{\alpha}x(t) =&I_{0^{+}}^{n-\alpha} \frac{d^{n}x(t)}{d t^{n}} \\ =&\frac{1}{\Gamma(n-\alpha)} \int_{0}^{t}(t-s)^{n-\alpha-1}x^{(n)}(s)\,ds, \end{aligned}$$

where n is the smallest integer greater than or equal to α, provided that the right side integral is pointwise defined on $(0,+\infty)$.

Lemma 2.1

[8]

Let $\alpha>0$. Assume that $x,D_{0^{+}}^{\alpha}x\in L([0,1],\mathbb{R})$. Then the following equality holds:

$$\begin{aligned} I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}x(t)=x(t)+c_{0}+c_{1}t+ \cdots+c_{n-1}t^{n-1}, \end{aligned}$$

where $c_{i}\in{\mathbb{R}}$, $i=0,1,\ldots,n-1$, and n is the smallest integer greater than or equal to α.

Lemma 2.2

[33]

For any $u,v\geq0$,

$$\begin{aligned}& \phi_{p}(u+v)\leq\phi_{p}(u)+\phi_{p}(v), \quad \textit{if } p< 2; \\& \phi_{p}(u+v)\leq2^{p-2}\bigl(\phi_{p}(u)+ \phi_{p}(v)\bigr), \quad \textit{if } p\geq2. \end{aligned}$$

Next we introduce an extension of Mawhin’s continuation theorem [34, 35] which allows us to deal with the more general abstract operator equations, such as BVPs of p-Laplacian equations.

Let X and Z be Banach spaces with norms $\Vert \cdot \Vert _{X}$ and $\Vert \cdot \Vert _{Z}$, respectively.

Definition 2.3

[35]

A continuous operator $M:\operatorname {dom}M\cap X\rightarrow Z$ is said to be a quasi-linear operator if

• (1) $\operatorname {Im}M=M(\operatorname {dom}M\cap X)$ is a closed subset of Z,

• (2) $\operatorname {Ker}M=\{x\in \operatorname {dom}M\cap X|Mx=0\}$ is linearly homeomorphic to $\mathbb{R}^{n}$ with $n<\infty$.

Definition 2.4

[35]

Let $Z_{1}$ be a subspace of Z. An operator $Q:Z\rightarrow Z_{1}$ is said to be a semi-projector provided that

• (1) $Q^{2}z=Qz$, $\forall z\in Z$,

• (2) $Q(\lambda z)=\lambda Qz$, $\forall z\in Z$, $\lambda\in\mathbb{R}$.

Set $X_{1}=\operatorname {Ker}M$ and let $X_{2}$ be the complement space of $X_{1}$ in X, then $X=X_{1}\oplus X_{2}$. Suppose $Z_{1}$ is a subspace of Z and $Z_{2}$ is the complement space of $Z_{1}$ in Z such that $Z=Z_{1}\oplus Z_{2}$. Let $P:X\rightarrow X_{1}$ be a projector and $Q:Z\rightarrow Z_{1}$ a semi-projector, and $\Omega\subset X$ an open bounded set with the origin $\theta\in\Omega$.

Definition 2.5

[35]

A continuous operator $N_{\lambda}:\overline{\Omega}\rightarrow Z$, $\lambda\in[0,1]$ is said to be M-compact in Ω̅ if there is a vector subspace $Z_{1}$ of Z with $\dim Z_{1}=\dim X_{1}$, and an operator $R:\overline {\Omega}\times[0,1]\rightarrow X_{2}$ being continuous and compact such that

$$\begin{aligned}& (I-Q)N_{\lambda}(\overline{\Omega})\subset \operatorname {Im}M \subset(I-Q)Z, \end{aligned}$$

(2.1)

$$\begin{aligned}& QN_{\lambda}x=\theta, \quad \lambda\in(0,1) \quad \Leftrightarrow\quad QNx=\theta, \end{aligned}$$

(2.2)

$$\begin{aligned}& R(\cdot,0) \mbox{ is the zero operator}\quad \mbox{and}\quad R(\cdot, \lambda)|_{\sum _{\lambda}}=(I-P)|_{\sum_{\lambda}}, \end{aligned}$$

(2.3)

$$\begin{aligned}& M\bigl(P+R(\cdot,\lambda)\bigr)=(I-Q)N_{\lambda}, \end{aligned}$$

(2.4)

where $\lambda\in[0,1]$, $N=N_{1}$, and $\sum_{\lambda}=\{x\in\overline{\Omega }|Mx=N_{\lambda}x\}$.

Lemma 2.3

[35]

Suppose $M:\operatorname {dom}M\cap X\rightarrow Z$ is a quasi-linear operator and $N_{\lambda}:\overline{\Omega}\rightarrow Z$, $\lambda\in[0,1]$ is M-compact in Ω̅. In addition, if

(C₁):

$Mx\neq N_{\lambda}x$ for every $(x,\lambda)\in[(\operatorname {dom}M\setminus \operatorname {Ker}M)\cap\partial\Omega]\times(0,1)$;

(C₂):

$QNx\neq0$ for every $x\in \operatorname {Ker}M\cap\partial\Omega$;

(C₃):

$\deg \{JQN,\Omega\cap \operatorname {Ker}M,0\}\neq0$,

where $N=N_{1}$ and $J:Z_{1}\rightarrow X_{1}$ is a homeomorphism with $J(\theta)=\theta$, then the abstract equation $Mx=Nx$ has at least one solution in $\operatorname {dom}M\cap\overline{\Omega}$.

We set $Z=C([0,1],\mathbb{R})$ with the norm $\Vert z\Vert _{0}=\max_{t\in [0,1]}\vert z(t)\vert$, and $X=\{x\in Z|D_{0^{+}}^{\alpha}{x\in Z}, x(0)=x(1),D_{0^{+}}^{\alpha}x(0)=0\}$, $X^{1}=\{x\in Z|D_{0^{+}}^{\alpha}x\in Z,x(0)=x(1),D_{0^{+}}^{\alpha}x(1)=0\}$ with the norm $\Vert x\Vert _{X}=\max\{\Vert x\Vert _{0},\Vert D_{0^{+}}^{\alpha}x \Vert _{0}\}$. By using linear functional analysis theory, we can prove X, $X^{1}$ are Banach spaces.

3 Related lemmas

We will give some lemmas that are useful in the proof of our main results.

Define the operator $M:\operatorname {dom}M\cap X\rightarrow Z$ by

$$\begin{aligned} Mx=D_{0^{+}}^{\beta}\phi_{p} \bigl(D_{0^{+}}^{\alpha}x\bigr), \end{aligned}$$

(3.1)

where $\operatorname {dom}M=\{x\in X|D_{0^{+}}^{\beta}\phi_{p}(D_{0^{+}}^{\alpha}x)\in Z\}$. For $\lambda\in[0,1]$, we define $N_{\lambda}:X\rightarrow Z$ by

$$\begin{aligned} N_{\lambda}x(t)=\lambda f\bigl(t,x(t),D_{0^{+}}^{\alpha}x(t)\bigr), \quad \forall t\in[0,1]. \end{aligned}$$

(3.2)

Then BVP (1.1)-(1.2) is equivalent to the equation

$$\begin{aligned} Mx=Nx,\quad x\in \operatorname {dom}M, \end{aligned}$$

where $N=N_{1}$.

Lemma 3.1

The operator M, defined by (3.1), is a quasi-linear operator.

Proof

The proof will be given in the following two steps.

Step 1. KerM is linearly homeomorphic to $\mathbb{R}$.

From Lemma 2.1, the homogeneous equation $D_{0^{+}}^{\beta}\phi _{p}(D_{0^{+}}^{\alpha}x(t))=0$ has the following solutions:

$$\begin{aligned} x(t)=d_{2}+\frac{\phi_{q}(d_{1})}{\Gamma(\alpha+1)}t^{\alpha}, \quad d_{1},d_{2}\in\mathbb{R}. \end{aligned}$$

Thus, by the boundary value condition $D_{0^{+}}^{\alpha}x(0)=0$, one has

$$\begin{aligned} \operatorname {Ker}M=\bigl\{ x\in X|x(t)=d, \forall t\in[0,1],d\in \mathbb{R}\bigr\} . \end{aligned}$$

Obviously, $\operatorname {Ker}M\simeq\mathbb{R}$.

Step 2. ImM is a closed subset of Z.

Take $x\in \operatorname {dom}M$ and consider the equation $D_{0^{+}}^{\beta}\phi _{p}(D_{0^{+}}^{\alpha}x(t))=z(t)$. Then we have $z\in Z$ and

$$\begin{aligned} \phi_{p}\bigl(D_{0^{+}}^{\alpha}x(t)\bigr) =d_{1}+I_{0^{+}}^{\beta}z(t), \quad d_{1}\in \mathbb{R}. \end{aligned}$$

By the condition $D_{0^{+}}^{\alpha}x(0)=0$, one has $d_{1}=0$. Thus we get

$$\begin{aligned} x(t) =d_{2}+I_{0^{+}}^{\alpha}\phi_{q} \bigl(I_{0^{+}}^{\beta}z\bigr) (t),\quad d_{2}\in \mathbb{R}, \end{aligned}$$

where $\phi_{q}$ is understood as the operator $\phi_{q}:Z\rightarrow Z$ defined by $\phi_{q}(x)(t)=\phi_{q}(x(t))$. Hence, from the condition $x(0)=x(1)$, we obtain

$$\begin{aligned} I_{0^{+}}^{\alpha}\phi_{q} \bigl(I_{0^{+}}^{\beta}z\bigr) (1)=0. \end{aligned}$$

(3.3)

Suppose $z\in Z$ and satisfies (3.3). Let $x(t)=I_{0^{+}}^{\alpha}\phi_{q}(I_{0^{+}}^{\beta}z)(t)$, then we have $x\in \operatorname {dom}M$ and

$$\begin{aligned} Mx(t)=D_{0^{+}}^{\beta}\phi_{p}\bigl[D_{0^{+}}^{\alpha}I_{0^{+}}^{\alpha}\phi _{q}\bigl(I_{0^{+}}^{\beta}z\bigr)\bigr](t)=z(t). \end{aligned}$$

Hence we obtain

$$\begin{aligned} \operatorname {Im}M= \biggl\{ z\in Z\Big| \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl( \int_{0}^{s}(s-\tau)^{\beta-1}z(\tau)\,d\tau \biggr)\,ds=0 \biggr\} . \end{aligned}$$

Obviously, $\operatorname {Im}M\subset Z$ is closed.

Therefore, by Definition 2.3, M is a quasi-linear operator. □

Let $X_{1}=\operatorname {Ker}M$ and define the continuous operators $P:X\rightarrow X$, $Q:Z\rightarrow Z$ by

$$\begin{aligned}& Px(t)=x(0), \quad \forall t\in[0,1], \\& Qz(t)=\phi_{p} \biggl[\frac{1}{\rho} \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl( \int_{0}^{s}(s-\tau)^{\beta-1}z(\tau)\,d\tau \biggr)\,ds \biggr], \quad \forall t\in[0,1], \end{aligned}$$

where $\rho=\frac{1}{\beta^{q-1}}\int_{0}^{1}(1-s)^{\alpha-1}s^{\beta (q-1)}\,ds>0$. It is easy to see that P is a projector and $Q^{2}z=Qz$, $Q(\lambda z)=\lambda Qz$, $\forall z\in Z$, $\lambda\in\mathbb{R}$, that is, Q is a semi-projector. Moreover, $X_{1}=\operatorname {Im}P$ and $\operatorname {Im}M=\operatorname {Ker}Q$.

Lemma 3.2

Let $\Omega\subset X$ be an open bounded set. Then the operator $N_{\lambda}$, defined by (3.2), is M-compact in Ω̅.

Proof

Choose $X_{2}=\operatorname {Ker}P$, $Z_{1}=\operatorname {Im}Q$ and define the operator $R:\overline{\Omega}\times[0,1]\rightarrow X_{2}$ by

$$\begin{aligned} R(x,\lambda) (t) =&I_{0^{+}}^{\alpha}\phi_{q} \bigl[I_{0^{+}}^{\beta}(I-Q)N_{\lambda}x\bigr](t) \\ =&\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}\phi_{q} \biggl[\frac{1}{\Gamma(\beta)} \\ &{}\cdot \int_{0}^{s}(s-\tau)^{\beta-1} \bigl(\lambda f \bigl(\tau,x(\tau),D_{0^{+}}^{\alpha}x(\tau)\bigr)-QN_{\lambda}x(\tau)\bigr)\,d\tau \biggr]\,ds. \end{aligned}$$

Obviously, $\dim Z_{1}=\dim X_{1}=1$. The remainder of the proof will be given in the following two steps.

Step 1. $R:\overline{\Omega}\times[0,1]\rightarrow X_{2}$ is continuous and compact.

By the definition of R, we obtain

$$\begin{aligned} D_{0^{+}}^{\alpha}Rx(t)=\phi_{q}\bigl[I_{0^{+}}^{\beta}(I-Q)N_{\lambda}x\bigr](t). \end{aligned}$$

Clearly, the operators R, $D_{0^{+}}^{\alpha}R$ are compositions of the continuous operators. So R, $D_{0^{+}}^{\alpha}R$ are continuous in Z. Hence R is a continuous operator, and $R(\overline{\Omega})$, $D_{0^{+}}^{\alpha}R(\overline{\Omega})$ are bounded in Z. Furthermore, there exists a constant $T>0$ such that $|I_{0^{+}}^{\beta}(I-Q)N_{\lambda}x(t)|\leq T$, $\forall x\in\overline{\Omega}$, $t\in[0,1]$. Thus, based on the Arzelà-Ascoli theorem, we need only to show $R(\overline {\Omega})\subset X$ is equicontinuous.

For $0\leq t_{1}< t_{2}\leq1$, $x\in\overline{\Omega}$, we have

$$\begin{aligned} & \bigl\vert Rx(t_{2})-Rx(t_{1})\bigr\vert \\ &\quad = \frac{1}{\Gamma(\alpha)} \biggl\vert \int_{0}^{t_{2}}(t_{2}-s)^{\alpha-1} \phi_{q}\bigl[I_{0^{+}}^{\beta}(I-Q)N_{\lambda}x(s) \bigr]\,ds \\ &\qquad{} - \int_{0}^{t_{1}}(t_{1}-s)^{\alpha-1} \phi_{q}\bigl[I_{0^{+}}^{\beta}(I-Q)N_{\lambda}x(s) \bigr]\,ds\biggr\vert \\ &\quad \leq \frac{T^{q-1}}{\Gamma(\alpha)} \biggl\{ \int_{0}^{t_{1}}\bigl[(t_{1}-s)^{\alpha-1}-(t_{2}-s)^{\alpha-1} \bigr]\,ds + \int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-1}\,ds \biggr\} \\ &\quad = \frac{T^{q-1}}{\Gamma(\alpha+1)}\bigl[t_{1}^{\alpha}-t_{2}^{\alpha} +2(t_{2}-t_{1})^{\alpha}\bigr]. \end{aligned}$$

As $t^{\alpha}$ is uniformly continuous in $[0,1]$, we obtain $R(\overline{\Omega})\subset Z$ is equicontinuous. A similar proof can show that $I_{0^{+}}^{\beta}(I-Q)N_{\lambda}(\overline{\Omega})\subset Z$ is equicontinuous. This, together with the uniformly continuity of $\phi _{q}(s)$ on $[-T,T]$, shows that $D_{0^{+}}^{\alpha}R(\overline{\Omega })\subset Z$ is equicontinuous. Thus we find R is compact.

Step 2. Equations (2.1)-(2.4) are satisfied.

For $x\in\overline{\Omega}$, it is easy to show that $Q(I-Q)N_{\lambda}x=QN_{\lambda}x-Q^{2}N_{\lambda}x=0$. So $(I-Q)N_{\lambda}x\in \operatorname {Ker}Q=\operatorname {Im}M$. Moreover, for $z\in \operatorname {Im}M\subset Z$, one has $Qz=0$. Thus $z=z-Qz=(I-Q)z\in(I-Q)Z$. Hence (2.1) holds. Since $QN_{\lambda}x=\lambda QNx$, (2.2) holds too.

For $x\in\sum_{\lambda}$, we have $Mx=N_{\lambda}x\in \operatorname {Im}M=\operatorname {Ker}Q$. So $QN_{\lambda}x=0$. From the condition $D_{0^{+}}^{\alpha }x(0)=0$, one has $I_{0^{+}}^{\beta}D_{0^{+}}^{\beta}\phi_{p}(D_{0^{+}}^{\alpha}x)=\phi_{p}(D_{0^{+}}^{\alpha}x)$. Thus we obtain

$$\begin{aligned} R(x,\lambda) (t) =&I_{0^{+}}^{\alpha}\phi_{q} \bigl(I_{0^{+}}^{\beta}N_{\lambda}x\bigr) (t) \\ =&I_{0^{+}}^{\alpha}\phi_{q} \bigl[I_{0^{+}}^{\beta}D_{0^{+}}^{\beta}\phi _{p}\bigl(D_{0^{+}}^{\alpha}x\bigr)\bigr](t) \\ =&x(t)-x(0) \\ =&(I-P)x(t). \end{aligned}$$

Furthermore, when $\lambda=0$, we have $N_{\lambda}x(t)\equiv0$, which yields $R(x,0)(t)\equiv0$, $\forall x\in\overline{\Omega}$. Hence (2.3) holds.

For $x\in\overline{\Omega}$, one has

$$\begin{aligned} M\bigl(Px+R(x,\lambda)\bigr) (t) =&D_{0^{+}}^{\beta}\phi_{p}\bigl[D_{0^{+}}^{\alpha}\bigl(Px+R(x,\lambda) \bigr)\bigr](t) \\ =&D_{0^{+}}^{\beta}\phi_{p}\bigl[D_{0^{+}}^{\alpha}I_{0^{+}}^{\alpha}\phi_{q}\bigl(I_{0^{+}}^{\beta}(I-Q)N_{\lambda}x\bigr)\bigr](t) \\ =&(I-Q)N_{\lambda}x(t), \end{aligned}$$

which implies that (2.4) holds.

Therefore, by Definition 2.5, $N_{\lambda}$ is M-compact in Ω̅. □

4 Solutions of BVP (1.1)-(1.2)

We will give a theorem on the existence of solutions for BVP (1.1)-(1.2).

Theorem 4.1

Let $f:[0,1]\times\mathbb{R}^{2}\rightarrow\mathbb {R}$ be continuous. Assume that:

(H₁):

there exist nonnegative functions $a,b,c\in Z$ such that

$$\begin{aligned} \bigl\vert f(t,x,y)\bigr\vert \leq a(t)+b(t)\vert x\vert ^{p-1}+c(t)\vert y\vert ^{p-1}, \quad \forall t\in [0,1],(x,y)\in\mathbb{R}^{2}; \end{aligned}$$

(H₂):

there exists a constant $A>0$ such that, for $\forall x\in \operatorname {dom}M\setminus \operatorname {Ker}M$ satisfying $\vert x(t)\vert >A$ for $\forall t\in [0,1]$, we have

$$\begin{aligned} \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl( \int_{0}^{s}(s-\tau)^{\beta-1}f\bigl(\tau,x( \tau),D_{0^{+}}^{\alpha}x(\tau)\bigr)\,d\tau \biggr)\,ds\neq0; \end{aligned}$$

(H₃):

there exists a constant $B>0$ such that, for $\forall r\in \mathbb{R}$ with $\vert r\vert >B$, we have either

$$\begin{aligned} \phi_{q}(r) \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl( \int_{0}^{s}(s-\tau)^{\beta-1}f(\tau,r,0)\,d\tau \biggr)\,ds>0 \end{aligned}$$

(4.1)

or

$$\begin{aligned} \phi_{q}(r) \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl( \int_{0}^{s}(s-\tau)^{\beta-1}f(\tau,r,0)\,d\tau \biggr)\,ds< 0. \end{aligned}$$

(4.2)

Then BVP (1.1)-(1.2) has at least one solution, provided that

$$ \begin{aligned} &\gamma_{1}:=\frac{1}{\Gamma(\beta+1)} \biggl[ \frac{2^{p-1}\Vert b\Vert _{0}}{(\Gamma(\alpha+1))^{p-1}}+\Vert c\Vert _{0} \biggr]< 1,\quad \textit{if } p< 2; \\ &\gamma_{2}:=\frac{1}{\Gamma(\beta+1)} \biggl[\frac{2^{2p-3}\Vert b\Vert _{0}}{(\Gamma(\alpha+1))^{p-1}}+\Vert c \Vert _{0} \biggr]< 1, \quad \textit{if } p\geq2. \end{aligned} $$

(4.3)

Proof

The proof will be given in the following four steps.

Step 1. $\Omega_{1}=\{x\in \operatorname {dom}M\setminus \operatorname {Ker}M|Mx=N_{\lambda}x,\lambda\in(0,1)\}$ is bounded.

For $x\in\Omega_{1}$, one has $Nx\in \operatorname {Im}M=\operatorname {Ker}Q$. Thus we have

$$\begin{aligned} \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl( \int_{0}^{s}(s-\tau)^{\beta-1}f\bigl(\tau,x( \tau),D_{0^{+}}^{\alpha}x(\tau)\bigr)\,d\tau \biggr)\,ds=0. \end{aligned}$$

From (H₂), there exists a constant $\xi\in[0,1]$ such that $|x(\xi )|\leq A$. By Lemma 2.1, one has

$$\begin{aligned} x(t)=x(\xi)-I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}x( \xi)+I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}x(t), \end{aligned}$$

which together with

$$\begin{aligned} \bigl\vert I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}x(t)\bigr\vert =&\frac{1}{\Gamma(\alpha)}\biggl\vert \int_{0}^{t}(t-s)^{\alpha-1}D_{0^{+}}^{\alpha}x(s)\,ds\biggr\vert \\ \leq&\frac{1}{\Gamma(\alpha)}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}\cdot\frac{1}{\alpha }t^{\alpha} \\ \leq&\frac{1}{\Gamma(\alpha+1)}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0},\quad \forall t\in[0,1] , \end{aligned}$$

(4.4)

and $\vert x(\xi)\vert \leq A$ yields

$$\begin{aligned} \Vert x\Vert _{0}\leq A+\frac{2}{\Gamma(\alpha+1)}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}. \end{aligned}$$

(4.5)

Then, from (H₁), we have

$$\begin{aligned} \bigl\vert I_{0^{+}}^{\beta}Nx(t)\bigr\vert =& \frac{1}{\Gamma(\beta)}\biggl\vert \int_{0}^{t}(t-s)^{\beta -1}f \bigl(s,x(s),D_{0^{+}}^{\alpha}x(s)\bigr)\,ds\biggr\vert \\ \leq&\frac{1}{\Gamma(\beta)} \int_{0}^{t}(t-s)^{\beta -1}\bigl(a(s)+b(s)\bigl\vert x(s)\bigr\vert ^{p-1} \\ &{}+c(s)\bigl\vert D_{0^{+}}^{\alpha}x(s)\bigr\vert ^{p-1}\bigr)\,ds \\ \leq&\frac{1}{\Gamma(\beta)}\bigl(\Vert a\Vert _{0}+\Vert b\Vert _{0}\Vert x\Vert _{0}^{p-1} +\Vert c\Vert _{0}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}^{p-1}\bigr)\cdot\frac{1}{\beta}t^{\beta} \\ \leq&\frac{1}{\Gamma(\beta+1)} \biggl[\Vert a\Vert _{0}+\Vert c\Vert _{0}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}^{p-1} \\ &{}+\Vert b\Vert _{0} \biggl(A+ \frac{2}{\Gamma(\alpha+1)}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0} \biggr)^{p-1} \biggr], \quad \forall t\in[0,1]. \end{aligned}$$

(4.6)

By $Mx=N_{\lambda}x$, $D_{0^{+}}^{\alpha}x(0)=0$, and Lemma 2.1, one has

$$\begin{aligned} \phi_{p}\bigl(D_{0^{+}}^{\alpha}x(t)\bigr)=\lambda I_{0^{+}}^{\beta}Nx(t), \end{aligned}$$

which, together with $\vert \phi_{p}(D_{0^{+}}^{\alpha}x(t))\vert =\vert D_{0^{+}}^{\alpha}x(t)\vert ^{p-1}$ and (4.6), implies

$$\begin{aligned} \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}^{p-1} \leq& \frac{1}{\Gamma(\beta+1)} \biggl[\Vert a\Vert _{0}+\Vert c\Vert _{0}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}^{p-1} \\ &{}+\Vert b\Vert _{0} \biggl(A+ \frac{2}{\Gamma(\alpha+1)}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0} \biggr)^{p-1} \biggr]. \end{aligned}$$

(4.7)

If $p<2$, from (4.7) and Lemma 2.2, we have

$$\begin{aligned} \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}^{p-1} \leq& \frac{1}{\Gamma(\beta+1)} \biggl[\Vert a\Vert _{0}+A^{p-1} \Vert b\Vert _{0} \\ &{}+ \biggl(\frac{2^{p-1}\Vert b\Vert _{0}}{(\Gamma(\alpha+1))^{p-1}} +\Vert c\Vert _{0} \biggr) \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}^{p-1} \biggr]. \end{aligned}$$

Then, based on (4.3), one has

$$\begin{aligned} \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0} \leq \biggl[\frac{\Vert a\Vert _{0}+A^{p-1}\Vert b\Vert _{0}}{ (1-\gamma_{1})\Gamma(\beta+1)} \biggr]^{q-1}:=K_{1}. \end{aligned}$$

(4.8)

Thus, from (4.5), we have

$$\begin{aligned} \Vert x\Vert _{0}\leq A+\frac{2K_{1}}{\Gamma(\alpha+1)}. \end{aligned}$$

(4.9)

Similarly, if $p\geq2$, we obtain

$$\begin{aligned}& \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0} \leq \biggl[\frac{\Vert a\Vert _{0}+2^{p-2}A^{p-1}\Vert b\Vert _{0}}{ (1-\gamma_{2})\Gamma(\beta+1)} \biggr]^{q-1}:=K_{2}, \end{aligned}$$

(4.10)

$$\begin{aligned}& \Vert x\Vert _{0}\leq A+\frac{2K_{2}}{\Gamma(\alpha+1)}. \end{aligned}$$

(4.11)

Therefore, combining (4.8), (4.10) with (4.9), (4.11), we have

$$\begin{aligned} \Vert x\Vert _{X} =&\max\bigl\{ \Vert x\Vert _{0}, \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}\bigr\} \\ \leq&\max \biggl\{ K_{1},K_{2},A+\frac{2K_{1}}{\Gamma(\alpha+1)},A+ \frac{2 K_{2}}{\Gamma(\alpha+1)} \biggr\} :=K. \end{aligned}$$

That is, $\Omega_{1}$ is bounded.

Step 2. $\Omega_{2}=\{x\in \operatorname {Ker}M|QNx=0\}$ is bounded.

For $x\in\Omega_{2}$, one has $x(t)=d$, $\forall d\in\mathbb{R}$. Then we have

$$\begin{aligned} \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl( \int_{0}^{s}(s-\tau)^{\beta-1}f(\tau,d,0)\,d\tau \biggr)\,ds=0, \end{aligned}$$

which together with (H₃) implies $\vert d\vert \leq B$. Thus we obtain

$$\begin{aligned} \Vert x\Vert _{X}\leq\max\{B,0\}=B. \end{aligned}$$

Hence $\Omega_{2}$ is bounded.

Step 3. If (4.1) holds, then

$$\begin{aligned} \Omega_{3}=\bigl\{ x\in \operatorname {Ker}M|\lambda Ix+(1-\lambda)JQNx=0, \lambda\in [0,1]\bigr\} \end{aligned}$$

is bounded, where $J:\operatorname {Im}Q\rightarrow \operatorname {Ker}M$ is a homeomorphism such that $J(d)=d$, $\forall d\in\mathbb{R}$. If (4.2) holds, then

$$\begin{aligned} \Omega'_{3}=\bigl\{ x\in \operatorname {Ker}M|{-}\lambda Ix+(1- \lambda)JQNx=0,\lambda\in [0,1]\bigr\} \end{aligned}$$

is bounded.

For $x\in\Omega_{3}$, we have $x(t)=d$, $\forall d\in\mathbb{R}$, and

$$\lambda d =-(1-\lambda)\phi_{p} \biggl[\frac{1}{\rho} \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl( \int_{0}^{s}(s-\tau)^{\beta-1}f(\tau,d,0)\,d\tau \biggr)\,ds \biggr]. $$

If $\lambda=1$, then $d=0$. If $\lambda\in[0,1)$, we can show $\vert d\vert \leq B$. Otherwise, if $\vert d\vert >B$, in view of (4.1), one has

$$\begin{aligned} 0\leq\lambda d^{2} =&-(1-\lambda)\phi_{p} \biggl[ \frac{\phi_{q}(d)}{\rho} \int_{0}^{1}(1-s)^{\alpha -1} \\ &{}\cdot\phi_{q} \biggl( \int_{0}^{s}(s-\tau)^{\beta-1}f(\tau,d,0)\,d\tau \biggr)\,ds \biggr]< 0, \end{aligned}$$

which is a contradiction. Hence $\Omega_{3}$ is bounded.

Similar to the above argument, we can show $\Omega'_{3}$ is also bounded.

Step 4. All conditions of Lemma 2.3 are satisfied.

Define

$$\begin{aligned} \Omega=\bigl\{ x\in X|\Vert x\Vert _{X}< \max\{K,B\}+1\bigr\} . \end{aligned}$$

Clearly, $(\Omega_{1}\cup\Omega_{2}\cup\Omega_{3})\subset\Omega$ (or $(\Omega _{1}\cup\Omega_{2}\cup\Omega'_{3})\subset\Omega$). From Lemma 3.1 and Lemma 3.2, M is a quasi-linear operator and $N_{\lambda}$ is M-compact in Ω̅. Moreover, by the above arguments, we see that the following two conditions are satisfied:

(C₁):

$Mx\neq N_{\lambda}x$ for every $(x,\lambda)\in[(\operatorname {dom}M\setminus \operatorname {Ker}M)\cap\partial\Omega]\times(0,1)$;

(C₂):

$QNx\neq0$ for every $x\in \operatorname {Ker}M\cap\partial\Omega$.

Now we verify the condition (C₃) of Lemma 2.3. Let us define the homotopy

$$\begin{aligned} H(x,\lambda)=\pm\lambda Ix+(1-\lambda)JQNx. \end{aligned}$$

According to the above argument, we know

$$\begin{aligned} H(x,\lambda)\neq0,\quad \forall x\in\partial\Omega\cap \operatorname {Ker}M. \end{aligned}$$

Thus we have

$$\begin{aligned} \deg \{JQN,\Omega\cap \operatorname {Ker}M,\theta\} =&\deg \bigl\{ H(\cdot,0), \Omega\cap \operatorname {Ker}M,\theta\bigr\} \\ =&\deg \bigl\{ H(\cdot,1),\Omega\cap \operatorname {Ker}M,\theta\bigr\} \\ =&\deg \{\pm I,\Omega\cap \operatorname {Ker}M,\theta\}\neq0. \end{aligned}$$

So the condition (C₃) of Lemma 2.3 is satisfied.

Therefore, the operator equation $Mx=Nx$ has at least one solution in $\operatorname {dom}M\cap\overline{\Omega}$. That is, BVP (1.1)-(1.2) has at least one solution in X. □

5 Solutions of BVP (1.1)-(1.3)

We will give a theorem on the existence of solutions for BVP (1.1)-(1.3).

Define the operator $M_{1}:\operatorname {dom}M_{1}\cap X^{1}\rightarrow Z$ by

$$\begin{aligned} M_{1}x=D_{0^{+}}^{\beta}\phi_{p}\bigl(D_{0^{+}}^{\alpha}x\bigr), \end{aligned}$$

(5.1)

where $\operatorname {dom}M_{1}=\{x\in X^{1}|D_{0^{+}}^{\beta}\phi_{p}(D_{0^{+}}^{\alpha}x)\in Z\}$. Then BVP (1.1)-(1.3) is equivalent to the operator equation

$$\begin{aligned} M_{1}x=Nx,\quad x\in \operatorname {dom}M_{1}, \end{aligned}$$

where $N=N_{1}$ and $N_{\lambda}:X^{1}\rightarrow Z$, $\lambda\in[0,1]$ is defined by (3.2).

By similar arguments to Section 3, we obtain

$$\begin{aligned}& \operatorname {Ker}M_{1}=\bigl\{ x\in X^{1}|x(t)=d, \forall t \in[0,1],d\in\mathbb{R}\bigr\} , \\& \operatorname {Im}M_{1}= \biggl\{ z\in Z\Big| \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl(- \int_{0}^{1}(1-\tau)^{\beta-1}z(\tau)\,d\tau \\& \hphantom{\operatorname {Im}M_{1}=} {} + \int_{0}^{s}(s-\tau)^{\beta-1}z(\tau)\,d\tau \biggr)\,ds=0 \biggr\} . \end{aligned}$$

Lemma 5.1

The operator $M_{1}$, defined by (5.1), is a quasi-linear operator.

Let $X_{1}^{1}=\operatorname {Ker}M_{1}$, define the projector $P_{1}:X^{1}\rightarrow X^{1}$ and the semi-projector $Q_{1}:Z\rightarrow Z$ by

$$\begin{aligned}& P_{1}x(t)=x(0), \quad \forall t\in[0,1], \\& Q_{1}z(t)=\phi_{p} \biggl[\frac{1}{\rho_{1}} \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl(- \int_{0}^{1}(1-\tau)^{\beta-1}z(\tau)\,d\tau \\& \hphantom{ Q_{1}z(t)=} {}+ \int_{0}^{s}(s-\tau)^{\beta-1}z(\tau)\,d\tau \biggr)\,ds \biggr], \quad \forall t\in[0,1], \end{aligned}$$

where $\rho_{1}=\frac{1}{\beta^{q-1}}\int_{0}^{1}(1-s)^{\alpha-1}\phi _{q}(-1+s^{\beta})\,ds<0$. Furthermore, let $\Omega^{1}\subset X^{1}$ be an open bounded set, choose $X_{2}^{1}=\operatorname {Ker}P_{1}$, $Z_{1}^{1}=\operatorname {Im}Q_{1}$ and define the operator $R_{1}:\overline{\Omega^{1}}\times[0,1]\rightarrow X_{2}^{1}$ by

$$\begin{aligned} R_{1}(x,\lambda) (t) =&I_{0^{+}}^{\alpha}\phi_{q}\bigl[I_{0^{+}}^{\beta}(I-Q)N_{\lambda}x+ \tilde {d}\bigl((I-Q)N_{\lambda}x\bigr)\bigr](t) \\ =&\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}\phi_{q} \biggl[\frac{1}{\Gamma(\beta)} \\ &{}\cdot \int_{0}^{s}(s-\tau)^{\beta-1} \bigl(\lambda f \bigl(\tau,x(\tau),D_{0^{+}}^{\alpha}x(\tau)\bigr)-QN_{\lambda}x(\tau)\bigr)\,d\tau \\ &{}-\frac{1}{\Gamma(\beta)} \int_{0}^{1}(1-\tau)^{\beta-1} \bigl((I-Q)N_{\lambda}x(\tau)\bigr)\,d\tau \biggr]\,ds, \end{aligned}$$

where $\tilde{d}:Z\rightarrow\mathbb{R}$ is defined by

$$\begin{aligned} \tilde{d}(z) =&-I_{0^{+}}^{\beta}z(1) \\ =&-\frac{1}{\Gamma(\beta)} \int_{0}^{1}(1-s)^{\beta-1}z(s)\,ds. \end{aligned}$$

Lemma 5.2

The operator $N_{\lambda}:X^{1}\rightarrow Z$, $\lambda\in [0,1]$, defined by (3.2), is M-compact in $\overline{\Omega^{1}}$.

Our second result, based on Lemma 5.1 and Lemma 5.2, is stated as follows.

Theorem 5.1

Let $f:[0,1]\times\mathbb{R}^{2}\rightarrow\mathbb {R}$ be continuous. Assume that:

(H₄):

there exists a constant $A_{1}>0$ such that, for $\forall x\in \operatorname {dom}M_{1}\setminus \operatorname {Ker}M_{1}$ satisfying $|x(t)|>A_{1}$ for $\forall t\in[0,1]$, we have

$$\begin{aligned}& \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl(- \int_{0}^{1}(1-\tau)^{\beta-1}f\bigl(\tau,x( \tau),D_{0^{+}}^{\alpha}x(\tau)\bigr)\,d\tau \\& \quad {}+ \int_{0}^{s}(s-\tau)^{\beta-1}f\bigl(\tau,x( \tau),D_{0^{+}}^{\alpha}x(\tau )\bigr)\,d\tau \biggr)\,ds\neq0; \end{aligned}$$

(H₅):

there exists a constant $B_{1}>0$ such that, for $\forall r_{1}\in \mathbb{R}$ with $|r_{1}|>B_{1}$, we have either

$$\begin{aligned}& \phi_{q}(r_{1}) \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl(- \int_{0}^{1}(1-\tau)^{\beta-1}f( \tau,r_{1},0)\,d\tau \\& \quad {} + \int_{0}^{s}(s-\tau)^{\beta-1}f( \tau,r_{1},0)\,d\tau \biggr)\,ds>0 \end{aligned}$$

or

$$\begin{aligned}& \phi_{q}(r_{1}) \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl(- \int_{0}^{1}(1-\tau)^{\beta-1}f( \tau,r_{1},0)\,d\tau \\& \quad {}+ \int_{0}^{s}(s-\tau)^{\beta-1}f( \tau,r_{1},0)\,d\tau \biggr)\,ds< 0, \end{aligned}$$

and (H₁) is true. Then BVP (1.1)-(1.3) has at least one solution, provided that

$$ \begin{aligned} &\delta_{1}:=\frac{2}{\Gamma(\beta+1)} \biggl[ \frac{2^{p-1}\Vert b\Vert _{0}}{(\Gamma(\alpha+1))^{p-1}}+\Vert c\Vert _{0} \biggr]< 1,\quad \textit{if } p< 2; \\ &\delta_{2}:=\frac{2}{\Gamma(\beta+1)} \biggl[\frac{2^{2p-3}\Vert b\Vert _{0}}{(\Gamma(\alpha+1))^{p-1}}+\Vert c \Vert _{0} \biggr]< 1,\quad \textit{if } p\geq2. \end{aligned} $$

(5.2)

Proof

Let

$$\begin{aligned} \Omega_{1}^{1}=\bigl\{ x\in \operatorname {dom}M_{1} \setminus \operatorname {Ker}M_{1}|M_{1}x=N_{\lambda}x,\lambda \in(0,1)\bigr\} . \end{aligned}$$

Now we prove $\Omega_{1}^{1}$ is bounded.

For $x\in\Omega_{1}^{1}$, one has $Nx\in \operatorname {Im}M_{1}=\operatorname {Ker}Q_{1}$. Thus we have

$$\begin{aligned}& \int_{0}^{1}(1-s)^{\alpha-1} \phi_{q} \biggl(- \int_{0}^{1}(1-\tau)^{\beta-1}f\bigl(\tau,x( \tau),D_{0^{+}}^{\alpha}x(\tau)\bigr)\,d\tau \\& \quad {}+ \int_{0}^{s}(s-\tau)^{\beta-1}f\bigl(\tau,x( \tau),D_{0^{+}}^{\alpha}x(\tau )\bigr)\,d\tau \biggr)\,ds=0. \end{aligned}$$

From (H₄), there exists a constant $\eta\in[0,1]$ such that $\vert x(\eta )\vert \leq A_{1}$. Hence, by (4.4), one has

$$\begin{aligned} \Vert x\Vert _{0}\leq A_{1}+ \frac{2}{\Gamma(\alpha+1)}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}. \end{aligned}$$

(5.3)

Since $M_{1}x=N_{\lambda}x$, $D_{0^{+}}^{\alpha}x(1)=0$, one has

$$\begin{aligned} \phi_{p}\bigl(D_{0^{+}}^{\alpha}x(t)\bigr)=-\lambda I_{0^{+}}^{\beta}Nx(1)+\lambda I_{0^{+}}^{\beta}Nx(t), \end{aligned}$$

which together with (4.6) and (5.3) implies

$$\begin{aligned} \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}^{p-1} \leq& \frac{2}{\Gamma(\beta+1)} \biggl[\Vert a\Vert _{0}+\Vert c\Vert _{0}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}^{p-1} \\ &{}+\Vert b\Vert _{0} \biggl(A_{1}+\frac{2}{\Gamma(\alpha+1)}\bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0} \biggr)^{p-1} \biggr]. \end{aligned}$$

(5.4)

If $p<2$, from (5.4) and Lemma 2.2, we have

$$\begin{aligned} \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}^{p-1} \leq& \frac{2}{\Gamma(\beta+1)} \biggl[\Vert a\Vert _{0}+A_{1}^{p-1} \Vert b\Vert _{0} \\ &{}+ \biggl(\frac{2^{p-1}\Vert b\Vert _{0}}{(\Gamma(\alpha+1))^{p-1}} +\Vert c\Vert _{0} \biggr) \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0}^{p-1} \biggr]. \end{aligned}$$

Then, in view of (5.2), one has

$$\begin{aligned} \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0} \leq \biggl[\frac{2(\Vert a\Vert _{0}+A_{1}^{p-1}\Vert b\Vert _{0})}{ (1-\delta_{1})\Gamma(\beta+1)} \biggr]^{q-1}:=T_{1}. \end{aligned}$$

(5.5)

Similarly, if $p\geq2$, we obtain

$$\begin{aligned} \bigl\Vert D_{0^{+}}^{\alpha}x\bigr\Vert _{0} \leq \biggl[\frac{2(\Vert a\Vert _{0}+2^{p-2}A_{1}^{p-1}\Vert b\Vert _{0})}{ (1-\delta_{2})\Gamma(\beta+1)} \biggr]^{q-1}:=T_{2}. \end{aligned}$$

(5.6)

Therefore, from (5.3), (5.5), and (5.6), we have

$$\begin{aligned} \Vert x\Vert _{X} \leq\max \biggl\{ T_{1},T_{2},A_{1}+ \frac{2T_{1}}{\Gamma(\alpha+1)},A_{1}+\frac{2 T_{2}}{\Gamma(\alpha+1)} \biggr\} . \end{aligned}$$

That is, $\Omega_{1}^{1}$ is bounded.

The remainder of proof are similar to the proof of Theorem 4.1, so we omit the details. □