• Research
• Open Access

# Existence results for $$(k,n-k)$$ conjugate boundary-value problems with integral boundary conditions at resonance with $$\dim\ker L=2$$

Boundary Value Problems20172017:29

https://doi.org/10.1186/s13661-017-0760-6

• Received: 16 November 2016
• Accepted: 21 February 2017
• Published:

## Abstract

We shall study the existence of solutions for a $$(k,n-k)$$ conjugate boundary-value problem at resonance with $$\dim\ker L =2$$ in this paper. The boundary-value problem is shown as follows:
\begin{aligned}& (-1)^{n-k}\varphi^{(n)}(x)=f \bigl(x,\varphi(x), \varphi'(x),\ldots,\varphi^{(n-1)}(x) \bigr), \quad x\in[0,1], \\ & \varphi^{(i)}(0)=\varphi^{(j)}(1)=0, \quad1\leq i\leq k-1, 1\leq j \leq n-k-1, \\ & \varphi(0)= \int^{1}_{0}\varphi(x)\,dA(x), \quad\quad \varphi(1)= \int^{1}_{0}\varphi(x)\,dB(x). \end{aligned}
We can obtain that this boundary-value problem has at least one solution under the conditions we provided through Mawhin’s continuation theorem, and an example is also provided for our new results.

## Keywords

• boundary value problem
• resonance
• Fredholm operator
• Mawhin continuation theorem

## 1 Introduction

Conjugate boundary-value problems at non-resonance have aroused considerable attention in recent years (see ), and there is also much research on boundary-value problems at resonance (see ). However, there are very few papers involving $$(k,n-k)$$ conjugate boundary-value problems at resonance, especially with $$\dim\ker L =2$$. For example, Jiang  investigated the following boundary-value problem at resonance with $$\dim\ker L =2$$:
\begin{aligned}& (-1)^{n-k}y^{(n)}(t)=f \bigl(t,y(t),y'(t), \ldots,y^{(n-1)}(t) \bigr)+\varepsilon(t), \quad \text{a.e. } t\in[0,1], \\ & y^{(i)}(0)=y^{(j)}(1)=0, \quad 0\leq i\leq k-1, 0\leq j\leq n-k-3, \\ & y^{(n-1)}(1)=\sum_{i=1}^{m} \alpha_{i}y^{(n-1)}(\xi_{i}), \quad\quad y^{(n-2)}(1)=\sum_{j=1}^{l} \beta_{j}y^{(n-2)}( \eta_{j}), \end{aligned}
where $$1\leq k\leq n-3$$, $$0<\xi_{1}<\xi_{2}<\cdots<\xi_{m}<1$$, $$0<\eta _{1}<\eta_{2}<\cdots<\eta_{l}<1$$.
Motivated by , we shall study the following $$(k,n-k)$$ conjugate boundary-value problem in the situation of resonance with $$\dim\ker L =2$$:
\begin{aligned}& (-1)^{n-k}\varphi^{(n)}(x)=f \bigl(x,\varphi(x), \varphi'(x),\ldots,\varphi^{(n-1)}(x) \bigr), \quad x\in[0,1], \end{aligned}
(1)
\begin{aligned}& \varphi^{(i)}(0)=\varphi^{(j)}(1)=0, \quad 1\leq i \leq k-1,1\leq j\leq n-k-1, \end{aligned}
(2)
\begin{aligned}& \varphi(0)= \int^{1}_{0}\varphi(x)\,dA(x), \quad\quad \varphi(1)= \int^{1}_{0}\varphi(x)\,dB(x), \end{aligned}
(3)
where $$1\leq k\leq n-1$$, $$n\geq2$$, $$A(x)$$, $$B(x)$$ are left continuous at $$x=1$$, right continuous on $$[0,1)$$; $$\int^{1}_{0}u(x)\,dA(x)$$ and $$\int^{1}_{0}u(x)\,dB(x)$$ denote the Riemann-Stieltjes integrals of u with respect to A and B, respectively.

However, there are great differences between this article and the above results, the boundary conditions we study are $$\varphi(0)=\int ^{1}_{0}\varphi(x)\,dA(x)$$ and $$\varphi(1)=\int^{1}_{0}\varphi(x)\,dB(x)$$. As is well known, it is an original case to study conjugate boundary-value problems with integral boundary conditions in the situation of resonance.

The organization of this paper is as follows. In Section 2, we provide a definition and a theorem which will be used to prove the main results. In Section 3, we will give some lemmas and prove the solvability of problem (1)-(3).

## 2 Preliminaries

For the convenience of the reader, we recall some definitions and a theorem to be used later.

### Definition 2.1



Suppose that X and Y are real Banach spaces, $$L: \operatorname {dom}L \subset X \rightarrow Y$$ is a Fredholm operator of index zero if: (1) ImL is a closed subspace of Y; (2) $$\dim\ker L$$ = $$\operatorname {codim}\operatorname{Im}L < \infty$$.

If X, Y are real Banach spaces, $$L:\operatorname {dom}L\subset X\rightarrow Y$$ is a Fredholm operator of index zero, and $$P:X\rightarrow X$$, $$Q:Y\rightarrow Y$$ are continuous projectors such that
$$\operatorname{Im}P = \ker L ,\quad\quad \ker Q = \operatorname{Im}L ,\quad\quad X =\ker L \oplus\ker P , \quad\quad Y = \operatorname{Im}L \oplus \operatorname{Im}Q ,$$
then we can conclude that
$$L|_{\operatorname {dom}L\cap\ker P }:\operatorname {dom}L\cap\ker P \rightarrow \operatorname{Im}L$$
is invertible. We denote the inverse of the mapping by $$K_{P}$$ (generalized inverse operator of L). Let Ω be an open bounded subset of X and $$\operatorname {dom}L \cap\Omega \neq\emptyset$$, then we say the mapping $$N :X\rightarrow Y$$ is L-compact on Ω̅ if $$K_{P}(I-Q)N :\overline{\Omega}\rightarrow X$$ is compact and $$QN(\overline{\Omega})$$ is bounded.

### Theorem 2.1

; Mawhin continuation theorem

$$L: \operatorname {dom}L \subset X \rightarrow Y$$ is a Fredholm operator of index zero, and N is L-compact on Ω̅. The equation $$L\varphi=N\varphi$$ has at least one solution in $$\operatorname {dom}L \cap\overline {\Omega}$$ if the following conditions are satisfied:
1. (1)

$$L\varphi\neq\lambda N\varphi$$ for every $$(\varphi,\lambda) \in [(\operatorname {dom}L\backslash\ker L )\cap\partial\Omega] \times(0,1)$$;

2. (2)

$$N\varphi\notin \operatorname{Im}L$$ for every $$\varphi\in\ker L \cap \partial\Omega$$;

3. (3)

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

Let $$X=C^{n-1}[0,1]$$ with norm $$\Vert u\Vert =\max\{\Vert u\Vert _{\infty}, \Vert u'\Vert _{\infty},\ldots ,\Vert u^{(n-1)}\Vert _{\infty}\}$$, in which $$\Vert u\Vert _{\infty}=\max_{x\in[0,1]}\vert u(x)\vert$$, and $$Y=L^{1}[0,1]$$ with norm $$\Vert x\Vert _{1}=\int_{0}^{1}\vert x(t)\vert \,dt$$. We define an operator L as follows:
$$(L\varphi) (x)=(-1)^{n-k}\varphi^{(n)}(x)$$
with
\begin{aligned} \operatorname {dom}L&= \biggl\{ \varphi\in X:\varphi^{(i)}(0)=\varphi ^{(j)}(1)=0, 1\leq i\leq k-1, 1\leq j\leq n-k-1, \\ &\quad \varphi(0)= \int_{0}^{1}\varphi(x)\,dA(x), \varphi (1)= \int^{1}_{0}\varphi(x)\,dB(x) \biggr\} . \end{aligned}
An operator $$N:X\rightarrow Y$$ is defined as
$$(N\varphi) (x)=f \bigl(x,\varphi(x),\varphi'(x),\ldots,\varphi ^{(n-1)}(x) \bigr).$$
So problem (1)-(3) becomes $$L\varphi=N\varphi$$.

## 3 Main results

Assume that the following conditions hold in this paper:
\begin{aligned} \mathrm{(H1)} \quad& \int^{1}_{0}\Phi_{1}(x)\,dA(x)=1,\quad \quad \int^{1}_{0}\Phi_{2}(x)\,dB(x)=1, \\ & \int^{1}_{0}\Phi_{1}(x)\,dB(x)=0,\quad \quad \int^{1}_{0}\Phi_{2}(x)\,dA(x)=0, \end{aligned}
where
\begin{aligned}& \Phi_{1}(x)=\frac{(n-1)!}{(k-1)!(n-k-1)!} \int_{x}^{1}t^{k-1}(1-t)^{n-k-1} \,dt, \\& \Phi_{2}(x)=\frac{(n-1)!}{(k-1)!(n-k-1)!} \int_{0}^{x}t^{k-1}(1-t)^{n-k-1} \,dt. \\ & \mathrm{(H2)} \quad e= \left \vert \textstyle\begin{array}{c@{\quad}c}e_{1}&e_{2}\\ e_{3}&e_{4} \end{array}\displaystyle \right \vert \neq0, \end{aligned}
where
\begin{aligned}& e_{1}= \int^{1}_{0} \int^{1}_{0}k(x,y)\Phi_{1}(x)\,dy\,dA(x), \quad \quad e_{2}= \int^{1}_{0} \int^{1}_{0}k(x,y)\Phi_{1}(x)\,dy\,dB(x), \\& e_{3}= \int^{1}_{0} \int^{1}_{0}k(x,y)\Phi_{2}(x)\,dy\,dA(x), \quad\quad e_{4}= \int^{1}_{0} \int^{1}_{0}k(x,y)\Phi_{2}(x)\,dy\,dB(x), \\& k(x,y)= \textstyle\begin{cases} \frac{1}{(k-1)!(n-k-1)!}\int _{0}^{x(1-y)}t^{k-1}(t+y-x)^{n-k-1}\,dt,&{0\leq x\leq y\leq1};\\ \frac{1}{(k-1)!(n-k-1)!}\int _{0}^{y(1-x)}t^{n-k-1}(t+x-y)^{k-1}\,dt,&{0\leq y\leq x\leq1}. \end{cases}\displaystyle \end{aligned}

(H3) $$f:[0,1]\times R^{n}\rightarrow R$$ satisfies Caratháodory conditions.

(H4) There exist functions $$r(x), q_{i}(x)\in L^{1}[0,1]$$ with $$\sum_{i=1}^{n}\Vert q_{i}\Vert _{1}<1$$ such that
$$\bigl\vert f (x,\varphi_{1},\varphi_{2},\ldots, \varphi_{n} ) \bigr\vert \leq\sum_{i=1}^{n}q_{i}(x) \vert \varphi_{i}\vert +r(x),$$
where $$x\in[0,1]$$, $$\varphi_{i}\in R$$.
(H5) There exists a constant $$M>0$$ such that if $$\vert \varphi (x)\vert +\vert \varphi^{(n-1)}(x)\vert >M$$ for all $$x\in[0,1]$$, then
$$\int_{0}^{1} \int_{0}^{1}k(x,y)f \bigl(y,\varphi(y),\varphi '(y),\varphi'(y),\ldots,\varphi^{(n-1)}(y) \bigr) \,dy\,dA(x)\neq0,$$
or
$$\int_{0}^{1} \int_{0}^{1}k(x,y)f \bigl(y,\varphi(y),\varphi '(y),\varphi''(y),\ldots, \varphi^{(n-1)}(y) \bigr)\,dy\,dB(x)\neq0.$$
(H6) There are constants $$a, b >0$$ such that one of the following two conditions holds:
\begin{aligned}& c_{1} \int_{0}^{1} \int_{0}^{1}k(x,y)N \bigl(c_{1} \Phi_{1}(y)+c_{2}\Phi_{2}(y) \bigr)\,dy\,dA(x)< 0, \end{aligned}
(4)
\begin{aligned}& c_{2} \int_{0}^{1} \int_{0}^{1}k(x,y)N \bigl(c_{1} \Phi_{1}(y)+c_{2}\Phi_{2}(y) \bigr)\,dy\,dB(x)< 0 \end{aligned}
(5)
if $$\vert c_{1}\vert >a$$ and $$\vert c_{2}\vert >b$$, or
\begin{aligned}& c_{1} \int_{0}^{1} \int_{0}^{1}k(x,y)N \bigl(c_{1} \Phi_{1}(y)+c_{2}\Phi_{2}(y) \bigr)\,dy\,dA(x)>0, \end{aligned}
(6)
\begin{aligned}& c_{2} \int_{0}^{1} \int_{0}^{1}k(x,y)N \bigl(c_{1} \Phi_{1}(y)+c_{2}\Phi_{2}(y) \bigr)\,dy\,dB(x)>0 \end{aligned}
(7)
if $$\vert c_{1}\vert >a$$ and $$\vert c_{2}\vert >b$$.

Then we can present the following theorem.

### Theorem 3.1

Suppose (H1)-(H6) are satisfied, then there must be at least one solution of problem (1)-(3) in X.

To prove the theorem, we need the following lemmas.

### Lemma 3.1

Assume that (H1) and (H2) hold, then $$L:\operatorname {dom}L \subset X\rightarrow Y$$ is a Fredholm operator with index zero. And a linear continuous projector $$Q:Y\rightarrow Y$$ can be defined by
$$(Qu) (x)=(Q_{1}u)\Phi_{1}(x)+(Q_{2}u) \Phi_{2}(x),$$
where
\begin{aligned}& Q_{1}u=\frac{1}{e}(e_{4}T_{1}u-e_{3}T_{2}u), \quad\quad Q_{2}u=\frac{1}{e}(-e_{2}T_{1}u+e_{1}T_{2}u), \\& T_{1}u= \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dA(x),\quad\quad T_{2}u= \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dB(x). \end{aligned}
Furthermore, define a linear operator $$K_{P}:\operatorname{Im}L \rightarrow \operatorname {dom}L\cap\ker P$$ as follows:
$$(K_{P}u) (x)= \int^{1}_{0}k(x,y)u(y)\,dy+\Phi_{1}(x)T_{1}u+ \Phi_{2}(x)T_{2}u$$
such that $$K_{P}=(L|_{\operatorname {dom}L\cap\ker P })^{-1}$$.

### Proof

It follows from (H1) that
\begin{aligned}& (-1)^{n-k}\Phi_{1}^{(n)}(x)=0,\quad\quad (-1)^{n-k} \Phi_{2}^{(n)}(x)=0,\quad x\in [0,1], \\& \Phi_{1}^{(i)}(0)=\Phi_{1}^{(j)}(1)=0, \quad \quad \Phi_{2}^{(i)}(0)=\Phi_{2}^{(j)}(1)=0, \quad 1\leq i\leq k-1, 1\leq j\leq n-k-1, \\& \Phi_{1}(0)=1,\quad\quad \Phi_{1}(1)=0, \quad\quad \Phi_{2}(0)=0, \quad \Phi_{2}(1)=1. \end{aligned}
It is obvious that
\begin{aligned}& \Phi_{1}(0)= \int^{1}_{0}\Phi_{1}(x)\,dA(x), \quad\quad \Phi _{2}(1)= \int^{1}_{0}\Phi_{2}(x)\,dB(x), \\& \Phi_{1}(1)= \int^{1}_{0}\Phi_{1}(x)\,dB(x)=0, \quad \quad \Phi _{2}(0)= \int^{1}_{0}\Phi_{2}(x)\,dA(x)=0. \end{aligned}
Thus we have
$$\ker L = \bigl\{ {c_{1}\Phi_{1}(x)+c_{2} \Phi_{2}(x)}, c_{1}, c_{2}\in R \bigr\} .$$
Moreover, we can obtain that
$$\operatorname{Im}L = \biggl\{ u\in Y: \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dA(x) = \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dB(x)=0 \biggr\} .$$
On the one hand, suppose $$u\in \operatorname{Im}L$$, then there exists $$\varphi\in \operatorname {dom}L$$ such that
$$u=L\varphi\in Y.$$
Then we have
$$\varphi(x) = \int_{0}^{1}k(x,y)u(y)\,dy+\varphi(0) \Phi_{1}(x) +\varphi(1)\Phi_{2}(x).$$
Furthermore, for $$\varphi\in \operatorname {dom}L$$, then
\begin{aligned} \varphi(0)&= \int_{0}^{1}\varphi(x)\,dA(x) \\ &= \int_{0}^{1} \biggl[ \int_{0}^{1}k(x,y)u(y)\,dy +\varphi(0) \Phi_{1}(x)+\varphi(1)\Phi_{2}(x) \biggr]\,dA(x) \\ &= \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dA(x) +\varphi(0) \int^{1}_{0}\Phi_{1}(x)\,dA(x)+\varphi(1) \int^{1}_{0}\Phi_{2}(x)\,dA(x). \end{aligned}
Using this together with (H1), we can get
$$\varphi(0)= \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dA(x)+\varphi(0),$$
it means $$\int_{0}^{1}\int_{0}^{1}k(x,y)u(y)\,dy\,dA(x)=0$$. And
\begin{aligned} \varphi(1)&= \int_{0}^{1}\varphi(x)\,dB(x) \\ &= \int_{0}^{1} \biggl[ \int_{0}^{1}k(x,y)u(y)\,dy+\varphi(0) \Phi_{1}(x) +\varphi(1)\Phi_{2}(x) \biggr]\,dB(x) \\ &= \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dB(x)+\varphi(0) \int^{1}_{0}\Phi_{1}(x)\,dB(x) +\varphi(1) \int^{1}_{0}\Phi_{2}(x)\,dB(x). \end{aligned}
So we obtain that
$$\int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dB(x)=0.$$
Thus
$$\operatorname{Im}L \subset \biggl\{ u: \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dA(x) = \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dB(x)=0 \biggr\} .$$
On the other hand, if $$u\in Y$$ satisfies
$$\int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dA(x) = \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dB(x)=0,$$
we let
$$\varphi(x)= \int_{0}^{1}k(x,y)u(y)\,dy+\Phi_{1}(x)+ \Phi_{2}(x),$$
then we conclude that
$$(L\varphi) (x)=(-1)^{n-k}\varphi^{(n)}(x)=u(x),$$
$$\varphi^{(i)}(0)=\varphi^{(j)}(1)=0, \quad 1\leq i\leq k-1, 1\leq j \leq n-k-1,$$
and
\begin{aligned}& \varphi(0)= \int_{0}^{1}k(0,y)u(y)\,dy+\Phi_{1}(0)+ \Phi_{2}(0)=1, \\& \varphi(1)= \int_{0}^{1}k(1,y)u(y)\,dy+\Phi_{1}(1)+ \Phi_{2}(1)=1. \end{aligned}
Besides,
$$\int_{0}^{1}\varphi(x)\,dA(x)= \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dA(x) + \int_{0}^{1}\Phi_{1}(x)\,dA(x)+ \int_{0}^{1}\Phi_{2}(x)\,dA(x)=1,$$
and
$$\int_{0}^{1}\varphi(x)\,dB(x)= \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dB(x) + \int_{0}^{1}\Phi_{1}(x)\,dB(x)+ \int_{0}^{1}\Phi_{2}(x)\,dB(x)=1.$$
Therefore
$$\varphi(0)= \int_{0}^{1}\varphi(x)\,dA(x), \quad\quad \varphi(1)= \int_{0}^{1}\varphi(x)\,dB(x).$$
That is, $$\varphi\in \operatorname {dom}L$$, hence, $$u\in \operatorname{Im}L$$. In conclusion,
$$\operatorname{Im}L = \biggl\{ u\in Y: \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dA(x)= \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dB(x)=0 \biggr\} .$$
We define a linear operator $$P:X\rightarrow X$$ as
$$(P\varphi) (x)=\Phi_{1}(x)\varphi(0)+\Phi_{2}(x)\varphi(1),$$
then
\begin{aligned} \bigl(P^{2}\varphi \bigr) (x)&= \bigl(P(P \varphi) \bigr) (x) \\ &=\Phi_{1}(x) \bigl[(P\varphi) (0) \bigr]+\Phi_{2}(x) \bigl[(P \varphi) (1) \bigr] \\ &=\Phi_{1}(x) \bigl[\Phi_{1}(0)\varphi(0)+ \Phi_{2}(0)\varphi(1) \bigr] +\Phi_{2}(x) \bigl[ \Phi_{1}(1)\varphi(0)+\Phi_{2}(1)\varphi(1) \bigr] \\ &=\Phi_{1}(x)\varphi(0)+\Phi_{2}(x)\varphi(1). \end{aligned}
It is obvious that $$P^{2}\varphi=P\varphi$$ and $$\operatorname{Im}P =\ker L$$. For any $$\varphi\in X$$, together with $$\varphi=(\varphi-P\varphi)+P\varphi$$, we have $$X=\ker P +\ker L$$. It is easy to obtain that $$\ker L \cap\ker P =\{0\}$$, which implies
$$X=\ker P \oplus\ker L .$$
Next, an operator $$Q: Y\rightarrow Y$$ is defined as follows:
$$(Qu) (x)=(Q_{1}u)\Phi_{1}(x)+(Q_{2}u) \Phi_{2}(x),$$
where
\begin{aligned}& Q_{1}u=\frac{1}{e}(e_{4}T_{1}u-e_{3}T_{2}u),\quad\quad Q_{2}u=\frac{1}{e}(-e_{2}T_{1}u+e_{1}T_{2}u), \\& T_{1}u= \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dA(x),\quad\quad T_{2}u= \int_{0}^{1} \int_{0}^{1}k(x,y)u(y)\,dy\,dB(x). \end{aligned}
Obviously, $$e_{1}=T_{1}(\Phi_{1}(x))$$, $$e_{2}=T_{2}(\Phi_{1}(x))$$, $$e_{3}=T_{1}(\Phi_{2}(x))$$, $$e_{4}=T_{2}(\Phi_{2}(x))$$. Noting that
\begin{aligned} \bigl(Q^{2}u \bigr) (x)&= \bigl(Q_{1}(Qu) \bigr) (x)\Phi_{1}(x)+ \bigl(Q_{2}(Qu) \bigr) (x)\Phi_{2}(x) \\ &= \bigl[Q_{1} \bigl((Q_{1}u)\Phi_{1}(x)+(Q_{2}u) \Phi_{2}(x) \bigr) \bigr]\Phi_{1}(x) \\ &+ \bigl[Q_{2} \bigl((Q_{1}u)\Phi_{1}(x)+(Q_{2}u) \Phi_{2}(x) \bigr) \bigr]\Phi_{2}(x), \end{aligned}
since
\begin{aligned}& \begin{aligned} Q_{1} \bigl((Q_{1}u) \Phi_{1}(x) \bigr)&=\frac{1}{e} \bigl(e_{4}T_{1} \bigl(\Phi_{1}(x) \bigr)-e_{3}T_{2} \bigl( \Phi_{1}(x) \bigr) \bigr)Q_{1}u \\ &=\frac{1}{e}(e_{4}e_{1}-e_{3}e_{2})Q_{1}u=Q_{1}u, \end{aligned} \\& \begin{aligned} Q_{1} \bigl((Q_{2}u) \Phi_{2}(x) \bigr)&=\frac{1}{e} \bigl(e_{4}T_{1} \bigl(\Phi_{2}(x) \bigr)-e_{3}T_{2} \bigl( \Phi_{2}(x) \bigr) \bigr)Q_{2}u \\ &=\frac{1}{e}(e_{4}e_{3}-e_{3}e_{4})Q_{2}u=0, \end{aligned} \\& \begin{aligned} Q_{2} \bigl((Q_{1}u) \Phi_{1}(x) \bigr)&=\frac{1}{e} \bigl(-e_{2}T_{1} \bigl(\Phi_{1}(x) \bigr)+e_{1}T_{2} \bigl( \Phi_{1}(x) \bigr) \bigr)Q_{1}u \\ &=\frac{1}{e}(-e_{2}e_{1}+e_{1}e_{2})Q_{1}u=0, \end{aligned} \\& \begin{aligned} Q_{2} \bigl((Q_{2}u) \Phi_{2}(x) \bigr)&=\frac{1}{e} \bigl(-e_{2}T_{1} \bigl(\Phi_{2}(x) \bigr)+e_{1}T_{2} \bigl( \Phi_{2}(x) \bigr) \bigr)Q_{2}u \\ &=\frac{1}{e}(-e_{2}e_{3}+e_{1}e_{4})Q_{2}u=Q_{2}u, \end{aligned} \end{aligned}
so
$$\bigl(Q^{2}u \bigr) (x)=(Q_{1}u)\Phi_{1}(x)+(Q_{2}u) \Phi_{2}(x)=(Qu) (x).$$
And since $$u\in\ker Q$$, we have $$e_{4}T_{1}u-e_{3}T_{2}u=0$$, $$-e_{2}T_{1}u+e_{1}T_{2}u=0$$, it follows from (H2) that $$T_{1}u=T_{2}u=0$$, so $$u\in \operatorname{Im}L$$, that is, $$\ker Q \subset \operatorname{Im}L$$, and obviously, $$\operatorname{Im}L \subset\ker Q$$. So $$\ker Q =\operatorname{Im}L$$. For any $$u\in Y$$, because $$u=(u-Qu)+Qu$$, we have $$Y=\operatorname{Im}L + \operatorname{Im}Q$$. Moreover, together with $$Q^{2}u=Qu$$, we can get $$\operatorname{Im}Q \cap \operatorname{Im}L =\{0\}$$. Above all, $$Y=\operatorname{Im}L \oplus \operatorname{Im}Q$$.

To sum up, we can get that ImL is a closed subspace of Y; $$\dim\ker L = \operatorname {codim}\operatorname{Im}L <+\infty$$; that is, L is a Fredholm operator of index zero.

We now define an operator $$K_{P}:Y\rightarrow X$$ as follows:
$$(K_{P}u) (x)= \int_{0}^{1}k(x,y)u(y)\,dy+\Phi_{1}(x)T_{1}u+ \Phi_{2}(x)T_{2}u.$$
For any $$u\in \operatorname{Im}L$$, we have $$T_{1}u=0$$, $$T_{2}u=0$$. Consequently,
$$(K_{P}u) (x)= \int_{0}^{1}k(x,y)u(y)\,dy, \quad\quad (K_{P}u) (0)=0, \quad\quad (K_{P}u) (1)=0.$$
So
\begin{aligned}& (K_{P}u) (x) \in\ker P , \quad \quad (K_{P}u) (0)= \int_{0}^{1}(K_{P}u) (x)\,dA(x), \\& (K_{P}u) (1)= \int_{0}^{1}(K_{P}u) (x)\,dB(x). \end{aligned}
In addition, it is easy to know that
$$(K_{P}u)^{(i)}(0)=0, \quad 1\leq i\leq k-1;\quad\quad (K_{P}u)^{(j)}(1)=0,\quad 1\leq j\leq n-k-1,$$
then $$(K_{P}u)(x)\in \operatorname {dom}L$$. Therefore
$$K_{P}u\in \operatorname {dom}L\cap\ker P ,\quad u\in \operatorname{Im}L .$$
Next we will prove that $$K_{P}$$ is the inverse of $$L|_{\operatorname {dom}L\cap\ker P }$$. It is clear that
$$(LK_{P}u) (x)=u(x),\quad u\in \operatorname{Im}L .$$
For each $$v\in \operatorname {dom}L\cap\ker P$$, we have
\begin{aligned} (K_{P}Lv) (x)&= \int_{0}^{1}k(x,y) (-1)^{n-k}v^{(n)}(y) \,dy+ \Phi_{1}(x) \int_{0}^{1} \int_{0}^{1}k(x,y) (-1)^{n-k}v^{(n)}(y) \,dy\,dA(x) \\ &\quad{} +\Phi_{2}(x) \int_{0}^{1} \int_{0}^{1}k(x,y) (-1)^{n-k}v^{(n)}(y) \,dy\,dB(x) \\ &= v(x)-v(0)\Phi_{1}(x)-v(1)\Phi_{2}(x)+ \Phi_{1}(x) \int_{0}^{1} \bigl(v(x)-v(0)\Phi_{1}(x) \\ & \quad{} -v(1)\Phi_{2}(x) \bigr)\,dA(x)+\Phi_{2}(x) \int_{0}^{1} \bigl(v(x)-v(0)\Phi_{1}(x)-v(1) \Phi_{2}(x) \bigr)\,dB(x) \\ &= v(x)+\Phi_{1}(x) \int_{0}^{1}v(x)\,dA(x)+\Phi_{2}(x) \int_{0}^{1}v(x)\,dB(x) \\ &= v(x)+v(0)\Phi_{1}(x)+v(1)\Phi_{2}(x) \\ &= v(x). \end{aligned}
It implies that $$K_{P}Lv=v$$. So $$K_{P}=(L|_{\operatorname {dom}L\cap\ker P })^{-1}$$. Thus the lemma holds. □

### Lemma 3.2

N is L-compact on Ω̅ if $$\operatorname {dom}L\cap\overline {\Omega}\neq0$$, where Ω is a bounded open subset of X.

### Proof

We can get easily that QN is bounded. From (H3) we know that there exists $$M_{0}(x)\in L^{1}$$ such that $$\vert (I-Q)N\varphi \vert \leq M_{0}(x)$$, a.e. $$x\in[0,1]$$, $$\varphi\in\overline{\Omega}$$. Hence $$K_{P}(I-Q)N(\overline{\Omega})$$ is bounded. By the Lebesgue dominated convergence theorem and condition (H3), we can obtain that $$K_{P}(I-Q)N(\overline{\Omega})$$ is continuous. In addition, for $$\{\int_{0}^{1}k(x,y)(I-Q)N\varphi(y)\,dy +\Phi_{1}(x)\int_{0}^{1}\int_{0}^{1}k(x,y)(I-Q)N\varphi(y)\,dy\,dA(x)+ \Phi_{2}(x)\int_{0}^{1}\int_{0}^{1}k(x,y)(I-Q)N\varphi (y)\,dy\,dB(x)\}$$ is equi-continuous, by the Ascoli-Arzela theorem, we get $$K_{P}(I-Q)N:\overline{\Omega}\rightarrow X$$ is compact. Thus, N is L-compact. The proof is completed. □

### Lemma 3.3

The set $$\Omega_{1}=\{\varphi\in \operatorname {dom}L \backslash\ker L:L\varphi=\lambda N\varphi,\lambda\in[0,1]\}$$ is bounded if (H1)-(H5) are satisfied.

### Proof

Take $$\varphi\in\Omega_{1}$$, then $$N\varphi\in \operatorname{Im}L$$, thus we have
$$\int_{0}^{1} \int_{0}^{1}k(x,y)f \bigl(y,\varphi(y),\varphi '(y),\ldots,\varphi^{(n-1)}(y) \bigr) \,dy\,dA(x)=0$$
(8)
and
$$\int_{0}^{1} \int_{0}^{1}k(x,y)f \bigl(y,\varphi(y),\varphi '(y),\ldots,\varphi^{(n-1)}(y) \bigr) \,dy\,dB(x)=0.$$
(9)
By this together with (H5) we know that there exists $$x_{0}\in[0,1]$$ such that
$$\bigl\vert \varphi(x_{0}) \bigr\vert + \bigl\vert \varphi ^{(n-1)}(x_{0}) \bigr\vert \leq M.$$
And $$\varphi^{(i)}(0)=\varphi^{(j)}(1)=0$$, $$1\leq i\leq k-1$$, $$1\leq j\leq n-k-1$$, hence there exists at least a point $$\theta_{i}\in[0,1]$$ such that $$\varphi^{(i)}(\theta_{i})=0$$, $$i=1,2,\ldots,n-2$$. Thus, we get $$\varphi^{(i)}(x)=\int_{\theta_{i}}^{x}\varphi ^{(i+1)}(t)\,dt$$, $$i=1,2,\ldots,n-2$$. So,
$$\bigl\Vert \varphi^{(i)} \bigr\Vert _{\infty} \leq \bigl\Vert \varphi^{(i+1)} \bigr\Vert _{1} \leq \bigl\Vert \varphi^{(i+1)} \bigr\Vert _{\infty}, \quad i=1,2,\ldots,n-2.$$
(10)
From
$$\bigl\Vert \varphi^{(n-1)}(x) \bigr\Vert _{\infty}=\max _{x\in[0,1]} \bigl\vert \varphi^{(n-1)}(x) \bigr\vert$$
and
\begin{aligned} \varphi^{(n-1)}(x)&=\varphi^{(n-1)}(x_{0})+ \int_{x_{0}}^{x}\varphi^{(n)}(t)\,dt \\ &=\varphi^{(n-1)}(x_{0})+ \int_{x_{0}}^{x}(-1)^{n-k}f \bigl(t,\varphi(t), \varphi'(t),\ldots,\varphi^{(n-1)}(t) \bigr)\,dt , \end{aligned}
it follows from (H4) and (10) that
\begin{aligned}[b] \bigl\vert \varphi^{(n-1)}(x) \bigr\vert &\leq \bigl\vert \varphi^{(n-1)}(x_{0}) \bigr\vert + \biggl\vert \int_{x_{0}}^{x} \bigl\vert \varphi^{(n)}(t) \bigr\vert \,dt \biggr\vert \\ &\leq M+\sum_{i=1}^{n}\Vert q_{i}\Vert _{1} \bigl\Vert \varphi^{(i-1)} \bigr\Vert _{\infty}+\Vert r\Vert _{1} \\ &\leq M_{1}+c'\Vert \varphi \Vert _{\infty}+c'' \bigl\Vert \varphi^{(n-1)} \bigr\Vert _{\infty}, \end{aligned}
(11)
where $$c'=\Vert q_{1}\Vert _{1}$$, $$c''=\sum_{i=2}^{n}\Vert q_{i}\Vert _{1}$$, $$M_{1}=M+\Vert r\Vert _{1}$$.
$$\varphi(x)=\varphi(x_{0})+ \int_{x_{0}}^{x}\varphi'(t)\,dt,$$
from (10) we have
$$\Vert \varphi \Vert _{\infty}\leq M+ \bigl\Vert \varphi^{(n-1)} \bigr\Vert _{\infty}.$$
(12)
Besides, $$\Vert \varphi \Vert =\max\{\Vert \varphi \Vert _{\infty },\Vert \varphi^{(n-1)}\Vert _{\infty}\}$$. If $$\Vert \varphi \Vert _{\infty}\geq \Vert \varphi ^{(n-1)}\Vert _{\infty}$$, by (11) and (12) we have
$$\bigl\Vert \varphi^{(n-1)} \bigr\Vert _{\infty}\leq \frac{M_{1}+c'\Vert \varphi \Vert _{\infty}}{1-c''}$$
and
$$\Vert \varphi \Vert _{\infty}\leq M+\frac{M_{1}+c'\Vert \varphi \Vert _{\infty}}{1-c''},$$
so $$\Vert \varphi \Vert _{\infty} \leq\frac{1}{1-c'-c''}[(1-c'')M+M_{1}]$$.
If $$\Vert \varphi^{(n-1)}\Vert _{\infty}>\Vert \varphi \Vert _{\infty}$$, then by (11) and (12) we have
\begin{aligned} \bigl\Vert \varphi^{(n-1)} \bigr\Vert _{\infty}&\leq M_{1}+c' \bigl(M+ \bigl\Vert \varphi^{(n-1)} \bigr\Vert _{\infty} \bigr)+c'' \bigl\Vert \varphi^{(n-1)} \bigr\Vert _{\infty} \\ &\leq M_{1}+c'M+ \bigl(c'+c'' \bigr) \bigl\Vert \varphi^{(n-1)} \bigr\Vert _{\infty}, \end{aligned}
so $$\Vert \varphi^{(n-1)}\Vert _{\infty}\leq \frac{1}{1-c'-c''}(M_{1}+c'M)$$. Above all, $$\Vert \varphi \Vert \leq M_{X}$$, where
$$M_{X}=\max \biggl\{ \frac{1}{1-c'-c''} \bigl[ \bigl(1-c'' \bigr)M+M_{1} \bigr],\frac {1}{1-c'-c''} \bigl(M_{1}+c'M \bigr) \biggr\} .$$
Above all, we know $$\Omega_{1}$$ is bounded. The proof of the lemma is completed. □

### Lemma 3.4

The set $$\Omega_{2}=\{\varphi:\varphi\in\ker L ,N\varphi\in \operatorname{Im}L \}$$ is bounded if (H1)-(H3), (H6) hold.

### Proof

Let $$\varphi\in\Omega_{2}$$, then $$\varphi(x)\equiv c_{1}\Phi _{1}(x)+c_{2}\Phi_{2}(x)$$, and $$N\varphi\in \operatorname{Im}L$$, so we can get
$$c_{1} \int_{0}^{1} \int_{0}^{1}k(x,y)f \bigl(y,c_{1} \Phi_{1}(y) +c_{2}\Phi_{2}(y), \ldots,c_{1}\Phi_{1}^{(n-1)}(y)+c_{2}\Phi _{2}^{(n-1)}(y) \bigr)\,dy\,dA(x)=0$$
and
$$c_{2} \int_{0}^{1} \int_{0}^{1}k(x,y)f \bigl(y,c_{1} \Phi_{1}(y) +c_{2}\Phi_{2}(y), \ldots,c_{1}\Phi_{1}^{(n-1)}(y)+c_{2}\Phi _{2}^{(n-1)}(y) \bigr)\,dy\,dB(x)=0.$$
According to (H6), we have $$\vert c_{1}\vert \leq a$$, $$\vert c_{2}\vert \leq b$$, that is to say, $$\Omega_{2}$$ is bounded. We complete the proof. □

### Lemma 3.5

The set $$\Omega_{3}=\{\varphi\in\ker L :\lambda J\varphi+\alpha (1-\lambda)QN\varphi=0, \lambda\in[0,1]\}$$ is bounded if conditions (H1)-(H3), (H6) are satisfied, where $$J :\ker L \rightarrow \operatorname{Im}L$$ is a linear isomorphism given by $$J(c_{1}\Phi_{1}(x)+c_{2}\Phi_{2}(x))=\frac {1}{e}(e_{4}c_{1}-e_{3}c_{2})\Phi_{1}(x) +\frac{1}{e}(-e_{2}c_{1}+e_{1}c_{2})\Phi_{2}(x)$$, and
$$\alpha= \textstyle\begin{cases} -1,&\textit{if }(4)\mbox{-}(5)\textit{ hold};\\ 1,&\textit{if }(6)\mbox{-}(7)\textit{ hold}. \end{cases}$$

### Proof

Suppose that $$\varphi\in\Omega_{3}$$, we have $$\varphi(x)=c_{1}\Phi_{1}(x)+c_{2}\Phi_{2}(x)$$, and
$$\lambda c_{1}=-\alpha(1-\lambda)T_{1}N\varphi, \quad\quad \lambda c_{2}=-\alpha(1-\lambda)T_{2}N\varphi.$$
If $$\lambda=0$$, by condition (H6) we have $$\vert c_{1}\vert \leq a$$, $$\vert c_{2}\vert \leq b$$. If $$\lambda=1$$, then $$c_{1}=c_{2}=0$$. If $$\lambda\in(0,1)$$, we suppose $$\vert c_{1}\vert \geq a$$ or $$\vert c_{2}\vert \geq b$$, then
$$\lambda c_{1}^{2}=-\alpha(1-\lambda)c_{1}T_{1}N \varphi< 0$$
or
$$\lambda c_{2}^{2}=-\alpha(1-\lambda)c_{2}T_{2}N \varphi< 0,$$
which contradicts with $$\lambda c_{1}^{2}>0$$, $$\lambda c_{2}^{2}>0$$. So the lemma holds. □

Then Theorem 3.1 can be proved now.

### Proof of Theorem 3.1

Suppose that $$\Omega\supset\bigcup_{i=1}^{3}\overline{\Omega_{i}}\cup\{ 0\}$$ is a bounded open subset of X. From Lemma 3.2 we know that N is L-compact on Ω̅. In view of Lemmas 3.3 and 3.4, we can get
1. (1)

$$L\varphi\neq\lambda N\varphi$$, for every $$(\varphi,\lambda ) \in[(\operatorname {dom}L \backslash\ker L ) \cap\partial\Omega] \times(0,1)$$;

2. (2)

$$N\varphi\notin \operatorname{Im}L$$, for every $$\varphi\in\ker L \cap\partial\Omega$$.

Set $$H(\varphi,\lambda)=\lambda J\varphi+\alpha(1-\lambda)QN\varphi$$. It follows from Lemma 3.5 that $$H(\varphi,\lambda)\neq0$$ for any $$\varphi\in\partial\Omega\cap\ker L$$. So, by the homotopy of degree, we have
$$\operatorname {deg}(QN|_{\operatorname {ker}L},\Omega\cap \operatorname {ker}L,0 )=\operatorname {deg}(\alpha J,\Omega\cap \operatorname {ker}L,0 ) \neq0.$$
All the conditions of Theorem 2.1 are satisfied. So there must be at least one solution of problem (1)-(3) in X. The proof of Theorem 3.1 is completed. □

## 4 Example

We now present an example to illustrate our main theorem. Consider the following boundary-value problem:
\begin{aligned}& \varphi^{(4)}(x)=\frac{\pi}{24} \bigl\vert \varphi(x) \bigr\vert +\frac{1}{12}\sin\varphi'(x) +\frac{1}{4}\sin \varphi''(x)+\frac{1}{6}\varphi'''(x) \arctan \biggl(\frac {1}{5}\varphi'''(x) \biggr)+ x, \\& \quad x\in[0,1], \\& \varphi'(0)=\varphi'(1)=0,\quad\quad \varphi(0)=- \frac{5}{11}\varphi \biggl(\frac{1}{2} \biggr)+\frac{16}{11} \varphi \biggl(\frac{1}{4} \biggr), \\& \varphi(1)=\frac{40}{13}\varphi \biggl(\frac{1}{2} \biggr)-\frac{27}{13}\varphi \biggl( \frac{1}{3} \biggr). \end{aligned}
Obviously, $$n=4$$, $$k=2$$, and
$$A(x)= \textstyle\begin{cases} 0,&{ x\leq\frac{1}{4}};\\ \frac{16}{11},&{\frac{1}{4}< x\leq \frac{1}{2}};\\ 1,&{\frac{1}{2}< x\leq1}; \end{cases}\displaystyle \quad\quad B(x)= \textstyle\begin{cases} 0,&{ x\leq\frac{1}{3}};\\ -\frac{27}{13},&{\frac{1}{3}< x\leq \frac{1}{2}};\\ 1,&{\frac{1}{2}< x\leq1}. \end{cases}$$
Let $$\Phi_{1}(x)=2x^{3}-3x^{2}+1$$, $$\Phi_{2}(x)=-2x^{3}+3x^{2}$$, then
\begin{aligned}& \int_{0}^{1}\Phi_{1}(x)\,dA(x)=- \frac{5}{11}\Phi_{1} \biggl(\frac {1}{2} \biggr)+ \frac{16}{11}\Phi_{1} \biggl(\frac{1}{4} \biggr)=1, \\& \int_{0}^{1}\Phi_{2}(x)\,dB(x)= \frac{40}{13}\Phi_{2} \biggl(\frac {1}{2} \biggr)- \frac{27}{13}\Phi_{2} \biggl(\frac{1}{3} \biggr)=1, \end{aligned}
and
\begin{aligned}& \int_{0}^{1}\Phi_{1}(x)\,dB(x)= \frac{40}{13}\Phi_{1} \biggl(\frac {1}{2} \biggr)- \frac{27}{13}\Phi_{1} \biggl(\frac{1}{3} \biggr)=0, \\& \int_{0}^{1}\Phi_{2}(x)\,dA(x)=- \frac{5}{11}\Phi_{2} \biggl(\frac {1}{2} \biggr)+ \frac{16}{11}\Phi_{2} \biggl(\frac{1}{4} \biggr)=0, \end{aligned}
thus (H1) is satisfied. By calculation, we can obtain that $$e=\bigl \vert {\scriptsize\begin{matrix}{}e_{1}&e_{2}\cr e_{3}&e_{4} \end{matrix}} \bigr \vert \neq0$$, so (H2) holds. Let
$$f \bigl(x,\varphi,\varphi',\varphi'', \varphi''' \bigr)=\frac{\pi }{24}\vert \varphi \vert +\frac{1}{12}\sin\varphi' +\frac{1}{4} \sin\varphi''+\frac{1}{6}\varphi''' \arctan \biggl(\frac {1}{5}\varphi''' \biggr)+ x,$$
then
$$\bigl\vert f \bigl(x,\varphi,\varphi',\varphi'', \varphi''' \bigr) \bigr\vert \leq \frac{\pi}{24} \vert \varphi \vert +\frac{1}{12} \bigl\vert \varphi' \bigr\vert +\frac{1}{4} \bigl\vert \varphi'' \bigr\vert +\frac{\pi}{12} \bigl\vert \varphi''' \bigr\vert +1,$$
where
$$q_{1}=\frac{\pi}{24}, \quad\quad q_{2}=\frac{1}{12},\quad\quad q_{3}=\frac{1}{4},\quad\quad q_{4}=\frac{\pi}{12}, \quad\quad r(x)=1.$$
Taking $$M=11$$, we have $$\vert \varphi'''(x)\vert +\vert \varphi(x)\vert >11$$,
$$\textstyle\begin{cases} f(x,\varphi,\varphi',\varphi'',\varphi''')\geq\frac{\pi }{24}\cdot5-\frac{1}{12}-\frac{1}{4}>0,&\text{if } \vert \varphi(x)\vert \geq5;\\ f(x,\varphi,\varphi',\varphi'',\varphi''')\geq-\frac {1}{12}-\frac{1}{4}+\frac{1}{6}\cdot5\cdot\frac{\pi }{4}>0,&\text{if }\vert \varphi'''(x)\vert \geq5, \end{cases}$$
for $$k(x,y)>0$$,
$$\int_{0}^{1} \int_{0}^{1}k(x,y)f \bigl(y,\varphi(y), \varphi'(y),\varphi''(y), \varphi'''(y) \bigr)\,dy\,dA(x)\neq0$$
and
$$\int_{0}^{1} \int_{0}^{1}k(x,y)f \bigl(y,\varphi(y), \varphi'(y),\varphi''(y), \varphi'''(y) \bigr)\,dy\,dB(x)\neq0.$$
Hence (H5) holds. Finally, taking $$a=\frac{8}{\pi}$$, $$b=\frac{8}{\pi}$$, when $$\vert c_{1}\vert >a$$, $$\vert c_{2}\vert >b$$,
$$\textstyle\begin{cases} f(x,\varphi,\varphi',\varphi'',\varphi''')>\frac{\pi }{24}\cdot(\frac{8}{\pi}\Phi_{1}(x) +\frac{8}{\pi}\Phi_{2}(x))-\frac {1}{12}-\frac{1}{4}=0,&\text{if }c_{1}\cdot c_{2}>0 ;\\ f(x,\varphi,\varphi',\varphi'',\varphi''')>-\frac {1}{12}-\frac{1}{4} +\frac{1}{6}\cdot12(\frac{16}{\pi})\cdot\arctan(\frac{1}{5}\cdot12\cdot \frac{16}{\pi})>0,&\text{if }c_{1}\cdot c_{2}< 0, \end{cases}$$
then we obtain
$$c_{1} \int_{0}^{1} \int_{0}^{1}k(x,y)f \bigl(y,c_{1}\Phi _{1}(y)+c_{2}\Phi_{2}(y), \ldots,c_{1} \Phi_{1}'''(y)+c_{2} \Phi_{2}'''(y) \bigr)\,dy \,dA(x)>0$$
and
$$c_{2} \int_{0}^{1} \int_{0}^{1}k(x,y)f \bigl(y,c_{1}\Phi _{1}(y)+c_{2}\Phi_{2}(y), \ldots,c_{1} \Phi_{1}'''(y)+c_{2} \Phi_{2}'''(y) \bigr)\,dy \,dB(x)>0,$$
then condition (H6) is satisfied. It follows from Theorem 3.1 that there must be at least one solution in $$C^{3}[0,1]$$.

## Declarations

### Acknowledgements

This work was supported by the National Natural Science Foundation of China (11571207) and the Tai’shan Scholar Engineering Construction Fund of Shandong Province of China and SDUST graduate innovation project SDKDY-170342. 