Solvability of fractional boundary value problem with p-Laplacian operator at resonance

Abstract

In this paper, a class of multi-point boundary value problems for nonlinear fractional differential equations at resonance with p-Laplacian operator is considered. By using the extension of Mawhin’s continuation theorem due to Ge, the existence of solutions is obtained, which enriches previous results.

MSC:34A08, 34B15.

1 Introduction

In the recent years, fractional differential equations played an important role in many fields such as physics, electrical circuits, biology, control theory, etc. (see [1–7]). Thus, many scholars have paid more attention to fractional differential equations and gained some achievements (see [8–22]). For example, Wang [21] considered a class of fractional multi-point boundary value problems at resonance by Mawhin’s continuation theorem (see [23]):

$$\{\begin{array}{c}{D}_{0^{+}}^{\alpha }u(t)=f(t,u(t),D_{0^{+}}^{\alpha -1}u(t)),t\in (0,1),\text{ a.e. }t\in (0,1),\hfill \\ u(0)=0,D_{0^{+}}^{\alpha -1}u(1)={\sum }_{i=1}^m{a}_i{D}_{0^{+}}^{\alpha -1}u({\xi }_i),\hfill \\ D_{0^{+}}^{\alpha -2}u(1)={\sum }_{i=1}^m{b}_i{D}_{0^{+}}^{\alpha -2}u({\eta }_i),\hfill \end{array}$$

(1.1)

where $1<\alpha \le 2$, $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_m<1$, $0<{\eta }_1<{\eta }_2<\cdots <{\eta }_n<1$, ${\sum }_{i=1}^m{a}_i=1$, ${\sum }_{i=1}^n{b}_i=1$, ${\sum }_{i=1}^n{b}_i{\eta }_i=1$, $D_{0^{+}}^{\alpha }$ is the standard fractional derivative, $f:[0,1]\times {\mathbf{R}}^2\to \mathbf{R}$ satisfies the Carathéodory condition.

But Mawhin’s continuation theorem is not suitable for quasi-linear operators. In [24], Ge and Ren had extended Mawhin’s continuation theorem, which was used to deal with more general abstract operator equations. In [25], Pang et al. considered a higher order nonlinear differential equation with a p-Laplacian operator at resonance:

$$\{\begin{array}{c}{({\phi }_p(u^{(n-1)}(t)))}^{\prime }=f(t,u(t),\dots ,u^{(n-1)}(t))+e(t),t\in (0,1),\hfill \\ u^{(i)}(0)=0,i=1,2,\dots ,n-1,\hfill \\ u(1)={\int }_0^{1}u(s)dg(s),\hfill \end{array}$$

(1.2)

where ${\phi }_p(s)={|s|}^{p-2}s$, $p>1$, $f:[0,1]\times {\mathbf{R}}^n\to {\mathbf{R}}^n$ and $e:[0,1]\to \mathbf{R}$ are continuous, $n\ge 2$ is an integer. $g:[0,1]\to \mathbf{R}$ is a nondecreasing function with ${\int }_0^{1}dg(s)=1$, the integral in the second part of (1.2) is meant in the Riemann-Stieltjes sense.

However, there are few articles which consider the fractional multi-point boundary value problem at resonance with p-Laplacian operator and $dimKerM=2$, because p-Laplacian operator is a nonlinear operator, and it is hard to construct suitable continuous projectors. In this paper, we will improve and generalize some known results.

Motivated by the work above, our article is to investigate the multi-point boundary value problem at resonance for a class of Riemanne-Liouville fractional differential equations with p-Laplacian operator and $dimKerM=2$ by constructing suitable continuous projectors and using the extension of Mawhin’s continuation theorem:

$$\{\begin{array}{c}{D}_{0^{+}}^{\beta }{\phi }_p(D_{0^{+}}^{\alpha }u(t))=f(t,u(t),D_{0^{+}}^{\alpha -2}u(t),D_{0^{+}}^{\alpha -1}u(t),D_{0^{+}}^{\alpha }u(t)),t\in (0,1),\hfill \\ u(0)=D_{0^{+}}^{\alpha }u(0)=0,u(1)={\sum }_{i=1}^m{a}_{i}u({\xi }_i),\hfill \\ D_{0^{+}}^{\alpha -1}u(1)={\sum }_{i=1}^m{b}_i{D}_{0^{+}}^{\alpha -1}u({\eta }_i),\hfill \end{array}$$

(1.3)

where $2<\alpha \le 3$, $0<\beta \le 1$, $3<\alpha +\beta \le 4$, $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_m<1$, $0<{\eta }_1<{\eta }_2<\cdots <{\eta }_m<1$, $a_i\in \mathbf{R}$, $b_i\in \mathbf{R}$, $1<m$, $m\in N$, ${\sum }_{i=1}^m{a}_i{\xi }_i^{\alpha -1}=1$, ${\sum }_{i=1}^m{a}_i{\xi }_i^{\alpha -2}=1$, ${\sum }_{i=1}^m{b}_i=1$, ${\phi }_p(s)={|s|}^{p-2}s$, ${\phi }_p(0)=0$, $1<p$, $1/p+1/q=1$, ${\phi }_p$ is invertible and its inverse operator is ${\phi }_q$, $D_{0^{+}}^{\alpha }$ is Riemann-Liouville standard fractional derivative, $f:[0,1]\times {\mathbf{R}}^4\to \mathbf{R}$ is continuous.

In order to investigate the problem, we need to suppose that the following conditions hold:

$$\mathrm{\Lambda }={\mathrm{\Lambda }}_1{\mathrm{\Lambda }}_4-{\mathrm{\Lambda }}_2{\mathrm{\Lambda }}_3\ne 0,$$

where

$$\begin{array}{c}{\mathrm{\Lambda }}_1=\frac{\mathrm{\Gamma }{(\alpha )}^q\mathrm{\Gamma }(\alpha q+\beta q-q-\alpha -\beta +2)}{\mathrm{\Gamma }{(\alpha +\beta )}^{q-1}\mathrm{\Gamma }(\alpha q+\beta q-q-\beta +2)}(1-\sum _{i=1}^m{a}_i{\xi }_i^{\alpha q+\beta q-q-\beta +1}),\hfill \\ {\mathrm{\Lambda }}_2=\frac{\mathrm{\Gamma }{(\alpha -1)}^{q-1}\mathrm{\Gamma }(\alpha )\mathrm{\Gamma }(\alpha q+\beta q-2q-\alpha -\beta +3)}{\mathrm{\Gamma }{(\alpha +\beta -1)}^{q-1}\mathrm{\Gamma }(\alpha q+\beta q-2q-\beta +3)}(1-\sum _{i=1}^m{a}_i{\xi }_i^{\alpha q+\beta q-2q-\beta +2}),\hfill \\ {\mathrm{\Lambda }}_3=\frac{\mathrm{\Gamma }{(\alpha )}^{q-1}}{\mathrm{\Gamma }{(\alpha +\beta )}^{q-1}(\alpha q+\beta q-q-\alpha -\beta +2)}(1-\sum _{i=1}^m{b}_i{\eta }_i^{\alpha q+\beta q-q-\alpha -\beta +2}),\hfill \\ {\mathrm{\Lambda }}_4=\frac{\mathrm{\Gamma }{(\alpha -1)}^{q-1}}{\mathrm{\Gamma }{(\alpha +\beta -1)}^{q-1}(\alpha q+\beta q-2q-\alpha -\beta +3)}(1-\sum _{i=1}^m{b}_i{\eta }_i^{\alpha q+\beta q-2q-\alpha -\beta +3}).\hfill \end{array}$$

The rest of this article is organized as follows: In Section 2, we give some notations, definitions and lemmas. In Section 3, based on the extension of Mawhin’s continuation theorem due to Ge, we establish a theorem on existence of solutions for BVP (1.3).

2 Preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions for fractional calculus theory that can be found in the recent literature (see [1, 3, 14, 21, 24, 25]).

Let X and Y be two Banach spaces with norms ${\parallel \cdot \parallel }_X$ and ${\parallel u\parallel }_Y$, respectively. A continuous operator

$$M{|}_{domM\cap X}:X\cap domM\to Y$$

is said to be quasi-linear if

• (i) $ImM:=M(X\cap domM)$ is a closed subset of Y,

• (ii) $KerM:=\{u\in X\cap domM:Mu=0\}$ is linearly homeomorphic to $R^n$, $n<\mathrm{\infty }$.

Let $X_1=KerM$ and $X_2$ be the complement space of $X_1$ in X, then $X=X_1\oplus X_2$. On the other hand, suppose that $Y_1$ is a subspace of Y, and $Y_2$ is the complement space of $Y_1$ in Y, so that $Y=Y_1\oplus Y_2$. Let $P:X\to X_1$ be a projector and $Q:Y\to Y_1$ a semi-projector, and $\mathrm{\Omega }\subset X$ an open and bounded set with origin $\theta \in \mathrm{\Omega }$. θ is the origin of a linear space.

Suppose that $N_{\lambda }:\overline{\mathrm{\Omega }}\to Y$, $\lambda \in [0,1]$ is a continuous operator. Denote $N_1$ by N. Let ${\mathrm{\Sigma }}_{\lambda }=\{u\in \overline{\mathrm{\Omega }}:Mu=N_{\lambda }u\}$. $N_{\lambda }$ is said to be M-compact in $\overline{\mathrm{\Omega }}$ if there is an $Y_1\subset Y$ with $dim{Y}_1$ = $dim{X}_1$ and an operator $R:\overline{\mathrm{\Omega }}\times [0,1]\to X$ continuous and compact such that for $\lambda \in [0,1]$,

$$(I-Q)N_{\lambda }(\overline{\mathrm{\Omega }})\subset ImM\subset (I-Q)Y,$$

(2.1)

$$Q{N}_{\lambda }x=\theta ,\lambda \in (0,1)\iff QNx=\theta ,$$

(2.2)

$$R(\cdot ,\lambda ){|}_{{\mathrm{\Sigma }}_{\lambda }}=(I-P){|}_{{\mathrm{\Sigma }}_{\lambda }}$$

(2.3)

and $R(\cdot ,0)$ is the zero operator,

$$M[P+R(\cdot ,\lambda )]=(I-Q)N_{\lambda }.$$

(2.4)

Lemma 2.1 (Ge-Mawhin’s continuation theorem [24])

Let $(X,{\parallel \cdot \parallel }_X)$ and $(Y,{\parallel \cdot \parallel }_Y)$ be two Banach spaces, and $\mathrm{\Omega }\subset X$ an open and bounded nonempty set. Suppose that $M:X\cap domM\to Y$ is a quasi-linear operator $N_{\lambda }:\overline{\mathrm{\Omega }}\to Y$, $\lambda \in [0,1]$ is M-compact in $\overline{\mathrm{\Omega }}$. In addition, if

• (i) $Lu\ne N_{\lambda }u$, $\mathrm{\forall }(u,\lambda )\in (domM\cap \partial \mathrm{\Omega })\times (0,1)$,

• (ii) $deg(JQN,KerM\cap \mathrm{\Omega },0)\ne 0$,

where $J:ImQ\to KerM$ is a homeomorphism with $J(\theta )=\theta$ and $N=N_1$, then the equation $Mu=Nu$ has at least one solution in $domM\cap \overline{\mathrm{\Omega }}$.

Definition 2.1 The Riemann-Liouville fractional integral of order $\alpha >0$ of a function u is given by

$$I_{0^{+}}^{\alpha }u(t)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^t{(t-s)}^{\alpha -1}u(s)ds,$$

provided the right-hand side integral is pointwise almost everywhere defined on $(0,+\mathrm{\infty })$.

Definition 2.2 The Riemann-Liouville fractional derivative of order $\alpha >0$ of a function u is given by

$$D_{0^{+}}^{\alpha }u(t)=\frac{1}{\mathrm{\Gamma }(n-\alpha )}{\left(\frac{d}{dt}\right)}^n{\int }_0^t\frac{u(s)}{{(t-s)}^{\alpha -n+1}}ds,$$

provided the right-hand side integral is pointwise everywhere defined on $(0,+\mathrm{\infty })$, where $n=[\alpha ]+1$.

Definition 2.3 Let X be a Banach space, and $X_1\subset X$ is a subspace. A mapping $Q:X\to X_1$ is a semi-projector if Q satisfies

• (i) $Q^{2}x=Qx$, $\mathrm{\forall }x\in X$,

• (ii) $Q(\mu x)=\mu Qx$, $\mathrm{\forall }x\in X$, $\mu \in \mathbf{R}$.

Lemma 2.2 Assume that $u\in C(0,1)\cap L^1(0,1)$ with a fractional derivative of order $\alpha >0$ that belongs to $C(0,1)\cap L^1(0,1)$. Then

$$I_{0+}^{\alpha }{D}_{0+}^{\alpha }u(t)=u(t)+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+\cdots +c_N{t}^{\alpha -N}$$

for some $c_i\in \mathbf{R}$, $i=1,2,\dots ,N$, where $N=[\alpha ]+1$.

Lemma 2.3 Assume that $u(t)\in C[0,1]$, $0\le p\le q$, then

$$D_{0^{+}}^q{I}_{0^{+}}^{p}u(t)=I_{0^{+}}^{p-q}u(t).$$

Lemma 2.4 Assume that $\alpha \ge 0$, then:

• (i) If $\lambda >-1$, $\lambda \ne \alpha -i$, $i=1,2,\dots ,[\alpha ]+1$, we have that

  $$D_{0^{+}}^{\alpha }{t}^{\lambda }=\frac{\mathrm{\Gamma }(\lambda +1)}{\mathrm{\Gamma }(\lambda -\alpha +1)}{t}^{\lambda -\alpha }.$$

• (ii) $D_{0^{+}}^{\alpha }{t}^{\alpha -i}=0$, $i=1,2,\dots ,[\alpha ]+1$.

In this paper, we take $X=\{u\mid u,D_{0^{+}}^{\alpha -2}u,D_{0^{+}}^{\alpha -1}u,D_{0^{+}}^{\alpha }u\in C[0,1]\}$ with the norm ${\parallel u\parallel }_X=max\{{\parallel u\parallel }_{\mathrm{\infty }},{\parallel D_{0^{+}}^{\alpha -2}u\parallel }_{\mathrm{\infty }}{\parallel D_{0^{+}}^{\alpha -1}u\parallel }_{\mathrm{\infty }},{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}\}$, where ${\parallel u\parallel }_{\mathrm{\infty }}={max}_{t\in [0,1]}|u(t)|$, and $Y=C[0,1]$ with the norm ${\parallel y\parallel }_Y={\parallel y\parallel }_{\mathrm{\infty }}$. By means of the linear functional analysis theory, it is easy to prove that X and Y are Banach spaces, so we omit it.

Define the operator $M:domM\to Y$ by

$$Mu=D_{0^{+}}^{\beta }{\phi }_p(D_{0^{+}}^{\alpha }u(t)),$$

(2.5)

$$\begin{array}{c}domM=\{u\in X|D_{0^{+}}^{\beta }{\phi }_p\left(D_{0^{+}}^{\alpha }u\right)\in Y,u(0)=D_{0^{+}}^{\alpha }u(0)=0,\hfill \\ \phantom{domM=}u(1)=\sum _{i=1}^m{a}_{i}u({\xi }_i),D_{0^{+}}^{\alpha -1}u(1)=\sum _{i=1}^m{b}_i{D}_{0^{+}}^{\alpha -1}u({\eta }_i)\}.\hfill \end{array}$$

(2.6)

Based on the definition of domM, it is easy to find that $domM\ne \mathrm{\varnothing }$ such as $u(t)=c{t}^{\alpha -1}\in domM$, $c\in \mathbf{R}$.

Define the operator $N_{\lambda }:X\to Y$, $\lambda \in [0,1]$,

$$N_{\lambda }u(t)=\lambda f(t,u(t),D_{0^{+}}^{\alpha -2}u(t),D_{0^{+}}^{\alpha -1}u(t),D_{0^{+}}^{\alpha }u(t)),t\in [0,1].$$

Then BVP (1.3) is equivalent to the operator equation $Mu=Nu$, where $N=N_1$.

3 Main result

In this section, a theorem on existence of solutions for BVP (1.3) will be given.

Define operators $T_j:Y\to Y$, $j=1,2$ as follows:

$$\begin{array}{c}{T}_{1}y={\int }_0^1{(1-s)}^{\alpha -1}{\phi }_q(\frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^s{(s-\tau )}^{\beta -1}y(\tau )d\tau )ds\hfill \\ \phantom{T_{1}y=}-\sum _{i=1}^m{a}_i{\int }_0^{{\xi }_i}{({\xi }_i-s)}^{\alpha -1}{\phi }_q(\frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^s{(s-\tau )}^{\beta -1}y(\tau )d\tau )ds,\hfill \\ T_{2}y={\int }_0^1{\phi }_q(\frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^s{(s-\tau )}^{\beta -1}y(\tau )d\tau )ds\hfill \\ \phantom{T_{2}y=}-\sum _{i=1}^m{b}_i{\int }_0^{{\eta }_i}{\phi }_q(\frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^s{(s-\tau )}^{\beta -1}y(\tau )d\tau )ds.\hfill \end{array}$$

Let us make some assumptions, which will be used in the sequel.

(H1) There exist nonnegative functions $r,d,e,h,k\in Y$ such that for all $t\in [0,1]$, $(u,v,w,z)\in {\mathbf{R}}^4$,

$$|f(t,u,v,w,z)|\le r(t)+d(t)|u{|}^{p-1}+e(t)|v{|}^{p-1}+h(t)|w{|}^{p-1}+k(t)|z{|}^{p-1}.$$

(H2) There exists a constant $A>0$ such that for $u\in domM$, if $|D_{0^{+}}^{\alpha -1}u(t)|>A$ for all $t\in [0,1]$, then

$$sgn\{D_{0^{+}}^{\alpha -1}u(t)\}\frac{1}{\mathrm{\Lambda }}({\mathrm{\Lambda }}_4{T}_{1}Nu(t)-{\mathrm{\Lambda }}_3{T}_{2}Nu(t))>0$$

or

$$sgn\{D_{0^{+}}^{\alpha -1}u(t)\}\frac{1}{\mathrm{\Lambda }}({\mathrm{\Lambda }}_4{T}_{1}Nu(t)-{\mathrm{\Lambda }}_3{T}_{2}Nu(t))<0.$$

(H3) There exists a constant $B>0$ such that for $u\in domM$, if $|D_{0^{+}}^{\alpha -2}u(t)|>B$ for all $t\in [0,1]$, then

$$sgn\{D_{0^{+}}^{\alpha -2}u(t)\}\frac{1}{\mathrm{\Lambda }}(-{\mathrm{\Lambda }}_2{T}_{1}Nu(t)+{\mathrm{\Lambda }}_1{T}_{2}Nu(t))>0$$

or

$$sgn\{D_{0^{+}}^{\alpha -2}u(t)\}\frac{1}{\mathrm{\Lambda }}(-{\mathrm{\Lambda }}_2{T}_{1}Nu(t)+{\mathrm{\Lambda }}_1{T}_{2}Nu(t))<0.$$

Theorem 3.1 Let $f:[0,1]\times {\mathbf{R}}^4\to \mathbf{R}$ be continuous and condition (H1)-(H3) hold, then BVP (1.3) has at least one solution, provided that

$$\frac{1}{\mathrm{\Gamma }(\beta +1)}(A_1{\parallel d\parallel }_{\mathrm{\infty }}+{\parallel e\parallel }_{\mathrm{\infty }}+{\parallel h\parallel }_{\mathrm{\infty }}+{\parallel k\parallel }_{\mathrm{\infty }})<1,$$

(3.1)

where $A_1=\frac{1}{\mathrm{\Gamma }(\alpha +1)}+\frac{2}{\mathrm{\Gamma }(\alpha )}+\frac{7}{2\mathrm{\Gamma }(\alpha -1)}$.

In order to prove Theorem 3.1, we need to prove some lemmas below.

Lemma 3.1 The operator $M:domM\cap X\to Y$ is quasi-linear.

$$KerM=\{u\in X\mid u(t)=c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2},c_1,c_2\in \mathbf{R}\},$$

(3.2)

$$ImM=\{y\in Y\mid T_{j}y=0,j=1,2\}.$$

(3.3)

Proof Suppose that $u(t)\in domM$, by $D_{0^{+}}^{\beta }{\phi }_p(D_{0^{+}}^{\alpha }u(t))=0$, we have

$$D_{0^{+}}^{\alpha }u(t)={\phi }_q\left(c_0{t}^{\beta -1}\right).$$

Based on $D_{0^{+}}^{\alpha }u(0)=0$, one has

$$u(t)=c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+c_3{t}^{\alpha -3},$$

which together with $u(0)=0$ yields that

$$KerM=\{u\in X\mid u(t)=c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2},c_1,c_2\in \mathbf{R}\}.$$

It is clear that $dimKerM=2$. So, KerM is linearly homeomorphic to ${\mathbf{R}}^2$.

If $y\in ImM$, then there exists a function $u\in domM$ such that $y(t)=D_{0^{+}}^{\beta }{\phi }_p(D_{0^{+}}^{\alpha }u(t))$. Based on Lemmas 2.2 and 2.3, we have

$$\begin{array}{c}u(t)=I_{0^{+}}^{\alpha }{\phi }_q(I_{0^{+}}^{\beta }y(s))+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2},\hfill \\ D_{0^{+}}^{\alpha -1}u(t)=D_{0^{+}}^{\alpha -1}{I}_{0^{+}}^{\alpha }{\phi }_q(I_{0^{+}}^{\beta }y(s))+c_1\mathrm{\Gamma }(\alpha ),\hfill \end{array}$$

which together with ${\sum }_{i=1}^m{a}_i{\xi }_i^{\alpha -1}=1$, ${\sum }_{i=1}^m{a}_i{\xi }_i^{\alpha -2}=1$ and ${\sum }_{i=1}^m{b}_i=1$ yields that $T_{j}y(t)=0$, $j=1,2$.

On the other hand, suppose that $y\in Y$ and satisfies (3.3), and let $u(t)=I_{0^{+}}^{\alpha }{\phi }_q(I_{0^{+}}^{\beta }y(t))$, then $u\in domM$ and $Mu(t)=D_{0^{+}}^{\beta }{\phi }_p(D_{0^{+}}^{\alpha }u(t))=y$, so $y\in ImM$ and $ImM:=M(domM)$ is a closed subset of Y. Thus, M is a quasi-linear operator. □

Lemma 3.2 Let $\mathrm{\Omega }\subset X$ be an open and bounded set, then $N_{\lambda }$ is M-compact in $\overline{\mathrm{\Omega }}$.

Proof Define the continuous projector $P:X\to X_1$ by

$$Pu(t)=\frac{1}{\mathrm{\Gamma }(\alpha )}{D}_{0^{+}}^{\alpha -1}u(0)t^{\alpha -1}+\frac{1}{\mathrm{\Gamma }(\alpha -1)}{D}_{0^{+}}^{\alpha -2}u(0)t^{\alpha -2},t\in [0,1].$$

Define the continuous projector $Q:Y\to Y_1$, by

$$Qy(t)=(Q_{1}y(t))t^{\alpha -1}+(Q_{2}y(t))t^{\alpha -2},t\in [0,1],$$

where

$$\begin{array}{c}{Q}_{1}y(t)={\phi }_p\left(\frac{1}{\mathrm{\Lambda }}({\mathrm{\Lambda }}_4{T}_{1}y(t)-{\mathrm{\Lambda }}_3{T}_{2}y(t))\right),\hfill \\ Q_{2}y(t)={\phi }_p\left(\frac{1}{\mathrm{\Lambda }}(-{\mathrm{\Lambda }}_2{T}_{1}y(t)+{\mathrm{\Lambda }}_1{T}_{2}y(t))\right).\hfill \end{array}$$

Obviously, $X_1=KerM=ImP$ and $Y_1=\text{ Im}Q$. Thus, we have $dim{Y}_1$ = $dim{X}_1=2$. For any $y\in Y$, we have

$$\begin{array}{rcl}{Q}_1(Q_{1}y(t)t^{\alpha -1})& =& {\phi }_p\left(\frac{1}{\mathrm{\Lambda }}({\mathrm{\Lambda }}_4{T}_1(Q_{1}y(t)t^{\alpha -1})-{\mathrm{\Lambda }}_3{T}_2(Q_{1}y(t)t^{\alpha -1}))\right)\\ =& Q_{1}y(t){\phi }_p(\frac{1}{\mathrm{\Lambda }}({\mathrm{\Lambda }}_4{\mathrm{\Lambda }}_1-{\mathrm{\Lambda }}_3{\mathrm{\Lambda }}_2))=Q_{1}y(t).\end{array}$$

Similarly, we can get

$$Q_1(Q_{2}y(t)t^{\alpha -2})=0,Q_2(Q_{1}y(t)t^{\alpha -1})=0,Q_2(Q_{2}y(t)t^{\alpha -2})=Q_{2}y(t).$$

Hence, the map Q is idempotent. Similarly, we can get $Q(\mu y)=\mu Qy$, for all $y\in Y$, $\mu \in \mathbf{R}$. Thus, Q is a semi-projector. For any $y\in ImM$, we can get that $Qy=0$ and $y\in KerQ$, conversely, if $y\in KerQ$, we can obtain that $Qy=0$, that is to say, $y\in ImM$. Thus, $KerQ=ImM$. Let $\mathrm{\Omega }\subset X$ be an open and bounded set with $\theta \in \mathrm{\Omega }$, for each $u\in \overline{\mathrm{\Omega }}$, we can get $Q[(I-Q)N_{\lambda }(u)]=0$. Thus, $(I-Q)N_{\lambda }(u)\in ImM=KerQ$. Take any $y\in ImM$ in the type $y=(y-Qy)+Qy$, since $Qy=0$, we can get $y\in (I-Q)Y$. So, (2.1) holds. It is easy to verify (2.2).

Define $R:\overline{\mathrm{\Omega }}\times [0,1]\to X_2$ by

$$R(u,\lambda )(t)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^t{(t-s)}^{\alpha -1}{\phi }_q\left(\frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^s{(s-\tau )}^{\beta -1}((I-Q)N_{\lambda }u(\tau ))d\tau \right)ds.$$

By the continuity of f, it is easy to get that $R(u,\lambda )$ is continuous on $\overline{\mathrm{\Omega }}\times [0,1]$. Moreover, for all $u\in \overline{\mathrm{\Omega }}$, there exists a constant $T>0$ such that $|I_{0^{+}}^{\beta }(I-Q)N_{\lambda }u(\tau )|\le T$, so, we can easily obtain that $R(\overline{\mathrm{\Omega }},\lambda )$, $D_{0^{+}}^{\alpha -2}R(\overline{\mathrm{\Omega }},\lambda )$, $D_{0^{+}}^{\alpha -1}R(\overline{\mathrm{\Omega }},\lambda )$ and $D_{0^{+}}^{\alpha }R(\overline{\mathrm{\Omega }},\lambda )$ are uniformly bounded. By Arzela-Ascoli theorem, we just need to prove that $R:\overline{\mathrm{\Omega }}\times [0,1]\to X_2$ is equicontinuous.

For $u\in \overline{\mathrm{\Omega }}$, $0<t_1<t_2\le 1$, $2<\alpha \le 3$, $0<\beta \le 1$, $3<\alpha +\beta \le 4$, we have

$$\begin{array}{c}|R(u,\lambda )(t_2)-R(u,\lambda )(t_1)|\hfill \\ =\frac{1}{\mathrm{\Gamma }(\alpha )}|{\int }_0^{t_2}{(t_2-s)}^{\alpha -1}{\phi }_q\left(I_{0^{+}}^{\beta }((I-Q)N_{\lambda }u(\tau ))\right)ds\hfill \\ -{\int }_0^{t_1}{(t_1-s)}^{\alpha -1}{\phi }_q\left(I_{0^{+}}^{\beta }((I-Q)N_{\lambda }u(\tau ))\right)ds|\hfill \\ \le \frac{{\phi }_q(L)}{\mathrm{\Gamma }(\alpha )}({\int }_0^{t_1}({(t_2-s)}^{\alpha -1}-{(t_1-s)}^{\alpha -1})ds+{\int }_{t_1}^{t_2}{(t_2-s)}^{\alpha -1}ds)\hfill \\ =\frac{{\phi }_q(T)}{\mathrm{\Gamma }(\alpha +1)}(t_2^{\alpha }-t_1^{\alpha }),\hfill \\ |D_{0^{+}}^{\alpha -2}R(u,\lambda )(t_2)-D_{0^{+}}^{\alpha -2}R(u,\lambda )(t_1)|\hfill \\ =|{\int }_0^{t_2}(t-s){\phi }_q\left(I_{0^{+}}^{\beta }((I-Q)N_{\lambda }u(\tau ))\right)ds-{\int }_0^{t_1}(t-s){\phi }_q\left(I_{0^{+}}^{\beta }((I-Q)N_{\lambda }u(\tau ))\right)ds|\hfill \\ \le {\phi }_q(T)({\int }_0^{t_1}(t_2-s)-(t_1-s)ds+{\int }_{t_1}^{t_2}(t_2-s)ds)\hfill \\ =\frac{{\phi }_q(T)}{2}(t_2^2-t_1^2)\hfill \end{array}$$

and

$$\begin{array}{c}|D_{0^{+}}^{\alpha -1}R(u,\lambda )(t_2)-D_{0^{+}}^{\alpha -1}R(u,\lambda )(t_1)|\hfill \\ =|{\int }_0^{t_2}{\phi }_q\left(I_{0^{+}}^{\beta }((I-Q)N_{\lambda }u(\tau ))\right)ds-{\int }_0^{t_1}{\phi }_q\left(I_{0^{+}}^{\beta }((I-Q)N_{\lambda }u(\tau ))\right)ds|\hfill \\ \le {\phi }_q(T)(t_2-t_1).\hfill \end{array}$$

Since $t^{\alpha }$ is uniformly continuous on $[0,1]$, so, $R(\overline{\mathrm{\Omega }},\lambda )$, $D_{0^{+}}^{\alpha -2}R(\overline{\mathrm{\Omega }},\lambda )$ and $D_{0^{+}}^{\alpha -1}R(\overline{\mathrm{\Omega }},\lambda )$ are equicontinuous. Similarly, we can get that $I_{0^{+}}^{\beta }((I-Q)N_{\lambda }u(\tau ))\subset C[0,1]$ is equicontinuous. Considering that ${\phi }_q(s)$ is uniformly continuous on $[-T,T]$, we have that $D_{0^{+}}^{\alpha }R(\overline{\mathrm{\Omega }},\lambda )=I_{0^{+}}^{\beta }((I-Q)N_{\lambda }(\overline{\mathrm{\Omega }}))$ is also equicontinuous. So, we can obtain that $R:\overline{\mathrm{\Omega }}\times [0,1]\to X_2$ is compact.

For each $u\in {\mathrm{\Sigma }}_{\lambda }$, we have $D_{0^{+}}^{\beta }{\phi }_p(D_{0^{+}}^{\alpha }u(t))=N_{\lambda }(u(t))\in ImM$. Thus,

$$\begin{array}{rcl}R(u,\lambda )(t)& =& \frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^t{(t-s)}^{\alpha -1}{\phi }_q\left(\frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^s{(s-\tau )}^{\beta -1}((I-Q)N_{\lambda }u(\tau ))d\tau \right)ds\\ =& \frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^t{(t-s)}^{\alpha -1}{\phi }_q\left(\frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^s{(s-\tau )}^{\beta -1}{D}_{0^{+}}^{\beta }{\phi }_p(D_{0^{+}}^{\alpha }u(\tau ))d\tau \right)ds,\end{array}$$

which together with $D_{0^{+}}^{\alpha }u(0)=u(0)=0$ yields that

$$R(u,\lambda )(t)=u(t)-\frac{1}{\mathrm{\Gamma }(\alpha )}{D}_{0^{+}}^{\alpha -1}u(0)t^{\alpha -1}-\frac{1}{\mathrm{\Gamma }(\alpha -1)}{D}_{0^{+}}^{\alpha -2}u(0)t^{\alpha -2}=(I-P)u(t).$$

It is easy to verify that $R(u,0)(t)$ is the zero operator. So, (2.3) holds. Besides, for any $u\in \overline{\mathrm{\Omega }}$,

$$\begin{array}{c}M[Pu+R(u,\lambda )](t)\hfill \\ =M[\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^t{(t-s)}^{\alpha -1}{\phi }_q\left(\frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^s{(s-\tau )}^{\beta -1}((I-Q)N_{\lambda }u(\tau ))d\tau \right)ds\hfill \\ +\frac{1}{\mathrm{\Gamma }(\alpha )}{D}_{0^{+}}^{\alpha -1}u(0)t^{\alpha -1}+\frac{1}{\mathrm{\Gamma }(\alpha -1)}{D}_{0^{+}}^{\alpha -2}u(0)t^{\alpha -2}]\hfill \\ =(I-Q)N_{\lambda }u(t),\hfill \end{array}$$

which implies (2.4). So, $N_{\lambda }$ is M-compact in $\overline{\mathrm{\Omega }}$. □

Lemma 3.3 Suppose that (H1), (H2) hold, then the set

$${\mathrm{\Omega }}_1=\{u\in domM\setminus KerM\mid Mu=N_{\lambda }u,\lambda \in (0,1)\}$$

is bounded.

Proof By Lemma 2.2, for each $u\in domM$, we have

$$u(t)=I_{0^{+}}^{\alpha }{D}_{0^{+}}^{\alpha }u(t)+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+c_3{t}^{\alpha -3}.$$

Combined with $u(0)=0$, we get $c_3=0$. Thus,

$$\begin{array}{c}u(t)=I_{0^{+}}^{\alpha }{D}_{0^{+}}^{\alpha }u(t)+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2},\hfill \\ D_{0^{+}}^{\alpha -1}u(t)=I_{0^{+}}^1{D}_{0^{+}}^{\alpha }u(t)+c_1\mathrm{\Gamma }(\alpha ),\hfill \\ D_{0^{+}}^{\alpha -2}u(t)=I_{0^{+}}^2{D}_{0^{+}}^{\alpha }u(t)+c_1\mathrm{\Gamma }(\alpha )t+c_2\mathrm{\Gamma }(\alpha -1).\hfill \end{array}$$

By simple calculation, we get

$$\begin{array}{c}{c}_1=\frac{1}{\mathrm{\Gamma }(\alpha )}(D_{0^{+}}^{\alpha -1}u(t)-{\int }_0^t{D}_{0^{+}}^{\alpha }u(s)ds),\hfill \\ c_2=\frac{1}{\mathrm{\Gamma }(\alpha -1)}(D_{0^{+}}^{\alpha -2}u(t)-{\int }_0^t(t-s)D_{0^{+}}^{\alpha -1}u(s)ds-(D_{0^{+}}^{\alpha -1}u(t)-{\int }_0^t{D}_{0^{+}}^{\alpha }u(s)ds)t).\hfill \end{array}$$

Take any $u\in {\mathrm{\Omega }}_1$, then $Nu\in ImM=KerQ$ and $QNu=0$. It follows from (H2) and (H3) that there exist ${\epsilon }_1,{\epsilon }_2\in [0,1]$ such that $|D_{0^{+}}^{\alpha -1}u({\epsilon }_1)|\le A$, $|D_{0^{+}}^{\alpha -2}u({\epsilon }_2)|\le B$. Thus,

$$\begin{array}{c}{D}_{0^{+}}^{\alpha -1}u(t)=D_{0^{+}}^{\alpha -1}u({\epsilon }_1)+{\int }_{{\epsilon }_1}^t{D}_{0^{+}}^{\alpha }u(t)dt,\hfill \\ D_{0^{+}}^{\alpha -2}u(t)=D_{0^{+}}^{\alpha -2}u({\epsilon }_2)+{\int }_{{\epsilon }_2}^t{D}_{0^{+}}^{\alpha -1}u(t)dt,\hfill \\ {\parallel D_{0^{+}}^{\alpha -1}u\parallel }_{\mathrm{\infty }}\le A+{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }},\hfill \\ {\parallel D_{0^{+}}^{\alpha -2}u\parallel }_{\mathrm{\infty }}\le B+{\parallel D_{0^{+}}^{\alpha -1}u\parallel }_{\mathrm{\infty }}\le A+B+{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}.\hfill \end{array}$$

So, we get

$$\begin{array}{c}|c_1|\le \frac{1}{\mathrm{\Gamma }(\alpha )}({\parallel D_{0^{+}}^{\alpha -1}u\parallel }_{\mathrm{\infty }}+{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }})\le \frac{1}{\mathrm{\Gamma }(\alpha )}(A+2{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}),\hfill \\ |c_2|\le \frac{1}{\mathrm{\Gamma }(\alpha -1)}({\parallel D_{0^{+}}^{\alpha -2}u\parallel }_{\mathrm{\infty }}+\frac{3}{2}{\parallel D_{0^{+}}^{\alpha -1}u\parallel }_{\mathrm{\infty }}+{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }})\hfill \\ \phantom{|c_2|}\le \frac{1}{\mathrm{\Gamma }(\alpha -1)}(\frac{5A}{2}+B+\frac{7}{2}{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}),\hfill \\ {\parallel u\parallel }_{\mathrm{\infty }}\le A_1{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}+B_1,\hfill \end{array}$$

where $A_1=\frac{1}{\mathrm{\Gamma }(\alpha +1)}+\frac{2}{\mathrm{\Gamma }(\alpha )}+\frac{7}{2\mathrm{\Gamma }(\alpha -1)}$, $B_1=\frac{A}{\mathrm{\Gamma }(\alpha )}+\frac{5A}{2\mathrm{\Gamma }(\alpha -1)}+\frac{B}{\mathrm{\Gamma }(\alpha -1)}$.

Based on $D_{0^{+}}^{\alpha }u(0)=0$, we have

$${\phi }_p(D_{0^{+}}^{\alpha }u(t))=\lambda I_{0^{+}}^{\beta }Nu(t).$$

From (H1) and $\lambda \in (0,1)$, we have

$$\begin{array}{c}\left|{\phi }_p(D_{0^{+}}^{\alpha }u(t))\right|\hfill \\ \le \frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^t{(t-s)}^{\beta -1}\left|f(s,u(s),D_{0^{+}}^{\alpha -2}u(t),D_{0^{+}}^{\alpha -1}u(s),D_{0^{+}}^{\alpha }u(s))\right|ds\hfill \\ \le \frac{1}{\mathrm{\Gamma }(\beta )}{\int }_0^t{(t-s)}^{\beta -1}(r(s)+d(s)|u(s){|}^{p-1}+e(s)|D_{0^{+}}^{\alpha -2}u(s){|}^{p-1}\hfill \\ +h(s)|D_{0^{+}}^{\alpha -1}u(s){|}^{p-1}+k(s)|D_{0^{+}}^{\alpha }u(s){|}^{p-1})ds\hfill \\ \le \frac{1}{\mathrm{\Gamma }(\beta +1)}({\parallel r\parallel }_{\mathrm{\infty }}+{\parallel d\parallel }_{\mathrm{\infty }}{\parallel u\parallel }_{\mathrm{\infty }}^{p-1}+{\parallel e\parallel }_{\mathrm{\infty }}{\parallel D_{0^{+}}^{\alpha -2}u\parallel }_{\mathrm{\infty }}^{p-1}\hfill \\ +{\parallel h\parallel }_{\mathrm{\infty }}{\parallel D_{0^{+}}^{\alpha -1}u\parallel }_{\mathrm{\infty }}^{p-1}+{\parallel k\parallel }_{\mathrm{\infty }}{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}^{p-1}),\hfill \end{array}$$

which together with $|{\phi }_p(D_{0^{+}}^{\alpha }u(t))|=|D_{0^{+}}^{\alpha }u(t){|}^{p-1}$, we can get

$$\begin{array}{rcl}{}_{\parallel D_{0^{+}}^{\alpha }u\parallel }^{{}_{\mathrm{\infty }}p-1}& \le & \frac{1}{\mathrm{\Gamma }(\beta +1)}({\parallel r\parallel }_{\mathrm{\infty }}+{\parallel d\parallel }_{\mathrm{\infty }}{\parallel u\parallel }_{\mathrm{\infty }}^{p-1}+{\parallel e\parallel }_{\mathrm{\infty }}{\parallel D_{0^{+}}^{\alpha -2}u\parallel }_{\mathrm{\infty }}^{p-1}\\ +{\parallel h\parallel }_{\mathrm{\infty }}{\parallel D_{0^{+}}^{\alpha -1}u\parallel }_{\mathrm{\infty }}^{p-1}+{\parallel k\parallel }_{\mathrm{\infty }}{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}^{p-1})\\ \le & \frac{1}{\mathrm{\Gamma }(\beta +1)}({\parallel r\parallel }_{\mathrm{\infty }}+{\parallel d\parallel }_{\mathrm{\infty }}(A_1{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}+B_1)+{\parallel e\parallel }_{\mathrm{\infty }}{(2A+{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }})}^{p-1}\\ +{\parallel h\parallel }_{\mathrm{\infty }}{(A+{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }})}^{p-1}+{\parallel k\parallel }_{\mathrm{\infty }}{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}^{p-1}).\end{array}$$

In view of (3.1), we can obtain that there exists a constant $M_1>0$ such that

$$\begin{array}{c}{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}\le M_1,{\parallel D_{0^{+}}^{\alpha -1}u\parallel }_{\mathrm{\infty }}\le A+M_1:=M_2,\hfill \\ {\parallel D_{0^{+}}^{\alpha -2}u\parallel }_{\mathrm{\infty }}\le 2A+M_1:=M_3,{\parallel u\parallel }_{\mathrm{\infty }}\le A_1{M}_1+B_1:=M_4.\hfill \end{array}$$

Thus, we have

$${\parallel u\parallel }_X=max\{{\parallel u\parallel }_{\mathrm{\infty }},{\parallel D_{0^{+}}^{\alpha -2}u\parallel }_{\mathrm{\infty }},{\parallel D_{0^{+}}^{\alpha -1}u\parallel }_{\mathrm{\infty }},{\parallel D_{0^{+}}^{\alpha }u\parallel }_{\mathrm{\infty }}\}\le max\{M_1,M_2,M_3,M_4\}:=M.$$

So, ${\mathrm{\Omega }}_1$ is bounded. □

Lemma 3.4 Suppose that (H2) holds, then the set

$${\mathrm{\Omega }}_2=\{u\mid u\in KerM,Nu\in ImM\}$$

is bounded.

Proof For each $u\in {\mathrm{\Omega }}_2$, we have that $u(t)=c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}$, $c_1,c_2\in \mathbf{R}$ and $QNu=0$. It follows from (H2) and (H3) that there exists an ${\epsilon }_1,{\epsilon }_2\in [0,1]$ such that $|D_{0^{+}}^{\alpha -1}u({\epsilon }_1)|\le A$, $|D_{0^{+}}^{\alpha -2}u({\epsilon }_2)|\le A$, which implies that $|c_1|\le \frac{A}{\mathrm{\Gamma }(\alpha )}$ and $|c_2|\le \frac{A+B}{\mathrm{\Gamma }(\alpha -1)}$. So, ${\mathrm{\Omega }}_2$ is bounded. □

Define the isomorphism $J^{-1}:KerM\to ImQ$ by $J^{-1}(c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2})=c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}$, $c_1,c_2\in \mathbf{R}$, $t\in [0,1]$. In fact, for each $c_1,c_2\in R$, suppose that $(Q_{1}y(t),Q_{2}y(t))=(c_1,c_2)$, we have

$$\{\begin{array}{c}{\mathrm{\Lambda }}_4{T}_{1}y(t)-{\mathrm{\Lambda }}_3{T}_{2}y(t)=\mathrm{\Lambda }{\phi }_q(c_1):=\widetilde{c_1},\hfill \\ -{\mathrm{\Lambda }}_2{T}_{1}y(t)+{\mathrm{\Lambda }}_1{T}_{2}y(t)=\mathrm{\Lambda }{\phi }_q(c_2):=\widetilde{c_2},\hfill \end{array}$$

(3.4)

where $\widetilde{c_1},\widetilde{c_2}\in \mathbf{R}$, by the condition ${\mathrm{\Lambda }}_4{\mathrm{\Lambda }}_1-{\mathrm{\Lambda }}_2{\mathrm{\Lambda }}_3\ne 0$, there exists a unique solution for (3.4), which is $(T_{1}y(t),T_{2}y(t))=(m_1,m_2)$, $m_1,m_2\in \mathbf{R}$. Now, we will prove that there exists $y\in Y$ such that $(T_{1}y(t),T_{2}y(t))=(m_1,m_2)$. Based on $y(t)\in C[0,1]$, we choose $y(t)=D_{0^{+}}^{\beta }\overline{y}(t)$, where $\overline{y}(t)={\phi }_p(l_1{t}^{(\alpha +\beta -1)(q-1)}+l_2{t}^{(\alpha +\beta -2)(q-1)})$, $l_1=\frac{{\mathrm{\Delta }}_1\mathrm{\Gamma }{(\alpha )}^{q-1}}{\mathrm{\Gamma }{(\alpha +\beta )}^{q-1}}$, $l_2=\frac{{\mathrm{\Delta }}_2\mathrm{\Gamma }{(\alpha -1)}^{q-1}}{\mathrm{\Gamma }{(\alpha +\beta -1)}^{q-1}}$, ${\mathrm{\Delta }}_1=\frac{m_1{\mathrm{\Lambda }}_4-m_2{\mathrm{\Lambda }}_2}{\mathrm{\Lambda }}$, ${\mathrm{\Delta }}_2=\frac{m_2{\mathrm{\Lambda }}_1-m_1{\mathrm{\Lambda }}_3}{\mathrm{\Lambda }}$, $t\in [0,1]$, which together with $2<\alpha \le 3$, $0<\beta \le 1$, $3<\alpha +\beta \le 4$, $q>1$, we have $(\alpha +\beta -1)(q-1)>0$ and $(\alpha +\beta -2)(q-1)>0$. So, we have $\overline{y}(0)=0$. Thus, we can obtain that

$$I_{0^{+}}^{\beta }y(t)=I_{0^{+}}^{\beta }{D}_{0^{+}}^{\beta }\overline{y}(t)=\overline{y}(t)+c{t}^{\beta -1},$$

where $c\in \mathbf{\text{R}}$, which together with $\overline{y}(0)=0$, we have $c=0$. So, we have

$$\{\begin{array}{c}{T}_{1}y(t)={\int }_0^1{(1-s)}^{\alpha -1}{\phi }_q(\overline{y}(s))ds-{\sum }_{i=1}^m{a}_i{\int }_0^{{\xi }_i}{({\xi }_i-s)}^{\alpha -1}{\phi }_q(\overline{y}(s))ds\hfill \\ \phantom{T_{1}y(t)}={\mathrm{\Delta }}_1{\mathrm{\Lambda }}_1+{\mathrm{\Delta }}_2{\mathrm{\Lambda }}_2=m_1,\hfill \\ T_{2}y(t)={\int }_0^1{\phi }_q(\overline{y}(s))ds-{\sum }_{i=1}^m{b}_i{\int }_0^{{\eta }_i}{\phi }_q(\overline{y}(s))ds={\mathrm{\Delta }}_1{\mathrm{\Lambda }}_3+{\mathrm{\Delta }}_2{\mathrm{\Lambda }}_4=m_2.\hfill \end{array}$$

Thus, there exists $y\in Y$ such that $(T_{1}y(t),T_{2}y(t))=(m_1,m_2)$. So $J^{-1}:KerL\to ImQ$ is well defined.

Lemma 3.5 Suppose that the first part of (H3) holds, then the set

$${\mathrm{\Omega }}_3=\{u\in KerM\mid \lambda J^{-1}u+(1-\lambda )QNu=0,\lambda \in [0,1]\}$$

is bounded.

Proof For each $u\in {\mathrm{\Omega }}_3$, we can get that $u(t)=c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}$, $c_1,c_2\in \mathbf{R}$. By the definition of the set ${\mathrm{\Omega }}_3$, we can obtain that

$$\lambda (c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2})+(1-\lambda )(Q_{1}N(c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2})t^{\alpha -1}+Q_{2}N(c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2})t^{\alpha -2})=0.$$

Thus,

$$\lambda c_1+(1-\lambda ){\phi }_p\left(\frac{1}{\mathrm{\Lambda }}({\mathrm{\Lambda }}_4{T}_{1}N(c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2})-{\mathrm{\Lambda }}_3{T}_{2}N(c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}))\right)=0,$$

(3.5)

$$\lambda c_2+(1-\lambda ){\phi }_p\left(\frac{1}{\mathrm{\Lambda }}(-{\mathrm{\Lambda }}_2{T}_{1}N(c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2})+{\mathrm{\Lambda }}_1{T}_{2}N(c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}))\right)=0.$$

(3.6)

If $\lambda =0$, then ${\mathrm{\Omega }}_3$ is bounded because of the first part of (H2) and (H3). If $\lambda =1$, we get $c_1=c_2=0$, obviously, ${\mathrm{\Omega }}_3$ is bounded. If $\lambda \in (0,1)$, by the first part of (H2) and (3.5), we can obtain that $|c_1|\le \frac{A}{\mathrm{\Gamma }(\alpha )}$, by the first part of (H3) and (3.6), we have $|c_2|\le \frac{A+B}{\mathrm{\Gamma }(\alpha -1)}$. So, ${\mathrm{\Omega }}_3$ is bounded. □

Remark 3.1 If the second part of (H3) holds, then the set

$${\mathrm{\Omega }}_3^{\prime }=\{u\in KerM\mid -\lambda J^{-1}u+(1-\lambda )QNu=0,\lambda \in [0,1]\}$$

is bounded.

Proof of Theorem 3.1 Assume that Ω is a bounded open set of X with ${\bigcup }_{i=1}^3{\overline{\mathrm{\Omega }}}_i\cup {\mathrm{\Omega }}_3^{\prime }\subset \mathrm{\Omega }$. By Lemmas 3.1 and 3.2, we can obtain that $M:domM\cap X\to Y$ is quasi-linear, and $N_{\lambda }$ is M-compact on $\overline{\mathrm{\Omega }}$. By the definition of Ω , we have

$$\begin{array}{c}Lu\ne N_{\lambda }u,\mathrm{\forall }(u,\lambda )\in (domM\cap \partial \mathrm{\Omega })\times (0,1),\hfill \\ H(u,\lambda )=\pm \lambda J^{-1}(u)+(1-\lambda )QN(u)\ne 0,(\partial \mathrm{\Omega }\cap KerM)\times [0,1].\hfill \end{array}$$

Thus, by the homotopic property of degree, we can get

$$\begin{array}{rcl}deg(JQN,\mathrm{\Omega }\cap KerM,0)& =& deg(H(\cdot ,0),\mathrm{\Omega }\cap KerM,0)\\ =& deg(H(\cdot ,1),\mathrm{\Omega }\cap KerM,0)\\ =& deg(\pm I,\mathrm{\Omega }\cap KerM,0)\ne 0.\end{array}$$

So Lemma 2.1 is satisfied, and $Mu=Nu$ has at least one solution in $domM\cap \overline{\mathrm{\Omega }}$. Namely, BVP (1.3) have at least one solution in the space X. □