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

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.

1 Introduction

Conjugate boundary-value problems at non-resonance have aroused considerable attention in recent years (see [1–11]), and there is also much research on boundary-value problems at resonance (see [12–21]). 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 [13] 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 [11–13], 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

[22]

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

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

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

• (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 $c'=\Vert q_{1}\Vert _{1}$, $c''=\sum_{i=2}^{n}\Vert q_{i}\Vert _{1}$, $M_{1}=M+\Vert r\Vert _{1}$.

In addition, for

$$\varphi(x)=\varphi(x_{0})+ \int_{x_{0}}^{x}\varphi'(t)\,dt, $$

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

• (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 $C^{3}[0,1]$.