A resonant boundary value problem for the fractional p-Laplacian equation

Abstract

The purpose of this paper is to study the solvability of a resonant boundary value problem for the fractional p-Laplacian equation. By using the continuation theorem of coincidence degree theory, we obtain a new result on the existence of solutions for the considered problem.

1 Introduction

In this paper, we establish an existence theorem of solutions for the following resonant boundary value problem with p-Laplacian operator:

$$\begin{aligned} \left \{ \textstyle\begin{array}{l} {}_{0}^{c}D_{t}^{\beta}\phi_{p}({}_{0}^{c}D_{t}^{\alpha}x)=f(t,x,{}_{0}^{c}D_{t}^{\alpha}x),\quad t\in[0,1],\\ x(0)=0, \qquad {}_{0}^{c}D_{t}^{\alpha}x(0)={}_{0}^{c}D_{t}^{\alpha}x(1), \end{array}\displaystyle \right . \end{aligned}$$

(1.1)

where $0<\alpha,\beta\leq1$ are constants, ${}_{0}^{c}D_{t}^{\alpha}$ is a Caputo fractional derivative, $f:[0,1]\times\mathbb{R}^{2}\rightarrow \mathbb{R}$ is a continuous function, $\phi_{p}:\mathbb{R}\rightarrow \mathbb{R}$ is a p-Laplacian operator defined by

$$\phi_{p}(s)=|s|^{p-2}s\quad (s\neq0), \qquad\phi_{p}(0)=0,\quad p>1. $$

Obviously, $\phi_{p}$ is invertible and its inverse operator is $\phi_{q}$, where $q>1$ is a constant such that $1/p+1/q=1$.

Fractional calculus is a generalization of ordinary differentiation and integration, and fractional differential equations appear in various fields (see [1–4]). Recently, because of the intensive development of fractional calculus theory and its applications, the initial and boundary value problems (BVPs for short) of fractional differential equations have gained popularity (see [5–15] and the references therein).

In [11], by using the coincidence degree theory for Fredholm operators, the authors considered the existence of solutions for BVP (1.1). Notice that ${}_{0}^{c}D_{t}^{\beta}\phi_{p}({}_{0}^{c}D_{t}^{\alpha})$ is nonlinear, and so it is not a Fredholm operator. Thus there is a gap in the proof of the main result, and we fix this gap in the present paper.

2 Preliminaries

For convenience of the reader, we will introduce some necessary basic knowledge about fractional calculus theory (see [2, 4]).

Definition 2.1

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

$${}_{0}I_{t}^{\alpha}u=\frac{1}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha-1}u(s)\,ds, $$

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

Definition 2.2

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

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

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

Lemma 2.1

See [1]

Let $\alpha>0$. Assume that $u,{}_{0}^{c}D_{t}^{\alpha}u\in L([0,T],\mathbb{R})$. Then the following equality holds:

$${}_{0}I_{t}^{\alpha}{}_{0}^{c}D_{t}^{\alpha}u(t)=u(t)+c_{0}+c_{1}t+ \cdots+c_{n-1}t^{n-1}, $$

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

Next we present some notations and an abstract existence result (see [16]).

Let X, Y be real Banach spaces, $L: \operatorname{dom}L\subset X\rightarrow Y$ be a Fredholm operator with index zero, and $P: X\rightarrow X$, $Q:Y\rightarrow Y$ be projectors such that

$$\begin{aligned}& \operatorname{Im}P=\operatorname{Ker}L,\qquad \operatorname{Ker}Q= \operatorname{Im}L, \\& X=\operatorname{Ker}L\oplus\operatorname{Ker}P,\qquad Y=\operatorname{Im}L\oplus \operatorname{Im}Q. \end{aligned}$$

It follows that

$$L|_{\operatorname{dom}L\cap\operatorname{Ker}P}: \operatorname{dom}L\cap\operatorname{Ker}P\rightarrow \operatorname{Im}L $$

is invertible. We denote the inverse by $K_{P}$.

If Ω is an open bounded subset of X such that $\operatorname{dom}L\cap \overline{\Omega}\neq\varnothing$, then the map $N:X\rightarrow Y$ will be called L-compact on Ω̅ if $QN(\overline{\Omega})$ is bounded and $K_{P}(I-Q)N:\overline{\Omega}\rightarrow X$ is compact.

Lemma 2.2

See [16]

Let $L:\operatorname{dom}L\subset X\rightarrow Y$ be a Fredholm operator of index zero and $N:X\rightarrow Y$ be L-compact on Ω̅. Assume that the following conditions are satisfied:

• (1) $Lx\neq\lambda Nx$ for every $(x,\lambda)\in[(\operatorname{dom}L\setminus \operatorname{Ker}L)\cap\partial\Omega]\times(0,1)$,

• (2) $Nx\notin\operatorname{Im}L$ for every $x\in\operatorname{Ker}L\cap\partial\Omega$,

• (3) $\operatorname{deg}(QN|_{\operatorname{Ker}L},\Omega\cap\operatorname{Ker}L,0)\neq0$, where $Q:Y\rightarrow Y$ is a projection such that $\operatorname{Im}L=\operatorname{Ker}Q$.

Then the equation $Lx=Nx$ has at least one solution in $\operatorname{dom}L\cap\overline{\Omega}$.

In this paper, we let $Z=C([0,1],\mathbb{R})$ with the norm $\|z\| _{\infty}=\max_{t\in[0,1]}|z(t)|$ and take

$$X= \bigl\{ x=(x_{1},x_{2})^{\top}|x_{1},x_{2} \in Z \bigr\} $$

with the norm

$$\|x\|_{X}=\max\bigl\{ \|x_{1}\|_{\infty}, \|x_{2} \|_{\infty}\bigr\} . $$

By means of the linear functional analysis theory, we can prove that X is a Banach space.

3 Main result

We will establish the existence theorem of solutions for BVP (1.1).

Theorem 3.1

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

$(H_{1})$ :

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

$$\big|f(t,u,v)\big|\leq a(t)+b(t)|u|^{p-1}+c(t)|v|^{p-1},\quad \forall(t,u,v)\in [0,1]\times\mathbb{R}^{2}, $$

$(H_{2})$ :

there exists a constant $B>0$ such that

$$vf(t,u,v)>0\ (\textit{or } < 0), \quad\forall t\in[0,1],u\in\mathbb{R},|v|>B. $$

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

$$\gamma:=\frac{2}{\Gamma(\beta+1)} \biggl(\frac{\|b\|_{\infty}}{(\Gamma(\alpha+1))^{p-1}}+\|c\|_{\infty}\biggr)< 1. $$

Consider BVP of the linear differential system as follows:

$$\begin{aligned} \left \{ \textstyle\begin{array}{l} {}_{0}^{c}D_{t}^{\alpha}x_{1}=\phi_{q} (x_{2}), \quad t\in[0,1],\\ {}_{0}^{c}D_{t}^{\beta}x_{2}=f(t,x_{1},\phi_{q} (x_{2})),\quad t\in[0,1],\\ x_{1}(0)=0, \qquad x_{2}(0)=x_{2}(1). \end{array}\displaystyle \right . \end{aligned}$$

(3.1)

Obviously, if $x=(x_{1},x_{2})^{\top}$ is a solution of BVP (3.1), then $x_{1}$ must be a solution of BVP (1.1). Therefore, to prove BVP (1.1) has solutions, it suffices to show that BVP (3.1) has solutions.

Define the operator $L:\operatorname{dom}L\subset X\rightarrow X$ by

$$ Lx=\binom{{}_{0}^{c}D_{t}^{\alpha}x_{1}}{{}_{0}^{c}D_{t}^{\beta}x_{2}}, $$

(3.2)

where

$$\operatorname{dom}L= \bigl\{ x\in X|{}_{0}^{c}D_{t}^{\alpha}x_{1},{}_{0}^{c}D_{t}^{\beta}x_{2}\in Z, x_{1}(0)=0, x_{2}(0)=x_{2}(1) \bigr\} . $$

Let $N:X\rightarrow X$ be the Nemytskii operator defined by

$$ Nx(t)=\binom{\phi_{q}(x_{2}(t))}{f(t,x_{1}(t),\phi_{q}(x_{2}(t)))}, \quad\forall t\in[0,1]. $$

(3.3)

Then BVP (3.1) is equivalent to the following operator equation:

$$Lx=Nx,\quad x\in\operatorname{dom}L. $$

Now, in order to prove Theorem 3.1, we give some lemmas.

Lemma 3.1

Let L be defined by (3.2), then

$$\begin{aligned}& \operatorname{Ker}L=\bigl\{ x\in X|x_{1}(t)=0, x_{2}(t)=c, \forall t\in[0,1],c\in \mathbb{R}\bigr\} , \end{aligned}$$

(3.4)

$$\begin{aligned}& \operatorname{Im}L= \bigl\{ y\in X|{_{0}I_{t}^{\beta}}y_{2}(1)=0 \bigr\} . \end{aligned}$$

(3.5)

Proof

By Lemma 2.1, the equation $Lx=0$ has solutions

$$x_{1}(t)=c_{1},\quad\quad x_{2}(t)=c_{2},\quad c_{1},c_{2}\in\mathbb{R}. $$

Thus, from the boundary value condition $x_{1}(0)=0$, one has that (3.4) holds.

Let $y\in\operatorname{Im}L$, then there exists a function $x\in\operatorname{dom}L$ such that $y_{2}={}_{0}^{c}D_{t}^{\beta}x_{2}$. So, by Lemma 2.1, we have

$$x_{2}(t)=c+{}_{0}I_{t}^{\beta}y_{2}(t),\quad c\in\mathbb{R}. $$

Hence, from the boundary value condition $x_{2}(0)=x_{2}(1)$, we get (3.5).

On the other hand, suppose that $y\in X$ satisfies ${}_{0}I_{t}^{\beta}y_{2}(1)=0$. Let $x_{1}={}_{0}I_{t}^{\alpha}y_{1}$, $x_{2}={}_{0}I_{t}^{\beta}y_{2}(t)$, then $x=(x_{1},x_{2})^{\top}\in\operatorname{dom}L$ and $Lx=y$. That is, $y\in\operatorname{Im}L$. The proof is complete. □

Lemma 3.2

Let L be defined by (3.2), then L is a Fredholm operator of index zero. And the projectors $P:X\rightarrow X$, $Q:X\rightarrow X$ can be defined as

$$\begin{aligned}& Px(t)=\binom{0}{x_{2}(0)}, \quad\forall t\in[0,1], \\& Qy(t)=\binom{0}{\Gamma(\beta+1){}_{0}I_{t}^{\beta}y_{2}(1)}, \quad\forall t\in[0,1]. \end{aligned}$$

Furthermore, the operator $K_{P}:\operatorname{Im}L\rightarrow\operatorname{dom}L\cap\operatorname{Ker}P$ can be written as

$$K_{P}y=\binom{{}_{0}I_{t}^{\alpha}y_{1}}{{}_{0}I_{t}^{\beta}y_{2}}. $$

Proof

For any $y \in X$, one has

$$\begin{aligned} Q^{2}y&=Q\binom{0}{\Gamma(\beta+1){}_{0}I_{t}^{\beta}y_{2}(1)} \\ &=\binom{0}{\Gamma(\beta+1){}_{0}I_{t}^{\beta}y_{2}(1)\cdot\Gamma(\beta +1){}_{0}I_{t}^{\beta}1(1)} \\ &=Qy. \end{aligned}$$

(3.6)

Let $y^{*}=y-Qy$, then we get from (3.6) that

$$\begin{aligned} {}_{0}I_{t}^{\beta}y^{*}_{2}(1)&={}_{0}I_{t}^{\beta}y_{2}(1)-{}_{0}I_{t}^{\beta}(Qy_{2}) (1) \\ &=\frac{1}{\Gamma(\beta+1)}\bigl((Qy_{2}) (t)-\bigl(Q^{2}y_{2} \bigr) (t)\bigr) \\ &=0, \end{aligned}$$

which yields $y^{*}\in\operatorname{Im}L$. So $X=\operatorname{Im}L+\operatorname{Im}Q$. Since $\operatorname{Im}L\cap\operatorname{Im}Q=\{(0,0)^{\top}\}$, we have $X=\operatorname{Im}L\oplus \operatorname{Im}Q$. Hence

$$\operatorname{dim}\operatorname{Ker}L=\operatorname{dim}\operatorname{Im}Q= \operatorname{codim}\operatorname{Im}L=1. $$

Thus L is a Fredholm operator of index zero.

For $y\in\operatorname{Im}L$, by the definition of operator $K_{P}$, we have

$$\begin{aligned} LK_{P}y&=\binom{{}_{0}^{c}D_{t}^{\alpha}{}_{0}I_{t}^{\alpha}y_{1}}{{}_{0}^{c}D_{t}^{\beta}{}_{0}I_{t}^{\beta}y_{2}} \\ &=y. \end{aligned}$$

(3.7)

On the other hand, for $x\in\operatorname{dom}L\cap\operatorname{Ker}P$, one has

$$x_{1}(0)=x_{2}(0)=x_{2}(1)=0. $$

Thus, from Lemma 2.1, we get

$$\begin{aligned} K_{P}Lx(t)&=\binom{{}_{0}I_{t}^{\alpha}{}_{0}^{c}D_{t}^{\alpha}x_{1}(t)}{{}_{0}I_{t}^{\beta}{}_{0}^{c}D_{t}^{\beta}x_{2}(t)} \\ &=\binom{x_{1}(t)-x_{1}(0)}{x_{2}(t)-x_{2}(0)} \\ &=x(t). \end{aligned}$$

(3.8)

Hence, combining (3.7) with (3.8), we know $K_{P}= (L|_{\operatorname{dom}L\cap\operatorname{Ker}P} )^{-1}$. The proof is complete. □

Lemma 3.3

Let N be defined by (3.3). Assume $\Omega\subset X$ is an open bounded subset such that $\operatorname{dom}L\cap \overline{\Omega}\neq\varnothing$, then N is L-compact on Ω̅.

Proof

From the continuity of $\phi_{q}$ and f, we obtain $K_{P}(I-Q)N$ is continuous in X and $QN(\overline{\Omega})$, $K_{P}(I-Q)N(\overline{\Omega})$ are bounded. Moreover, there exists a constant $T>0$ such that

$$ \big\| (I-Q)Nx\big\| _{X}\leq T, \quad\forall x\in\overline{\Omega}. $$

(3.9)

Thus, in view of the Arzelà-Ascoli theorem, we need only to prove $K_{P}(I-Q)N(\overline{\Omega})\subset X$ is equicontinuous.

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

$$\begin{aligned} &\big|K_{P}(I-Q)Nx(t_{2})-K_{P}(I-Q)Nx(t_{1})\big| \\ &\quad=\binom{{}_{0}I_{t}^{\alpha}((I-Q)Nx)_{1}(t_{2})-{}_{0}I_{t}^{\alpha}((I-Q)Nx)_{1}(t_{1})}{ {}_{0}I_{t}^{\beta}((I-Q)Nx)_{2}(t_{2})-{}_{0}I_{t}^{\beta}((I-Q)Nx)_{2}(t_{1})}. \end{aligned}$$

From (3.9), we have

$$\begin{aligned} &\big|{}_{0}I_{t}^{\alpha}\bigl((I-Q)Nx \bigr)_{1}(t_{2})-{}_{0}I_{t}^{\alpha}\bigl((I-Q)Nx\bigr)_{1}(t_{1})\big| \\ &\quad=\frac{1}{\Gamma(\alpha)} \biggl\vert \int_{0}^{t_{2}}(t_{2}-s)^{\alpha-1} \bigl((I-Q)Nx\bigr)_{1}(s)\,ds \\ &\qquad - \int_{0}^{t_{1}}(t_{1}-s)^{\alpha-1} \bigl((I-Q)Nx\bigr)_{1}(s)\,ds \biggr\vert \\ &\quad\leq\frac{T}{\Gamma(\alpha)} \biggl\{ \int_{0}^{t_{1}}\bigl[(t_{1}-s)^{\alpha-1}-(t_{2}-s)^{\alpha-1} \bigr]\,ds + \int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-1}\,ds \biggr\} \\ &\quad=\frac{T}{\Gamma(\alpha+1)}\bigl[t_{1}^{\alpha}-t_{2}^{\alpha} +2(t_{2}-t_{1})^{\alpha}\bigr]. \end{aligned}$$

Since $t^{\alpha}$ is uniformly continuous on $[0,1]$, we get $(K_{P}(I-Q)N(\overline{\Omega}))_{1}\subset Z$ is equicontinuous. A similar proof can show that $(K_{P}(I-Q)N(\overline{\Omega}))_{2}\subset Z$ is also equicontinuous. Hence, we obtain $K_{P}(I-Q)N:\overline{\Omega }\rightarrow X$ is compact. The proof is complete. □

Finally, we give the proof of Theorem 3.1.

Proof of Theorem 3.1

Let

$$\Omega_{1}=\bigl\{ x\in\operatorname{dom}L\backslash \operatorname{Ker}L|Lx=\lambda Nx, \lambda \in(0,1)\bigr\} . $$

For $x\in\Omega_{1}$, we have $x_{1}(0)=0$ and $Nx\in\operatorname{Im}L$. So, by Lemma 2.1, we get

$$x_{1}={}_{0}I_{t}^{\alpha}{}_{0}^{c}D_{t}^{\alpha}x_{1}. $$

Thus one has

$$\big|x_{1}(t)\big|\leq\frac{1}{\Gamma(\alpha+1)}\big\| {}_{0}^{c}D_{t}^{\alpha}x_{1} \big\| _{\infty}, \quad\forall t\in[0,1]. $$

That is,

$$\begin{aligned} \|x_{1}\|_{\infty}\leq\frac{1}{\Gamma(\alpha+1)} \big\| {}_{0}^{c}D_{t}^{\alpha}x_{1}\big\| _{\infty}. \end{aligned}$$

(3.10)

From $Nx\in\operatorname{Im}L$ and (3.5), we obtain

$$\begin{aligned} 0&={}_{0}I_{t}^{\beta}(Nx)_{2}(1) \\ &=\frac{1}{\Gamma(\beta)} \int_{0}^{1}(1-s)^{\beta-1}f \bigl(s,x_{1}(s),\phi_{q}\bigl(x_{2}(s)\bigr) \bigr)\,ds. \end{aligned}$$

Then, by the integral mean value theorem, there exists a constant $\xi \in(0,1)$ such that

$$f\bigl(\xi,x_{1}(\xi),\phi_{q}\bigl(x_{2}(\xi) \bigr)\bigr)=0. $$

So, by $(H_{2})$, we have $|x_{2}(\xi)|\leq B^{p-1}$. From Lemma 2.1, we get

$$x_{2}(t)=x_{2}(\xi)-{}_{0}I_{t}^{\beta}{}_{0}^{c}D_{t}^{\beta}x_{2}( \xi)+{}_{0}I_{t}^{\beta}{}_{0}^{c}D_{t}^{\beta}x_{2}(t), $$

which together with

$$\bigl\vert {}_{0}I_{t}^{\beta}{}_{0}^{c}D_{t}^{\beta}x_{2}(t) \bigr\vert \leq\frac{1}{\Gamma(\beta +1)} \bigl\Vert {}_{0}^{c}D_{t}^{\beta}x_{2} \bigr\Vert _{\infty}, \quad\forall t\in[0,1] $$

yields

$$ \|x_{2}\|_{\infty}\leq B^{p-1}+ \frac{2}{\Gamma(\beta+1)} \bigl\Vert {}_{0}^{c}D_{t}^{\beta}x_{2} \bigr\Vert _{\infty}. $$

(3.11)

From $Lx=\lambda Nx$, one has

$$\begin{aligned}& {}_{0}^{c}D_{t}^{\alpha}x_{1}= \lambda\phi_{q}(x_{2}), \end{aligned}$$

(3.12)

$$\begin{aligned}& {}_{0}^{c}D_{t}^{\beta}x_{2}= \lambda f\bigl(t,x_{1},\phi_{q}(x_{2})\bigr). \end{aligned}$$

(3.13)

By (3.12), we have

$$\big\| {}_{0}^{c}D_{t}^{\alpha}x_{1} \big\| _{\infty}\leq\|x_{2}\|_{\infty}^{q-1}, $$

which together with (3.10) yields

$$ \|x_{1}\|_{\infty}\leq\frac{1}{\Gamma(\alpha+1)} \|x_{2}\|_{\infty}^{q-1}. $$

(3.14)

By (3.13) and $(H_{1})$, we obtain

$$\bigl\Vert {}_{0}^{c}D_{t}^{\beta}x_{2} \bigr\Vert _{\infty}\leq \|a\|_{\infty}+\|b\|_{\infty}\|x_{1}\|_{\infty}^{p-1}+\|c\|_{\infty}\|x_{2}\|_{\infty}, $$

which together with (3.11) and (3.14) yields

$$\begin{aligned} \bigl\Vert {}_{0}^{c}D_{t}^{\beta}x_{2} \bigr\Vert _{\infty}&\leq\|a\|_{\infty}+\frac{\Gamma(\beta+1)\gamma}{2} \|x_{2}\|_{\infty} \\ &\leq\|a\|_{\infty}+\frac{\Gamma(\beta+1)\gamma B^{p-1}}{2}+\gamma \bigl\Vert {}_{0}^{c}D_{t}^{\beta}x_{2} \bigr\Vert _{\infty}. \end{aligned}$$

(3.15)

Since $\gamma<1$, we get from (3.15) that there exists a constant $M_{0}>0$ such that

$$\bigl\Vert {}_{0}^{c}D_{t}^{\beta}x_{2} \bigr\Vert _{\infty}\leq M_{0}. $$

Thus, combining (3.11) with (3.14), we have

$$\begin{aligned}& \|x_{2}\|_{\infty}\leq B^{p-1}+\frac{2M_{0}}{\Gamma(\beta+1)}:=M_{1}, \\& \|x_{1}\|_{\infty}\leq\frac{M_{1}^{q-1}}{\Gamma(\alpha+1)}:=M_{2}. \end{aligned}$$

Hence

$$\|x\|_{X}\leq\max\{M_{1}, M_{2}\}:=M, $$

which means $\Omega_{1}$ is bounded.

Let

$$\Omega_{2}=\{x\in\operatorname{Ker}L|Nx\in\operatorname{Im}L\}. $$

For $x\in\Omega_{2}$, we have ${}_{0}I_{t}^{\beta}(Nx)_{2}(1)=0$ and $x_{1}(t)=0$, $x_{2}(t)=c$, $c\in\mathbb{R}$. Thus one has

$$\int_{0}^{1}(1-s)^{\beta-1}f\bigl(s,0, \phi_{q}(c)\bigr)\,ds=0, $$

which together with $(H_{2})$ yields $|c|\leq B^{p-1}$. Hence

$$\|x\|_{X}\leq\max\bigl\{ 0, B^{p-1}\bigr\} =B^{p-1}, $$

which means $\Omega_{2}$ is bounded.

By $(H_{2})$, one has

$$ \phi_{p}(v)f(t,u,v)>0,\quad \forall t\in[0,1], u\in \mathbb{R}, |v|>B $$

(3.16)

or

$$ \phi_{p}(v)f(t,u,v)< 0, \quad\forall t\in[0,1], u\in \mathbb{R}, |v|>B. $$

(3.17)

When (3.16) is true, let

$$\Omega_{3}=\bigl\{ x\in\operatorname{Ker}L|\lambda x+(1-\lambda)QNx=0, \lambda\in [0,1]\bigr\} . $$

For $x\in\Omega_{3}$, we have $x_{1}(t)=0$, $x_{2}(t)=c$, $c\in\mathbb{R}$ and

$$ \lambda c +(1-\lambda)\beta \int_{0}^{1}(1-s)^{\beta-1}f\bigl(s,0, \phi_{q}(c)\bigr)\,ds=0. $$

(3.18)

If $\lambda=0$, we get from (3.16) that $|c|\leq B^{p-1}$. If $\lambda\in(0,1]$, we assume $|c|>B^{p-1}$. Thus, by (3.16), we obtain

$$\lambda c^{2} +(1-\lambda)\beta \int_{0}^{1}(1-s)^{\beta-1}\phi_{p} \bigl(\phi_{q}(c)\bigr)f\bigl(s,0,\phi_{q}(c)\bigr)\,ds>0, $$

which contradicts (3.18). Hence, $\Omega_{3}$ is bounded.

When (3.17) is true, let

$$\Omega'_{3}=\bigl\{ x\in\operatorname{Ker}L|{-}\lambda x+(1-\lambda)QNx=0, \lambda\in [0,1]\bigr\} . $$

A similar proof can show $\Omega'_{3}$ is also bounded.

Set

$$\Omega=\bigl\{ x\in X|\|x\|_{X}< \max\bigl\{ M,B^{p-1}\bigr\} +1 \bigr\} . $$

Clearly, $\Omega_{1}\cup\Omega_{2}\cup\Omega_{3}\subset\Omega$ (or $\Omega _{1}\cup\Omega_{2}\cup\Omega'_{3}\subset\Omega$). It follows from Lemma 3.2 and 3.3 that L (defined by (3.2)) is a Fredholm operator of index zero and N (defined by (3.3)) is L-compact on Ω̅. Moreover, based on the above proof, the conditions (1) and (2) of Lemma 2.2 are satisfied. Define the operator $H:\overline{\Omega}\times[0,1]\rightarrow X$ by

$$H(x,\lambda)=\pm\lambda x+(1-\lambda)QNx. $$

Then, from the above proof, we have

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

Thus, by the homotopy property of degree, we get

$$\begin{aligned} \operatorname{deg}(QN|_{\operatorname{Ker}L},\Omega\cap\operatorname{Ker}L,0) &= \operatorname{deg}\bigl(H(\cdot,0),\Omega\cap\operatorname{Ker}L,0\bigr) \\ &=\operatorname{deg}\bigl(H(\cdot,1),\Omega\cap\operatorname{Ker}L,0\bigr) \\ &=\operatorname{deg}(\pm I,\Omega\cap\operatorname{Ker}L,0) \\ &\neq0. \end{aligned}$$

Hence, condition (3) of Lemma 2.2 is also satisfied.

Therefore, by using Lemma 2.2, the operator equation $Lx=Nx$ has at least one solution in $\operatorname{dom}L\cap\overline{\Omega}$. Namely, BVP (1.1) has at least one solution in X. The proof is complete. □