# Existence of solutions for functional boundary value problems of second-order nonlinear differential equations system at resonance

## Abstract

In this paper, by using the coincidence degree theory due to Mawhin and constructing suitable operators, we study the solvability for functional boundary value problems of second-order nonlinear differential equations system at resonance with $$\dim\operatorname{Ker} L=3$$ and 4, respectively.

## Introduction

The existence of solutions for integer order differential equations with specific boundary conditions and resonance scenarios have been studied by many authors (see  and the references therein). Recently, attention has shifted to problems with linear functional conditions. The differential operator $$L:C^{1}[0,1]\rightarrow L^{1}[0,1]$$, $$Lx=x''$$ known to us is done in  for a resonant problem, where the authors studied the existence of solutions to the problem of second-order nonlinear differential equation

$$\textstyle\begin{cases} x''(t)=f(t,x(t),x'(t)), & 0< t< 1, \\ \Gamma_{1}(x)=0,\qquad\Gamma_{2}(x)=0, \end{cases}$$

which generalizes recent work on multi-point and integral boundary value problems. Although it excellently generalizes and extends many results for nonlocal second-order problems at resonance, it does not contain a complete analysis for this problem. For example, in , to see this, set $$B_{1}(t)=\alpha b, B_{1}(1)=\alpha a, B_{2}(t)=b, B_{2}(1)=a$$, where $$a,b,\alpha\in R$$ and $$a,b\neq0$$, then $$B_{1}(t)B_{2}(1)=B_{1}(1)B_{2}(t)$$ with $$\operatorname{Ker} L=\{c(at-b):c\in R\}$$, $$\operatorname{dim}\operatorname{Ker} L=1$$. This case cannot be derived from the results of  pertaining to the cases of resonance. And in , the authors also make the unnecessary artificial assumptions $$\Gamma _{1}(t^{2})\neq0,\Gamma_{1}(t^{3})\neq0$$, notably, for these assumptions, some interesting results have been obtained in  for a resonant problem that allow us to bypass above minor technical difficulty (see Lemma 1.1 below). Thus, we improve the results of  and  in that respect as well. In addition, it clearly can also be used for higher order problems with functional conditions see [18, 19]. Inspired by the above literature, we will study the existence of solutions to functional boundary value problems of differential equations system. To the best of our knowledge, this subject has not been studied. In the present paper, we investigate the following equations:

$$\textstyle\begin{cases} x''(t)=f(t,x(t),y(t),x'(t),y'(t)), & t\in[0,1],\\ y''(t)=g(t,x(t),y(t),x'(t),y'(t)), & t\in[0,1],\\ \Gamma_{1}(x)=0,\qquad\Gamma_{2}(x)=0,\\ \Gamma_{3}(y)=0,\qquad\Gamma_{4}(y)=0, \end{cases}$$
(1.1)

where $$\Gamma_{i}:C^{1}[0,1]\rightarrow\mathbb{R}$$, $$i=1,2,3,4$$, are continuous linear functionals. we will always suppose that the following condition holds:

$$(H)$$ :

Let $$f,g:[0,1]\times\mathbb{R}^{4}\rightarrow\mathbb{R}$$ satisfy Carathéodory conditions, i.e., $$f(\cdot,u)$$ and $$g(\cdot ,u)$$ are measurable for each fixed $$u\in\mathbb{R}^{4}$$, $$f(t,\cdot)$$ and $$g(t,\cdot)$$ are continuous for a.e. $$t\in[0,1]$$ and $$\sup\{ \vert f(t,x) \vert : x\in D_{0}\}, \sup\{ \vert g(t,x) \vert : x\in D_{0}\} \in L^{1}([0, 1])$$ for any compact set $$D_{0} \in\mathbb{R}^{4}$$.

### Lemma 1.1



There must exist $$h_{1}\in L^{1}[0,1]$$ such that $$(\Gamma_{1}-\alpha_{1}\Gamma_{2})(\int_{0}^{t}(t-s)h_{1}(s)\,ds)=1$$.

### Definition 1.1

We say $$(x,y)\in C^{1}[0,1]\times C^{1}[0,1]$$ is a solution of functional boundary value problems (FBVPs) (1.1) which means that $$(x,y)$$ satisfies (1.1).

## Preliminaries

We present some necessary definitions and lemmas. Consider the following conditions:

$$(A_{1})$$ :

$$\frac{\Gamma_{1}(t)}{\Gamma_{2}(t)}=\frac{\Gamma _{1}(1)}{\Gamma_{2}(1)}$$, $$\Gamma_{3}(1)=0$$, $$\Gamma_{3}(t)=0$$, $$\Gamma _{4}(1)=0$$, $$\Gamma_{4}(t)=0$$,

$$(A_{2})$$ :

$$\frac{\Gamma_{3}(t)}{\Gamma_{4}(t)}=\frac{\Gamma _{3}(1)}{\Gamma_{4}(1)}$$, $$\Gamma_{1}(1)=0$$, $$\Gamma_{1}(t)=0$$, $$\Gamma _{2}(1)=0$$, $$\Gamma_{2}(t)=0$$,

$$(A_{3})$$ :

$$\Gamma_{1}(1)=\Gamma_{1}(t)=\Gamma_{2}(1)=\Gamma _{2}(t)=0$$, $$\Gamma_{3}(1)=\Gamma_{3}(t)=\Gamma_{4}(1)=\Gamma_{4}(t)=0$$.

We should prove the following. If $$(A_{1})$$ or $$(A_{2})$$ holds, then

\begin{aligned}& \operatorname{Ker} L=\bigl\{ \bigl(c_{1}(at-b),ct+d \bigr)|c_{1},c,d\in\mathbb{R},a^{2}+b^{2}\neq0\bigr\} \quad\text{or} \\& \operatorname{Ker} L=\bigl\{ at+b,c_{2}(ct-d)|a,b,c_{2}\in \mathbb{R},c^{2}+d^{2}\neq0\bigr\} . \end{aligned}

If $$(A_{3})$$ holds, then $$\operatorname{Ker} L=\{(at+b,ct+d)| a,b,c,d\in\mathbb{R}\}$$. In fact, if exchange the places of $$\Gamma_{1}$$ and $$\Gamma_{3}$$, $$\Gamma _{2}$$ and $$\Gamma_{4}$$ in the boundary value conditions, respectively, condition $$(A_{1})$$ just becomes $$(A_{2})$$. So we only need to focus on the FBVPs (1.1) under conditions $$(A_{1})$$, $$(A_{3})$$.

As usual, we shall use the classical spaces $$C^{1}[0,1]$$ and $$L^{1}[0,1]$$. For $$(x,y)\in C^{1}[0,1]\times C^{1}[0,1]$$, we define the norm $$\Vert (x,y) \Vert =\max\{ \Vert x \Vert , \Vert y \Vert \}$$, where $$\Vert x \Vert =\max\{ \Vert x \Vert _{\infty}, \Vert x' \Vert _{\infty}\}$$, $$\Vert x \Vert _{\infty}=\max_{t\in[0,1]}|x(t)|$$. We denote the norm in $$L^{1}[0,1]$$ by $$\Vert \cdot \Vert _{1}$$. Similarly, for $$(u,v)\in L^{1}[0,1]\times L^{1}[0,1]$$, we denote the norm $$\Vert (u,v) \Vert _{1}$$ and define the norm $$\Vert (u,v) \Vert _{1}=\max\{ \Vert u \Vert _{1}, \Vert v \Vert _{1}\}$$, where $$\Vert u \Vert _{1}=\int _{0}^{1}|u(t)|\,dt,u\in L^{1}[0,1]$$. We also use the Sobolev space $$W^{2,1}(0,1)$$ defined by

$$W^{2,1}(0,1)=\bigl\{ (x,y)\in C[0,1]\times C[0,1]| x,x',y,y' \text{ are absolutely continuous on }[0,1]\bigr\} .$$

Let $$Y=C^{1}[0,1]\times C^{1}[0,1]$$ with norm $$\Vert (x,y) \Vert$$, $$Z=L^{1}[0,1]\times L^{1}[0,1]$$ with norm $$\Vert (x,y) \Vert _{1}$$. Clearly, $$Y,Z$$ are Banach spaces.

Let the linear operator $$L:\operatorname{dom} L\subset Y \rightarrow Z$$ be defined by $$L(x,y)=(x'',y'')$$, where

$$\operatorname{dom} L=\bigl\{ (x,y)\in W^{2,1}(0,1):\Gamma_{1}(x)=0, \Gamma_{2}(x)=0,\Gamma _{3}(y)=0,\Gamma_{4}(y)=0 \bigr\} .$$

Let the nonlinear operator $$N:Y\rightarrow Z$$ be defined by

$$\bigl(N(x,y)\bigr) (t)=\bigl(f\bigl(t,x(t),y(t),x'(t),y'(t) \bigr), g\bigl(t,x(t),y(t),x'(t),y'(t)\bigr)\bigr).$$

Then FBVPs (1.1) can be written as $$L(x,y)=N(x,y)$$.

### Definition 2.1

Let Y, Z be real Banach spaces, $$L:\operatorname{dom} L\subset Y\rightarrow Z$$ be a linear operator. Y is said to be the Fredholm operator of index zero provided that:

1. (i)

ImL is a closed subset of Z;

2. (ii)

$$\dim \operatorname{Ker} L=\operatorname{co}\dim \operatorname{Im} L<+\infty$$.

Let Y, Z be real Banach spaces, $$L:\operatorname{dom} L \subset Y\rightarrow Z$$ be a linear operator. L is said to be the Fredholm operator of index zero. $$P:Y\rightarrow Y$$, $$Q:Z\rightarrow Z$$ are continuous projectors such that $$\operatorname{Im}P=\operatorname{Ker}L$$, $$\operatorname{Ker}Q=\operatorname{Im}L$$, $$Y=\operatorname{Ker}L\oplus \operatorname{Ker}P$$ and $$Z=\operatorname{Im}L\oplus \operatorname{Im}Q$$. It follows that $$L|_{\operatorname{dom}L\cap \operatorname{Ker}P}:\operatorname{dom}L\cap \operatorname{Ker}P\rightarrow \operatorname{Im}L$$ is reversible. We denote the inverse of the mapping by $$K_{P}$$ (generalized inverse operator of L). If Ω is an open bounded subset of Y such that $$\operatorname{dom} L\cap\Omega\neq\emptyset$$, the mapping $$N:Y\rightarrow Z$$ will be called L-compact on Ω̅, if $$QN(\overline{\Omega})$$ and $$K_{P}(I-Q)N:\overline{\Omega }\rightarrow Y$$ are continuous and compact.

The following is the Kolmogorov-Riesz criterion (see, for example, )

### Lemma 2.1

For $$1\leq p <\infty,E\subset L^{P}[0,1]$$ is compact if

1. (a)

E is bounded;

2. (b)

the limit $$\lim_{\varepsilon \rightarrow0}\int_{0}^{1}|g(s+\varepsilon)-g(s)|^{p}\,ds=0$$ is uniform in E.

### Lemma 2.2



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

1. (i)

$$Lu \neq \lambda Nu$$ for every $$(u,\lambda)\in[(\operatorname{dom} L\setminus\operatorname{Ker} L) \cap\partial\Omega]\times(0,1)$$;

2. (ii)

$$Nu \notin \operatorname{Im} L$$ for every $$u\in\operatorname{Ker} L \cap\partial\Omega$$;

3. (iii)

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

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

Now, we give $$\operatorname{Ker} L, \operatorname{Im} L$$ and some necessary operators under conditions $$(A_{1})$$ and $$(A_{3})$$, respectively.

### Lemma 2.3

There exist $$m_{i},n_{i}\in\mathbb {N}^{+}, m_{i},n_{i}>1, m_{i}\neq n_{i},i=1,2$$ such that $$\Gamma_{1}(t^{n_{1}})\Gamma _{2}(t^{m_{1}})-\Gamma_{1}(t^{m_{1}})\Gamma_{2}(t^{n_{1}})\neq0$$, $$\Gamma _{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma_{4}(t^{n_{2}})\neq0$$.

### Proof

For convenience, assume, by way of contradiction, that $$\frac{\Gamma_{1}(t^{m_{1}})}{\Gamma_{2}(t^{m_{1}})}=\frac{\Gamma _{1}(t^{n_{1}})}{\Gamma_{2}(t^{n_{1}})}=k$$ for all $$m_{1},n_{1}\in\mathbb{N}^{*}$$, so we have

$$\Gamma_{1}\bigl(t^{m_{1}}\bigr)=k\Gamma_{2} \bigl(t^{m_{1}}\bigr)\quad \text{or}\quad \Gamma_{1}\bigl(t^{n_{1}}\bigr)=k \Gamma_{2}\bigl(t^{n_{1}}\bigr).$$

By $$(A_{2})$$, $$(\Gamma_{1}-k\Gamma_{2} )(1)= (\Gamma_{1}-k\Gamma _{2} )(t)=0$$. Thus, $$\Gamma_{1}(p(t))=k\Gamma_{2}(p(t))$$ for every polynomial p.

Since $$\Gamma_{1}(x)-k\Gamma_{2}(x)\neq0$$ on all of $$x\in C^{1}[0,1]$$, there exists $$v_{0}\in C^{1}[0,1]$$ such that $$\Gamma_{1}(v_{0})-k\Gamma_{2}(v_{0})\neq0$$. Choose a sequence of polynomials $$\{p_{m}\}$$ such that $$\Vert v_{0}-p_{m} \Vert <\frac {1}{m}$$. Then $$0\neq| (\Gamma_{1}-k\Gamma_{2} )(v_{0})|=| (\Gamma _{1}-k\Gamma_{2} )(v_{0}-p_{m})+ (\Gamma_{1}-k\Gamma_{2} )(p_{m})|=| (\Gamma_{1}-k\Gamma_{2} )(v_{0}-p_{m})| \leq \Vert (\Gamma_{1}-k\Gamma_{2} ) \Vert \Vert v_{0}-p_{m} \Vert <(\beta_{1}+|\alpha |\beta_{2})\frac{1}{m}$$ for all $$m\in\mathbb{N}$$, which is a contradiction. Similarly, for $$\Gamma_{3}$$ and $$\Gamma_{4}$$, we omit the corresponding details as straightforward. □

For convenience, we denote

$$(B_{1})$$ :

The linear functionals $$\Gamma_{1},\Gamma_{2}:Y\rightarrow\mathbb {R}$$ satisfy $$\Gamma_{2}(t)=b,\Gamma_{2}(1)=a,\Gamma_{1}(t)=\alpha_{1}b,\Gamma _{1}(1)=\alpha_{1}a$$, where $$a^{2}+b^{2}\neq0,\alpha_{1},a,b\in\mathbb{R}$$.

$$(B_{2})$$ :

The functionals $$\Gamma_{1},\Gamma_{2},\Gamma_{3},\Gamma _{4}:Y\rightarrow\mathbb{R}$$ are linear continuous with respective norms $$\beta_{1},\beta_{2},\beta_{3},\beta_{4}$$, that is, $$|\Gamma_{i}(x)|\leq\beta_{i} \Vert x \Vert , |\Gamma_{j}(y)|\leq\beta_{j} \Vert y \Vert ,i=1,2,j=3,4$$.

### Lemma 2.4

Assume $$(A_{1})$$ holds, then $$L:\operatorname{dom}L\subset Y\rightarrow Z$$ is a Fredholm mapping of index zero, $$\dim\operatorname{Ker} L=\operatorname{co}\dim \operatorname{Im}L=3$$.

### Proof

If $$(x,y)\in \operatorname{Ker}L$$ and $$L(x,y)=(x'',y'')=(0,0)$$, we have $$(x(t),y(t))=(k_{1}t+k_{2},k_{3}t+k_{4})$$.

Based on the condition $$(A_{1})$$, we have

$$x=c_{1}(at-b),\qquad y=ct+d,$$

where $$c_{1},c,d\in\mathbb{R}$$. So,

$$\operatorname{Ker}L=\bigl\{ \bigl(c_{1}(at-b),ct+d\bigr)|a^{2}+b^{2} \neq0,c_{1},c,d\in\mathbb{R}\bigr\} ,\qquad \dim \operatorname{Ker}L=3.$$

Now, we verify

\begin{aligned} \operatorname{Im} L =& \biggl\{ (u,v)\in Z:(\Gamma_{1}- \alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)u(s)\,ds \biggr)=0, \\ & {}\Gamma_{j} \biggl( \int_{0}^{t}(t-s)v(s)\,ds \biggr)=0,j=3,4\biggr\} . \end{aligned}
(2.1)

Let $$(u,v)\in \operatorname{Im}L$$, then there exists $$(x,y)\in \operatorname{dom}L$$ such that $$L(x,y)=(u,v)$$, that is,

$$\textstyle\begin{cases} x(t)= \int_{0}^{t}(t-s)u(s)\,ds+x(0)+x'(0)t,\\ y(t)= \int_{0}^{t}(t-s)v(s)\,ds+y(0)+y'(0)t, \end{cases}$$

and $$\Gamma_{i}(x)=0,\Gamma_{j}(y)=0,i=1,2,j=3,4$$. Hence,

$$\textstyle\begin{cases} \Gamma_{i}(x)=\Gamma_{i} (\int_{0}^{t}(t-s)u(s)\,ds )+\Gamma _{i}(1)x(0)+x'(0)\Gamma_{i}(t)=0, & i=1,2,\\ \Gamma_{j}(y)=\Gamma_{j} (\int_{0}^{t}(t-s)v(s)\,ds )+\Gamma _{j}(1)y(0)+y'(0)\Gamma_{j}(t)=0, & j=3,4. \end{cases}$$

Considering the resonance condition $$(B_{1})$$, we have

$$(\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)u(s)\,ds \biggr)=0,\qquad \Gamma_{j} \biggl( \int_{0}^{t}(t-s)v(s)\,ds \biggr)=0, \quad j=3,4.$$

That is,

\begin{aligned} \operatorname{Im}L \subseteq & \biggl\{ (u,v)\in Z:(\Gamma_{1}- \alpha_{1}\Gamma_{2}) \biggl( \int_{0}^{t}(t-s)u(s)\,ds \biggr)=0,\\ &{} \Gamma_{j} \biggl( \int _{0}^{t}(t-s)v(s)\,ds \biggr)=0,j=3,4\biggr\} . \end{aligned}

If

\begin{aligned} (u,v) \in& \biggl\{ (u,v)\in Z:(\Gamma_{1}-\alpha_{1} \Gamma_{2}) \biggl( \int_{0}^{t}(t-s)u(s)\,ds \biggr)=0, \\ &{}\Gamma_{j} \biggl( \int_{0}^{t}(t-s)v(s)\,ds \biggr)=0,j=3,4\biggr\} , \end{aligned}

take

$$\bigl(x(t),y(t)\bigr)= \biggl(-\frac{bt+a}{a^{2}+b^{2}}\Gamma _{2} \biggl( \int_{0}^{t}(t-s)u(s)\,ds \biggr)+ \int_{0}^{t}(t-s)u(s)\,ds, \int _{0}^{t}(t-s)v(s)\,ds \biggr).$$

It is clear that $$L(x,y)=(x'',y'')=(u,v)$$ and $$\Gamma_{i}(x)=0, \Gamma_{j}(y)=0,i=1,2,j=3,4$$.

That is, $$(u,v)\in \operatorname{Im}L$$, i.e.

\begin{aligned}& \biggl\{ (u,v)\in Z:(\Gamma_{1}-\alpha_{1} \Gamma_{2}) \biggl( \int_{0}^{t}(t-s)u(s)\,ds \biggr)=0, \\ & \quad{}\Gamma_{j} \biggl( \int _{0}^{t}(t-s)v(s)\,ds \biggr)=0,j=3,4\biggr\} \subseteq \operatorname{Im}L. \end{aligned}

Combining the above we obtain (2.1).

Define $$Q:Z\rightarrow Z$$ as follows: $$Q(u,v)=(Q_{1}u,(T_{1}v)t^{n_{2}-2}+(T_{2}v)t^{m_{2}-2})$$, where

\begin{aligned}& Q_{1}u= (\Gamma_{1}-\alpha_{1} \Gamma_{2}) \biggl( \int _{0}^{t}(t-s)u(s)\,ds \biggr)h_{1}(t), \\& T_{1}v=\frac{n_{2}(n_{2}-1) [\Gamma_{4}(t^{m_{2}})\Gamma_{3} (\int_{0}^{t}(t-s)v(s)\,ds ) -\Gamma_{3}(t^{m_{2}})\Gamma_{4} (\int_{0}^{t}(t-s)v(s)\,ds ) ]}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma _{4}(t^{n_{2}})}, \\& T_{2}v=-\frac{m_{2}(m_{2}-1) [\Gamma_{4}(t^{n_{2}}) \Gamma_{3} (\int_{0}^{t}(t-s)v(s)\,ds )-\Gamma_{3}(t^{n_{2}})\Gamma_{4} (\int_{0}^{t}(t-s)v(s)\,ds ) ]}{\Gamma_{3}(t^{n_{2}})\Gamma _{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma_{4}(t^{n_{2}})}, \end{aligned}

and $$h_{1}$$ is introduced in Lemma 1.1, $$m_{2}$$ and $$n_{2}$$ are the same as in Lemma 2.3.

By Lemma 2.3, $$(B_{1})$$, and the property of $$h_{1}$$ in Lemma 1.1, we have

\begin{aligned}& Q_{1}^{2}u=(\Gamma_{1}-\alpha_{1} \Gamma_{2}) \biggl( \int _{0}^{t}(t-s)u(s)\,ds \biggr)h_{1}(t)=Q_{1}u, \\& T_{1}\bigl((T_{1}v)t^{n_{2}-2}\bigr) \\& \quad = \frac{n_{2}(n_{2}-1) [\Gamma _{4}(t^{m_{2}})\Gamma_{3} (\int_{0}^{t}(t-s)s^{n_{2}-2}\,ds ) -\Gamma_{3}(t^{m_{2}})\Gamma_{4} (\int_{0}^{t}(t-s)s^{n_{2}-2}\,ds ) ]}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma _{4}(t^{n_{2}})}T_{1}v \\& \quad= \frac{n_{2}(n_{2}-1) [\Gamma_{4}(t^{m_{2}})\Gamma_{3} (\frac {t^{n_{2}}}{n_{2}(n_{2}-1)} ) -\Gamma_{3}(t^{m_{2}})\Gamma_{4} (\frac{t^{n_{2}}}{n_{2}(n_{2}-1)} ) ]}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma _{4}(t^{n_{2}})}T_{1}v \\& \quad =T_{1}v,\\& T_{1}\bigl((T_{2}v)t^{m_{2}-2}\bigr) \\& \quad =\frac{n_{2}(n_{2}-1) [\Gamma _{4}(t^{m_{2}})\Gamma_{3} (\int_{0}^{t}(t-s)s^{m_{2}-2}\,ds ) -\Gamma_{3}(t^{m_{2}})\Gamma_{4} (\int_{0}^{t}(t-s)s^{m_{2}-2}\,ds ) ]}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma _{4}(t^{n_{2}})}T_{2}v \\& \quad=\frac{n_{2}(n_{2}-1) [\Gamma_{4}(t^{m_{2}})\Gamma_{3} (\frac {t^{m_{2}}}{m_{2}(m_{2}-1)} ) -\Gamma_{3}(t^{m_{2}})\Gamma_{4} (\frac{t^{m_{2}}}{m_{2}(m_{2}-1)} ) ]}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma _{4}(t^{n_{2}})}T_{2}v \\& \quad=0, \\& T_{2}\bigl((T_{1}v)t^{n_{2}-2}\bigr) \\& \quad=-\frac{{m_{2}}({m_{2}}-1) [\Gamma _{4}(t^{n_{2}})\Gamma_{3} (\int_{0}^{t}(t-s)s^{n_{2}-2}\,ds ) -\Gamma_{3}(t^{n_{2}})\Gamma_{4} (\int_{0}^{t}(t-s)s^{n_{2}-2}\,ds ) ]}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma _{4}(t^{n_{2}})}T_{1}v \\& \quad=-\frac{m_{2}(m_{2}-1) [\Gamma_{4}(t^{n_{2}})\Gamma_{3} (\frac {t^{n_{2}}}{n_{2}(n_{2}-1)} ) -\Gamma_{3}(t^{n_{2}})\Gamma_{4} (\frac{t^{n_{2}}}{n_{2}(n_{2}-1)} ) ]}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma _{4}(t^{n_{2}})}T_{1}v \\& \quad=0, \\& T_{2}\bigl((T_{2}v)t^{m_{2}-2}\bigr) \\& \quad=-\frac{m_{2}(m_{2}-1) [\Gamma _{4}(t^{n_{2}})\Gamma_{3} (\int_{0}^{t}(t-s)s^{m_{2}-2}\,ds ) -\Gamma_{3}(t^{n_{2}})\Gamma_{4} (\int_{0}^{t}(t-s)s^{m_{2}-2}\,ds ) ]}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma _{4}(t^{n_{2}})}T_{2}v \\& \quad=-\frac{m_{2}(m_{2}-1) [\Gamma_{4}(t^{n_{2}})\Gamma_{3} (\frac {t^{m_{2}}}{m_{2}(m_{2}-1)} ) -\Gamma_{3}(t^{n_{2}})\Gamma_{4} (\frac{t^{m_{2}}}{m_{2}(m_{2}-1)} ) ]}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma _{4}(t^{n_{2}})}T_{2}v \\& \quad=T_{2}v. \end{aligned}

We have, for each $$(u,v)\in Z$$,

\begin{aligned} Q^{2}(u,v) = & \bigl(Q^{2}_{1}u, T_{1} \bigl[(T_{1}v)t^{n_{2}-2}+(T_{2}v)t^{m_{2}-2} \bigr]t^{n_{2}-2}+T_{2} \bigl[(T_{1}v)t^{n_{2}-2}+(T_{2}v)t^{m_{2}-2} \bigr]t^{m_{2}-2} \bigr) \\ =& \bigl(Q_{1}u,(T_{1}v)t^{n_{2}-2}+(T_{2}v)t^{m_{2}-2} \bigr) \\ =& Q(u,v). \end{aligned}

So $$Q:Z\rightarrow Z$$ is a continuous linear projector such that $$\operatorname{Im}L=\operatorname{Ker}Q$$ and $$\operatorname{Im}Q=\{(c_{1}h_{1}(t),ct^{n_{2}-2}+dt^{m_{2}-2})|c_{1},c,d\in \mathbb{R}\}$$. It is clear that $$Z=\operatorname{Im}L\oplus \operatorname{Im}Q$$ and $$\dim \operatorname{Ker}L=\operatorname{co}\dim \operatorname{Im}L=3$$, that is, L is a Fredholm mapping of index zero. □

Define an operator $$P:Y\rightarrow Y$$ as follows:

$$P(x,y) (t)= \biggl(\frac{1}{a^{2}+b^{2}}\bigl(ax'(0)-bx(0)\bigr) (at-b),y'(0)t+y(0) \biggr), \quad t\in[0,1].$$

It is easy to check that $$P^{2}(x,y)=P(x,y),(x,y)\in Y$$, it is also elementary to confirm the identity $$\operatorname{Im} P=\operatorname{Ker}L$$. So, $$Y=\operatorname{Ker}L\oplus \operatorname{Ker}P$$.

The mapping $$K_{P}:\operatorname{Im}L\rightarrow \operatorname{dom}L\cap \operatorname{Ker}P$$ defined by

$$K_{P}(u,v) (t)= \biggl(-\frac{bt+a}{a^{2}+b^{2}}\Gamma_{2} \biggl( \int _{0}^{t}(t-s)u(s)\,ds \biggr)+ \int_{0}^{t}(t-s)u(s)\,ds, \int_{0}^{t}(t-s)v(s)\,ds \biggr)$$

is the inverse of L. In fact, $$LK_{P}(u,v)=(u,v)$$ for all $$(u,v)\in \operatorname{Im}L$$. For $$(x,y)\in \operatorname{dom}L\cap \operatorname{Ker}P$$,

\begin{aligned} K_{P}L(x,y) (t) =& \bigl(K_{P}\bigl(x'',y'' \bigr)\bigr) (t) \\ =& \biggl(-\frac{bt+a}{a^{2}+b^{2}}\Gamma _{2}\biggl( \int_{0}^{t}(t-s)x''(s) \,ds\biggr)+ \int_{0}^{t}(t-s)x''(s) \,ds, \int _{0}^{t}(t-s)y''(s) \,ds \biggr) \\ =& \biggl(-\frac{bt+a}{a^{2}+b^{2}}\Gamma _{2}\bigl(x(t)-x'(0)t-x(0) \bigr)+x(t)-x'(0)t\\ &{}-x(0),y(t)-y'(0)t-y(0) \biggr) \\ =& \biggl(-\frac {bt+a}{a^{2}+b^{2}}\bigl(bx'(0)+ax(0) \bigr)+x(t)-x'(0)t-x(0),y(t)-y'(0)t-y(0) \biggr) \\ =& \biggl(x(t)-\frac {at-b}{a^{2}+b^{2}}\bigl(ax'(0)-bx(0) \bigr),y(t)-y'(0)t-y(0) \biggr) \\ =& \bigl(x(t),y(t)\bigr)-P(x,y) (t) \\ =& \bigl(x(t),y(t)\bigr). \end{aligned}

Thus, $$K_{P}=(L|_{\operatorname{dom}L\cap \operatorname{Ker}P})^{-1}$$.

### Lemma 2.5

If $$(A_{3})$$ holds, Then $$L:\operatorname{dom}L\subset Y\rightarrow Z$$ is a Fredholm mapping of index zero, $$\dim\operatorname{Ker} L=\operatorname{co}\dim \operatorname{Im}L=4$$.

### Proof

Considering $$(A_{3})$$, for every $$a,b,c,d\in\mathbb{R}$$, $$\Gamma_{i}(at+b)=a\Gamma_{i}(t)+b\Gamma_{i}(1)=0,\Gamma_{j}(ct+d)=c\Gamma _{j}(t)+d\Gamma_{j}(1)=0$$, $$i=1,2,j=3,4$$.

So it is easy to obtain

$$\operatorname{Ker}L=\bigl\{ (at+b,ct+d)|a,b,c,d\in\mathbb{R}\bigr\} ,\qquad \dim \operatorname{Ker}L=4.$$

For each $$(u,v)\in \operatorname{Im}L$$, there exists $$(x,y)\in \operatorname{dom}L$$ such that $$L(x,y)=(x'',y'')=(u,v)$$. Hence,

$$\textstyle\begin{cases} x(t)= \int_{0}^{t}(t-s)u(s)\,ds+x(0)+x'(0)t,\\ y(t)= \int_{0}^{t}(t-s)v(s)\,ds+y(0)+y'(0)t. \end{cases}$$

From the above equations, we have

\begin{aligned}& \Gamma_{i}(x)=\Gamma_{i} \biggl( \int_{0}^{t}(t-s)u(s)\,ds \biggr)=0, \\& \Gamma_{j}(y)=\Gamma_{j} \biggl( \int_{0}^{t}(t-s)v(s)\,ds \biggr)=0,\quad i=1,2,j=3,4. \end{aligned}

Therefore,

$$\operatorname{Im}L\subseteq\biggl\{ (u,v)\in Z:\Gamma_{i} \biggl( \int_{0}^{t}(t-s)u(s)\,ds \biggr)=0, \Gamma_{j} \biggl( \int_{0}^{t}(t-s)v(s)\,ds \biggr)=0,i=1,2,j=3,4 \biggr\} .$$

For each $$(u,v)\in Z$$ satisfying $$\Gamma_{i} (\int _{0}^{t}(t-s)u(s)\,ds )=0,\Gamma_{j} (\int_{0}^{t}(t-s)v(s)\,ds )=0,i=1,2,j=3,4$$, let

$$x(t)= \int_{0}^{t}(t-s)u(s)\,ds, \qquad y(t)= \int_{0}^{t}(t-s)v(s)\,ds.$$

We have $$L(x,y)=(u(t),v(t)),t\in(0,1)$$ and

\begin{aligned}& \Gamma_{i}(x)=\Gamma_{i} \biggl( \int_{0}^{t}(t-s)u(s)\,ds \biggr)=0,\quad i=1,2, \\& \Gamma_{j}(y)=\Gamma_{j} \biggl( \int_{0}^{t}(t-s)v(s)\,ds \biggr)=0,\quad j=3,4. \end{aligned}

That is, $$(u,v)\in \operatorname{Im}L$$, i.e.,

$$\biggl\{ (u,v)\in Z:\Gamma_{i} \biggl( \int_{0}^{t}(t-s)u(s)\,ds \biggr)=0, \Gamma_{j} \biggl( \int_{0}^{t}(t-s)v(s)\,ds \biggr)=0,i=1,2,j=3,4 \biggr\} \subseteq \operatorname{Im}L.$$

From the above two aspects, we have

$$\operatorname{Im}L= \biggl\{ (u,v)\in Z:\Gamma_{i} \biggl( \int_{0}^{t}(t-s)u(s)\,ds \biggr)=0,i=1,2, \Gamma_{j}\biggl( \int_{0}^{t}(t-s)v(s)\,ds\biggr)=0,j=3,4 \biggr\} .$$

By Lemma 2.3, define $$Q:Z\rightarrow Z$$ as follows:

\begin{aligned}& Q(u,v)\\& \quad= \biggl(\frac {n_{1}(n_{1}-1)[\Gamma_{2}(t^{m_{1}})\Gamma_{1} (\int_{0}^{t}(t-s)u(s)\,ds) -\Gamma_{1}(t^{m_{1}})\Gamma_{2}(\int_{0}^{t}(t-s)u(s)\,ds)]}{\Gamma _{1}(t^{n_{1}})\Gamma_{2}(t^{m_{1}})-\Gamma_{1}(t^{m_{1}})\Gamma _{2}(t^{n_{1}})}t^{{n_{1}}-2} \\& \qquad{}-\frac{{m_{1}}({m_{1}}-1)[\Gamma_{4}(t^{n_{1}}) \Gamma_{1} (\int_{0}^{t}(t-s)u(s)\,ds)-\Gamma_{1}(t^{n_{1}})\Gamma _{2}(\int_{0}^{t}(t-s)u(s)\,ds)]}{\Gamma_{1}(t^{n_{1}})\Gamma_{2}(t^{m_{1}})-\Gamma _{1}(t^{m_{1}})\Gamma_{2}(t^{n_{1}})}t^{m_{1}-2}, \\& \qquad{}\frac{n_{2}(n_{2}-1)[\Gamma_{4}(t^{m_{2}})\Gamma_{3}(\int _{0}^{t}(t-s)v(s)\,ds) -\Gamma_{3}(t^{m_{2}})\Gamma_{4} (\int_{0}^{t}(t-s)v(s)\,ds)]}{\Gamma _{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma_{3}(t^{m_{2}})\Gamma _{4}(t^{n_{2}})}t^{{n_{2}}-2} \\& \qquad{}-\frac{{m_{2}}({m_{2}}-1)[\Gamma_{4}(t^{n_{2}}) \Gamma_{3} (\int_{0}^{t}(t-s)v(s)\,ds)-\Gamma_{3}(t^{n_{2}})\Gamma _{4}(\int_{0}^{t}(t-s)v(s)\,ds)]}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma _{3}(t^{m_{2}})\Gamma_{4}(t^{n_{2}})}t^{m_{2}-2} \biggr). \end{aligned}

Similarly, we can get $$Q^{2}(u,v)=Q(u,v)$$, so $$Q:Z\rightarrow Z$$ is a well-defined projector. Now, it is obvious that $$\operatorname{Im}L=\operatorname{Ker}Q$$. Noting that Q is a linear projector, we have $$Z= \operatorname{Im}Q\oplus \operatorname{Ker}Q$$. So, $$Z=\operatorname{Im}L\oplus \operatorname{Im}Q$$ and $$\dim \operatorname{Ker}L=\dim \operatorname{Im}Q=\operatorname{co}\dim \operatorname{Im}L=4$$. So, L is a Fredholm mapping of index zero. □

Let the mapping $$P:Y\rightarrow Y$$ be defined by

$$P(x,y) (t)= \bigl(x(0)+x'(0)t,y(0)+y'(0)t \bigr),\quad t\in[0,1].$$

Noting that P is a continuous linear projector and $$\operatorname{Ker}P=\{(x,y)\in Y:x(0)=0,x'(0)=0,y(0)=0,y'(0)=0\}$$, it is easy to know that $$Y=\operatorname{Ker}L\oplus \operatorname{Ker}P$$.

The generalized inverse operator of L, $$K_{P}:\operatorname{Im}L\rightarrow \operatorname{dom}L\cap \operatorname{Ker}P$$ can be defined by

$$K_{P}(u,v) (t)= \biggl( \int_{0}^{t}(t-s)u(s)\,ds, \int_{0}^{t}(t-s)v(s)\,ds \biggr)$$

is the inverse of L. In fact, if $$(u,v)\in \operatorname{Im}L$$, then

$$LK_{P}(u,v)= \biggl( \biggl[ \int_{0}^{t}(t-s)u(s)\,ds \biggr]'', \biggl[ \int_{0}^{t}(t-s)v(s)\,ds \biggr]'' \biggr)=(u,v).$$

If $$(x,y)\in \operatorname{dom}L\cap \operatorname{Ker}P$$, then $$L(x,y)=(x'',y''), x(0)+x'(0)t=0$$ and $$y(0)+y'(0)t=0$$. We have

\begin{aligned} K_{P}L(x,y) =& \biggl( \int _{0}^{t}(t-s)x''(s) \,ds, \int_{0}^{t}(t-s)y''(s) \,ds \biggr) \\ = & \bigl(x(t)-x'(0)t-x(0),y(t)-y'(0)t-y(0) \bigr) \\ =& \bigl(x(t),y(t) \bigr). \end{aligned}

Thus, $$K_{P}=(L|_{\operatorname{dom}L\cap \operatorname{Ker}P})^{-1}$$.

## Main results

By making use of Lemmas 2.2, 2.3 and 2.4, we can obtain the following existence theorem for FBVPs (1.1) at $$\dim \operatorname{Ker}L=3$$.

### Theorem 3.1

Assume $$(A_{1})$$, $$(H)$$ and the following conditions hold:

$$(D_{1})$$. There exist constants $$M_{1}>0,M_{2}>0$$ such that, for $$(x,y)\in \operatorname{dom}L$$, if $$|x(t)|+|x'(t)|>M_{1}$$, for $$t\in[0,1]$$, then

$$(\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0,$$

if $$|y(t)|+|y'(t)|>M_{2}$$, for $$t\in[0,1]$$,

$$\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0,$$

or

$$\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0.$$

$$(D_{2})$$. There exist nonnegative functions $$a_{i},b_{i},e_{i},d_{i},\rho_{i}\in L^{1}[0,1],i=1,2$$ such that

\begin{aligned}& \bigl\vert f(t,x_{1},x_{2},y_{1},y_{2}) \bigr\vert < \rho _{1}(t)+a_{1}(t) \vert x_{1} \vert +b_{1}(t) \vert x_{2} \vert +e_{1}(t) \vert y_{1} \vert +d_{1}(t) \vert y_{2} \vert , \\& \bigl\vert g(t,x_{1},x_{2},y_{1},y_{2}) \bigr\vert \\& \quad < \rho _{2}(t)+a_{2}(t) \vert x_{1} \vert +b_{2}(t) \vert x_{2} \vert +e_{2}(t) \vert y_{1} \vert +d_{2}(t) \vert y_{2} \vert , \quad t\in [0,1],x_{i},y_{i}\in\mathbb{R},i=1,2. \end{aligned}

$$(D_{3})$$. There exist constants $$E_{i}>0,i=1,2,3$$, such that either for each $$(c_{1},b_{3},b_{4})\in\mathbb{R}^{3}$$:

$$|c_{1}|>E_{1}$$, then

$$c_{1}(\Gamma_{1}-\alpha_{1} \Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)>0,$$
(3.1)

$$|b_{3}|>E_{2}$$, then

$$b_{3}\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)>0,$$
(3.2)

$$|b_{4}|>E_{3}$$, then

$$b_{4}\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)>0,$$
(3.3)

or $$(c_{1},b_{3},b_{4})\in\mathbb{R}^{3}:|c_{1}|>E_{1}$$, then

$$c_{1}(\Gamma_{1}-\alpha_{1} \Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0,$$
(3.4)

$$|b_{3}|>E_{2}$$, then

$$b_{3}\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0,$$
(3.5)

$$|b_{4}|>E_{3}$$, then

$$b_{4}\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0.$$
(3.6)

Then FBVPs (1.1) has at least one solution in $$C^{1}[0,1]\times C^{1}[0,1]$$ provided that

$$B_{1}+\frac{C_{1}B_{2}}{1-C_{2}}< 1, \qquad C_{2}+\frac{C_{1}B_{2}}{1-B_{1}}< 1,$$

where $$B_{1}= \Vert a_{1} \Vert _{1}+ \Vert e_{1} \Vert _{1}, B_{2}= \Vert a_{2} \Vert _{1}+ \Vert e_{2} \Vert _{1}, C_{1}= \Vert b_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1},C_{2}= \Vert b_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1}$$.

The proof of Theorem 3.1 will be based on the next two lemmas.

### Lemma 3.1

Assume that $$(A_{1}),(H),(D_{1}),(D_{2})$$ and $$(D_{3})$$ hold. Then

$$\Omega_{1}=\bigl\{ (x,y)\in \operatorname{dom}L\setminus \operatorname{Ker}L:L(x,y)=\lambda N(x,y), \textit{ for some } \lambda\in[0,1] \bigr\} ,$$

and

$$\Omega_{2}=\bigl\{ (x,y)\in \operatorname{Ker}L:N(x,y)\in \operatorname{Im}L\bigr\}$$

are bounded.

### Proof

For $$(x,y)\in\Omega_{1}$$, we have $$(x,y)\notin \operatorname{Ker}L, \lambda\neq0$$ and $$N(x,y)\in \operatorname{Im}L$$.

So

\begin{aligned}& (\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)= 0, \\& \Gamma_{3} \biggl( \int _{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)=0, \end{aligned}

and

$$\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)= 0.$$

By $$(D_{1})$$, there exist constants $$t_{i}\in[0,1],i=1,2$$ such that $$|x(t_{1})|\leq M_{1},|x'(t_{1})|\leq M_{1},|y(t_{2})|\leq M_{2},|y'(t_{2})|\leq M_{2}$$.

Since $$x(t)=x(t_{1})+ \int _{t_{1}}^{t}x'(s)\,ds$$, $$y(t)=y(t_{2})+ \int_{t_{2}}^{t}y'(s)\,ds$$, we get

$$\bigl\vert x(t) \bigr\vert \leq \bigl\Vert x' \bigr\Vert _{\infty}+M_{1}, \qquad \bigl\vert y(t) \bigr\vert \leq \bigl\Vert y' \bigr\Vert _{\infty}+M_{2},\quad t\in[0,1].$$
(3.7)

Thus,

$$\bigl\Vert (x,y) \bigr\Vert \leq\max\bigl\{ \bigl\Vert x' \bigr\Vert _{\infty}, \bigl\Vert y' \bigr\Vert _{\infty}\bigr\} +\max\{M_{1},M_{2}\}.$$
(3.8)

By $$L(x,y)=\lambda N(x,y)$$, we obtain

\begin{aligned} \bigl(x'(t),y'(t) \bigr) =& \biggl(\lambda \int_{t_{1}}^{t}f \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds+x'(t_{1}),\\ &{}\lambda \int_{t_{2}}^{t}g \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds+y'(t_{2}) \biggr), \end{aligned}

thus, $$|x'(t)|< \Vert N_{1}x \Vert _{1}+M_{1},|y'(t)|< \Vert N_{2}y \Vert _{1}+M_{2}$$, where $$N(x,y)=(N_{1}x,N_{2}y)$$,

$$N_{1}x=f \bigl(s,x(s),x'(s),y(s),y'(s) \bigr), \qquad N_{2}y= g \bigl(s,x(s),x'(s),y(s),y'(s) \bigr).$$

That is, $$\max\{ \Vert x' \Vert _{\infty}, \Vert y' \Vert _{\infty}\}< \Vert N(x,y) \Vert _{1}+\max\{M_{1},M_{2}\}$$.

By $$(D_{2})$$ and (3.7), we have

\begin{aligned} \bigl\vert x'(t) \bigr\vert < & \Vert \rho_{1} \Vert _{1}+ \Vert a_{1} \Vert _{1} \Vert x \Vert _{\infty}+ \Vert b_{1} \Vert _{1} \Vert y \Vert _{\infty}+ \Vert e_{1} \Vert _{1} \bigl\Vert x' \bigr\Vert _{\infty}+ \Vert d_{1} \Vert _{1} \bigl\Vert y' \bigr\Vert _{\infty}+M_{1} \\ < & \Vert \rho_{1} \Vert _{1}+ \Vert a_{1} \Vert _{1}M_{1}+ \Vert b_{1} \Vert _{1}M_{2}+\bigl( \Vert a_{1} \Vert _{1}+ \Vert e_{1} \Vert _{1}\bigr) \bigl\Vert x' \bigr\Vert _{\infty} \\ &{}+\bigl( \Vert b_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1}\bigr) \bigl\Vert y' \bigr\Vert _{\infty}+M_{1}, \end{aligned}
(3.9)
\begin{aligned} \bigl\vert y'(t) \bigr\vert < & \Vert \rho_{2} \Vert _{1}+ \Vert a_{2} \Vert _{1}M_{1}+ \Vert b_{2} \Vert _{1}M_{2}+\bigl( \Vert a_{2} \Vert _{1}+ \Vert e_{2} \Vert _{1}\bigr) \bigl\Vert x' \bigr\Vert _{\infty} \\ &{}+\bigl( \Vert b_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1}\bigr) \bigl\Vert y' \bigr\Vert _{\infty}+M_{2}, \end{aligned}
(3.10)

for the sake of brevity, let $$A_{1}= \Vert \rho_{1} \Vert _{1}+ \Vert a_{1} \Vert _{1}M_{1}+ \Vert b_{1} \Vert _{1}M_{2}+M_{1}, A_{2}= \Vert \rho_{2} \Vert _{1}+ \Vert a_{2} \Vert _{1}M_{1}+ \Vert b_{2} \Vert _{1}M_{2}+M_{2}$$, then by (3.10) and (3.9), we have $$\Vert y' \Vert _{\infty}<\frac{A_{2}+B_{2} \Vert x' \Vert _{\infty}}{1-C_{2}}$$, $$\Vert x' \Vert _{\infty}<\frac{A_{1}+\frac{C_{1}A_{2}}{1-C_{2}}}{1-B_{1}-\frac {C_{1}B_{2}}{1-C_{2}}}$$.

Similarly, $$\Vert y' \Vert _{\infty}<\frac{A_{2}+\frac {B_{2}A_{1}}{1-B_{1}}}{1-C_{2}-\frac{C_{1}B_{2}}{1-B_{1}}}$$.

By (3.8), $$\Vert (x,y) \Vert <\infty$$. Therefore $$\Omega_{1}$$ is bounded.

For $$(x,y)\in \Omega_{2},(x,y)= (c_{1}(at-b),b_{3}t+b_{4} ),c_{1},b_{3},b_{4}\in\mathbb{R}$$ and $$N(x,y)\in \operatorname{Im}L$$. So,

$$(\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)=0$$

and

$$b_{j}\Gamma_{j} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)=0,\quad j=3,4.$$

Considering $$(D_{3}),|c_{1}|\leq E_{1},|b_{3}|\leq E_{2},|b_{4}|\leq E_{3}$$, we have $$\Vert x \Vert \leq E_{1} \Vert at-b \Vert$$, $$\Vert y \Vert \leq E_{2}+E_{3}$$. Therefore $$\Omega_{2}$$ is bounded. □

### Lemma 3.2

Assume that $$(A_{1})$$, $$(H)$$ and $$(D_{3})$$ hold. Then

$$\Omega_{3}=\bigl\{ (x,y)\in \operatorname{Ker}L:\lambda J(x,y)+(1- \lambda)QN(x,y)=0,\lambda\in [0,1]\bigr\}$$

is bounded, where $$J:\operatorname{Ker}L\rightarrow \operatorname{Im}Q$$ is homeomorphous: $$(x,y)= (c_{1}(at-b),b_{3}+b_{4}t ),c_{1},b_{3}, b_{4}\in\mathbb{R}$$,

\begin{aligned}& J(x,y)\\& \hspace{6pt}= \biggl(c_{1}h_{1},\\& \hspace{12pt}\frac{n_{2}(n_{2}-1)[\Gamma_{4}(t^{m_{2}})b_{3} -\Gamma_{3}(t^{m_{2}})b_{4}]t^{n_{2}-2}-m_{2}({m_{2}}-1)[\Gamma_{4}(t^{n_{2}})b_{3}-\Gamma _{3}(t^{n_{2}})b_{4}]t^{m_{2}-2}}{\Gamma_{3}(t^{n_{2}})\Gamma_{4}(t^{m_{2}})-\Gamma _{3}(t^{m_{2}})\Gamma_{4}(t^{n_{2}})} \biggr). \end{aligned}

### Proof

For $$(x,y)\in\Omega_{3}$$, $$\lambda J(x,y)+(1-\lambda )QN(x,y)=0$$. If $$\lambda=1$$, then $$c_{1}=0,b_{3}=0,b_{4}=0$$. That is, $$(x,y)=0$$. If $$\lambda\neq1$$, we can have

\begin{aligned}& \lambda c_{1}h_{1}=-(1-\lambda) ( \Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)h_{1}, \\& \Gamma_{4}\bigl(t^{m_{2}}\bigr) \biggl(\lambda b_{3} +(1-\lambda )\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g \bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr) \biggr) \\& \quad{}-\Gamma_{3}\bigl(t^{m_{2}}\bigr) \biggl(\lambda b_{4} +(1-\lambda)\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g \bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr) \biggr)=0, \end{aligned}
(3.11)

and

\begin{aligned}& \Gamma_{4}\bigl(t^{n_{2}}\bigr) \biggl(\lambda b_{3} +(1-\lambda )\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr) \biggr) \\& \quad {}-\Gamma_{3}\bigl(t^{n_{2}}\bigr) \biggl(\lambda b_{4} +(1-\lambda)\Gamma _{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr) \biggr)=0. \end{aligned}

From Lemma 2.3,

$$\begin{vmatrix} \Gamma_{4}\bigl(t^{m_{2}}\bigr) & \Gamma_{3}\bigl(t^{m_{2}}\bigr) \\ \Gamma_{4}\bigl(t^{n_{2}}\bigr) & \Gamma_{3} \bigl(t^{n_{2}}\bigr) \end{vmatrix} \neq0,$$

it yields

$$\textstyle\begin{cases} \lambda b_{3} +(1-\lambda)\Gamma_{3} (\int_{0}^{t}(t-s)g (s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} )\,ds )=0,\\ \lambda b_{4} +(1-\lambda)\Gamma_{4} (\int_{0}^{t}(t-s)g (s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} )\,ds )=0, \end{cases}$$

if $$|c_{1}|>E_{1},|b_{3}|>E_{2},|b_{4}|>E_{3}$$, considering above equalities, (3.11) and (3.1)-(3.3), we have

\begin{aligned}& \lambda c^{2}_{1}h_{1} =-(1-\lambda)c_{1}( \Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int_{0}^{t}(t-s)f\bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0, \\& \lambda b^{2}_{3} =-(1-\lambda)b_{3} \Gamma_{3} \biggl( \int _{0}^{t}(t-s)g \bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0, \\& \lambda b^{2}_{4} =-(1-\lambda)b_{4} \Gamma_{4} \biggl( \int _{0}^{t}(t-s)g \bigl(s,c_{1}(as-b),b_{3}s+b_{4},c_{1}a,b_{3} \bigr)\,ds \biggr)< 0. \end{aligned}

Thus $$|c_{1}|\leq E_{1},|b_{3}|\leq E_{2},|b_{4}|\leq E_{3}$$. So, $$\Omega_{3}$$ is bounded.

If (3.4)-(3.6) hold, then let

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

By the same method we can also see that $$\Omega_{3}$$ is bounded. □

### Proof of Theorem 3.1

Let Ω be a bounded open subset of Y such that $$\bigcup_{j=1}^{3}\overline{\Omega}_{j} \subset\Omega$$. The compactness of $$K_{P}(I-Q)N:\overline{\Omega}\rightarrow Y$$ and $$QN(\overline{\Omega})$$ will follow from the Arzela-Ascoli theorem and the Kolmogorov-Riesz criterion, respectively. Thus N is L-compact on Ω̅.

Then from above arguments, we have

1. (i)

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

2. (ii)

$$N(x,y) \notin\operatorname{Im} L$$, for every $$(x,y)\in\operatorname{Ker} L \cap \partial\Omega$$.

At last we will prove that (iii) of Lemma 2.2. is satisfied.

Let $$H ((x,y),\lambda )=\pm\lambda J(x,y)+(1-\lambda)QN(x,y)=0$$, noting that $$\Omega_{3}\subset\Omega$$, we know $$H ((x,y), \lambda )\neq0$$ for every $$((x,y),\lambda)\in\partial\Omega\cap \operatorname{Ker} L$$. Thus, by the homotopic property of degree

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

Then by Lemma 2.2, $$L(x,y)=N(x,y)$$ has at least one solution in $$\operatorname{dom}L\cap\overline{\Omega}$$. The proof of Theorem 3.1 is completed. □

### Theorem 3.2

Assume $$(A_{3}),(D_{2}),(H)$$ and the following conditions hold:

$$(D_{4})$$. There exist constants $$M_{3}>0,M_{4}>0$$ such that, for $$(x,y)\in \operatorname{dom}L$$, if $$|x(t)|+|x'(t)|>M_{3}$$, for $$t\in[0,1]$$, then

$$\Gamma_{1} \biggl( \int_{0}^{t}(t-s)f \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0,$$

or

$$\Gamma_{2} \biggl( \int_{0}^{t}(t-s)f \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0,$$

if $$|y(t)|+|y'(t)|>M_{4}$$, for $$t\in[0,1]$$, then

$$\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0,$$

or

$$\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g \bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0.$$

$$(D_{5})$$. There exist constants $$E_{i}>0,i=4,5$$, such that either for each $$(a_{1},a_{2},b_{3},b_{4})\in\mathbb{R}^{4}$$:

$$\vert a_{1} \vert >E_{4}$$, then

$$a_{1}\Gamma_{1} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)>0,$$
(3.12)

$$|a_{2}|>E_{5}$$, then

$$a_{2}\Gamma_{2} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)>0,$$
(3.13)

$$|b_{3}|>E_{6}$$, then

$$b_{3}\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)>0,$$
(3.14)

$$|b_{4}|>E_{7}$$, then

$$b_{4}\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)>0,$$
(3.15)

or for each $$(a_{1},a_{2},b_{3},b_{4})\in\mathbb{R}^{4}$$:

$$\vert a_{1} \vert >E_{4}$$, then

$$a_{1}\Gamma_{1} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)< 0,$$
(3.16)

$$|a_{2}|>E_{5}$$, then

$$a_{2}\Gamma_{2} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)< 0,$$
(3.17)

$$|b_{3}|>E_{6}$$, then

$$b_{3}\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)< 0,$$
(3.18)

$$|b_{4}|>E_{7}$$, then

$$b_{4}\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}s+a_{2},b_{3}s+b_{4},a_{1},b_{3} )\,ds \biggr)< 0.$$
(3.19)

Then FBVP (1.1) has at least one solution in $$C^{1}[0,1]\times C^{1}[0,1]$$ provided that

$$B_{1}+\frac{C_{1}B_{2}}{1-C_{2}}< 1,\qquad C_{2}+\frac{C_{1}B_{2}}{1-B_{1}}< 1,$$

where $$B_{1}= \Vert a_{1} \Vert _{1}+ \Vert e_{1} \Vert _{1}, B_{2}= \Vert a_{2} \Vert _{1}+ \Vert e_{2} \Vert _{1},C_{1}= \Vert b_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1},C_{2}= \Vert b_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1}$$.

The proof of Theorem 3.2 will also be based on the next two lemmas.

### Lemma 3.3

Assume that $$(A_{3}),(B_{2}),(H),(D_{2}),(D_{4})$$ and $$(D_{5})$$ hold. Then

$$\Omega_{1}=\bigl\{ (x,y)\in \operatorname{dom}L\setminus \operatorname{Ker}L:L(x,y)=\lambda N(x,y), \textit{ for some }\lambda\in[0,1]\bigr\}$$

and

$$\Omega_{2}=\bigl\{ (x,y)\in \operatorname{Ker}L:N(x,y)\in \operatorname{Im}L\bigr\}$$

are bounded.

### Proof

For $$(x,y)\in\Omega_{1}$$, we have $$(x,y)\notin \operatorname{Ker}L,\lambda\neq0$$ and $$N(x,y)\in \operatorname{Im}L$$.

So

\begin{aligned}& \Gamma_{1} \biggl( \int_{0}^{t}(t-s)f \bigl(s,x(s),x'(s),y(s),y'(s) \bigr)\,ds \biggr)=0,\\& \Gamma_{2} \biggl( \int _{0}^{t}(t-s)f \bigl(s,x(s),x'(s),y(s),y'(s) \bigr)\,ds \biggr)=0, \\& \Gamma_{3} \biggl( \int_{0}^{t}(t-s)g \bigl(s,x(s),x'(s),y(s),y'(s) \bigr)\,ds \biggr)=0, \\& \Gamma_{4} \biggl( \int _{0}^{t}(t-s)g \bigl(s,x(s),x'(s),y(s),y'(s) \bigr)\,ds \biggr)= 0. \end{aligned}

By $$(D_{4})$$, there exist constants $$t_{i}\in[0,1],i=3,4$$ such that

$$\bigl\vert x(t_{3}) \bigr\vert \leq M_{3},\qquad \bigl\vert x'(t_{3}) \bigr\vert \leq M_{3}, \qquad \bigl\vert y(t_{4}) \bigr\vert \leq M_{4},\qquad \bigl\vert y'(t_{4}) \bigr\vert \leq M_{4}.$$

Since

$$x(t)=x(t_{3})+ \int _{t_{3}}^{t}x'(s)\,ds,\qquad y(t)=y(t_{4})+ \int_{t_{4}}^{t}y'(s)\,ds,$$

we get

$$\bigl\vert x(t) \bigr\vert \leq \bigl\Vert x' \bigr\Vert _{\infty}+M_{3},\qquad \bigl\vert y(t) \bigr\vert \leq \bigl\Vert y' \bigr\Vert _{\infty}+M_{4}, \quad t\in [0,1].$$
(3.20)

Thus,

$$\bigl\Vert (x,y) \bigr\Vert \leq\max\bigl\{ \bigl\Vert x' \bigr\Vert _{\infty}, \bigl\Vert y' \bigr\Vert _{\infty}\bigr\} +\max\{M_{3},M_{4}\}.$$
(3.21)

By the proof of method in Lemma 3.1, we obtain $$\Vert x' \Vert _{\infty}<\frac{A_{3}+\frac{C_{1}A_{4}}{1-C_{2}}}{1-B_{1}-\frac {C_{1}B_{2}}{1-C_{2}}}$$, $$\Vert y' \Vert _{\infty}<\frac{A_{4}+\frac {B_{2}A_{3}}{1-B_{1}}}{1-C_{2}-\frac{C_{1}B_{2}}{1-B_{1}}}$$, where $$A_{3}= \Vert \rho_{1} \Vert _{1}+ \Vert a_{1} \Vert _{1}M_{3}+ \Vert b_{1} \Vert _{1}M_{4}+M_{3}, A_{4}= \Vert \rho_{2} \Vert _{1}+ \Vert a_{2} \Vert _{1}M_{3}+ \Vert b_{2} \Vert _{1}M_{4}+M_{4}$$, by (3.21), $$\Vert (x,y) \Vert <\infty$$. Therefore $$\Omega_{1}$$ is bounded.

For $$(x,y)\in\Omega_{2}, (x,y)(t)=(a_{1}+a_{2}t,b_{3}+b_{4}t),a_{i},b_{j}\in\mathbb {R},i=1,2,j=3,4, t\in[0,1]$$ and $$N(x,y)\in \operatorname{Im}L$$.

So

\begin{aligned}& \Gamma_{1} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr)= 0, \\& \Gamma_{2} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr)= 0, \\& \Gamma_{3} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr)=0, \end{aligned}

and

$$\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr)= 0.$$

Considering $$(D_{5})$$, $$|a_{1}|\leq E_{4},|a_{2}|\leq E_{5},|b_{3}|\leq E_{6},|b_{4}|\leq E_{7}$$, so $$\Vert x \Vert \leq E_{4}+E_{5}$$, $$\Vert y \Vert \leq E_{6}+E_{7}$$.

Therefore, $$\Omega _{2}$$ is bounded. □

### Lemma 3.4

Assume that $$(A_{3}),(B_{2}),(H)$$ and $$(D_{5})$$ hold. Then

$$\Omega_{3}=\bigl\{ (x,y)\in \operatorname{Ker}L:\lambda J(x,y)+(1- \lambda)QN(x,y)=0,\lambda\in [0,1]\bigr\}$$

is bounded, where $$J:\operatorname{Ker}L\rightarrow \operatorname{Im}Q$$ is homeomorphous: $$(x,y)(t)=(a_{1}+a_{2}t,b_{3}+b_{4}t)$$, $$a_{1},a_{2},b_{3},b_{4}\in\mathbb{R}$$,

\begin{aligned}& J(x,y) (t)\\& \hspace{6pt}= \biggl(\frac{n_{1}(n_{1}-1)[\Gamma _{2}(t^{m_{1}})a_{1}-\Gamma_{1}(t^{m_{1}})a_{2}]t^{n_{1}-2}-{m_{1}}({m_{1}}-1)[\Gamma _{2}(t^{n_{1}})a_{1}-\Gamma_{1}(t^{n_{1}})a_{2}]t^{m_{1}-2}}{\Gamma_{1}(t^{n_{1}})\Gamma _{2}(t^{m_{1}})-\Gamma_{1}(t^{m_{1}})\Gamma_{2}(t^{n_{1}})} , \\& \hspace{12pt}{}\frac{n_{2}(n_{2}-1)[\Gamma _{4}(t^{m_{2}})b_{3}-\Gamma_{3}(t^{m_{2}})b_{4}]t^{n_{2}-2}-{m_{2}}({m_{2}}-1)[\Gamma _{4}(t^{n_{2}})b_{3}-\Gamma_{3}(t^{n_{2}})b_{4}]t^{m_{2}-2}}{\Gamma_{3}(t^{n_{2}})\Gamma _{4}(t^{m_{2}}) -\Gamma_{3}(t^{m_{2}})\Gamma_{4}(t^{n_{2}})} \biggr). \end{aligned}

### Proof

For every $$(x,y)\in\Omega_{3}$$, $$\lambda J(x,y)+(1-\lambda )QN(x,y)=0$$. If $$\lambda=1$$, then $$a_{1}=0,a_{2}=0,b_{3}=0,b_{4}=0$$. That is, $$(x,y)=0$$. If $$\lambda\neq1$$, we can have

\begin{aligned}& \Gamma_{2}\bigl(t^{m_{1}}\bigr) \biggl(\lambda a_{1} +(1-\lambda)\Gamma_{1} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr) \\& \quad{}-\Gamma _{1}\bigl(t^{m_{1}}\bigr) \biggl(\lambda a_{2} +(1-\lambda)\Gamma_{2} \biggl( \int _{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr)=0, \\& \Gamma_{2}\bigl(t^{n_{1}}\bigr) \biggl(\lambda a_{1} +(1-\lambda)\Gamma_{1} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr) \\& \quad {}-\Gamma_{1}\bigl(t^{n_{1}}\bigr) \biggl(\lambda a_{2} +(1-\lambda)\Gamma _{2} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr)=0, \\& \Gamma_{4}\bigl(t^{m_{2}}\bigr) \biggl(\lambda b_{3} +(1-\lambda )\Gamma_{3} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr) \\& \quad{}-\Gamma_{3}\bigl(t^{m_{2}}\bigr) \biggl(\lambda b_{4} +(1-\lambda)\Gamma_{4} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr)=0, \end{aligned}

and

\begin{aligned}& \Gamma_{4}\bigl(t^{n_{2}}\bigr) \biggl(\lambda b_{3} +(1-\lambda )\Gamma_{3} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr) \\& \quad{}-\Gamma_{3}\bigl(t^{n_{2}}\bigr) \biggl(\lambda b_{4} +(1-\lambda)\Gamma_{4} \biggl( \int_{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds \biggr) \biggr)=0. \end{aligned}

From Lemma 2.3,

$$\begin{vmatrix} \Gamma_{2} \bigl(t^{m_{1}}\bigr) & \Gamma_{1}\bigl(t^{m_{1}}\bigr) \\ \Gamma_{2}\bigl(t^{n_{1}}\bigr) & \Gamma_{1} \bigl(t^{n_{1}}\bigr) \end{vmatrix} \neq0 \quad \mbox{and} \quad \begin{vmatrix} \Gamma_{4} \bigl(t^{m_{2}}\bigr) & \Gamma_{3}\bigl(t^{m_{2}}\bigr) \\ \Gamma_{4}\bigl(t^{n_{2}}\bigr) & \Gamma_{3} \bigl(t^{n_{2}}\bigr) \end{vmatrix} \neq0,$$

it yields

$$\textstyle\begin{cases} \lambda a^{2}_{1} +(1-\lambda)a_{1}\Gamma_{1} (\int _{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds )=0,\\ \lambda a^{2}_{2} +(1-\lambda)a_{2}\Gamma_{2} (\int _{0}^{t}(t-s)f (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds )=0,\\ \lambda b^{2}_{3} +(1-\lambda)b_{3}\Gamma_{3} (\int _{0}^{t}(t-s)g (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds )=0,\\ \lambda b^{2}_{4} +(1-\lambda)b_{4}\Gamma_{4} (\int _{0}^{t}(t-s)g (s,a_{1}+a_{2}s,b_{3}+b_{4}s,a_{2},b_{4} )\,ds )=0, \end{cases}$$

if $$|a_{1}|>E_{4},|a_{2}|>E_{5},|b_{3}|>E_{6},|b_{4}|>E_{7}$$, considering the above equalities and (3.12)-(3.15), we have $$\Vert x \Vert \leq E_{4}+E_{5}$$, $$\Vert y \Vert \leq E_{6}+E_{7}$$. So, $$\Omega_{3}$$ is bounded.

If (3.16)-(3.19) hold, then let $$\Omega_{3}=\{(x,y)\in \operatorname{Ker}L:-\lambda J(x,y)+(1-\lambda)QN(x,y)=0,\lambda\in[0,1]\}$$. Similar to the above arguments, we can show that $$\Omega_{3}$$ is bounded, too. □

### Proof of Theorem 3.2

Let Ω be a bounded open subset of Y such that $$\bigcup_{j=1}^{3}\overline{\Omega}_{j} \subset\Omega$$. The compactness of $$K_{P}(I-Q)N:\overline{\Omega}\rightarrow Y$$ and $$QN(\overline{\Omega})$$ will follow from the Arzela-Ascoli theorem and the Kolmogorov-Riesz criterion, respectively. Thus N is L-compact on Ω̅. Then from the above arguments, we have

1. (i)

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

2. (ii)

$$N(x,y) \notin\operatorname{Im} L$$, for every $$(x,y)\in\operatorname{Ker} L \cap \partial\Omega$$.

At last we will prove that (iii) of Lemma 2.2 is satisfied.

Let $$H ((x,y),\lambda )=\pm\lambda J(x,y)+(1-\lambda )QN(x,y)=0$$, noting that $$\Omega_{3}\subset\Omega$$, we know $$H ((x,y), \lambda )\neq0$$ for every $$((x,y),\lambda)\in\partial\Omega \cap \operatorname{Ker} L$$. Thus, by the homotopic property of degree

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

Then by Lemma 2.2, $$L(x,y)=N(x,y)$$ has at least one solution in $$\operatorname{dom}L\cap\overline{\Omega}$$. The proof of Theorem 3.2 is completed. □

The next lemma provides norm estimates needed for the following result.

### Lemma 3.5

For $$(u,v)\in Z, K_{P}(u,v)=(K_{P_{1}}u,K_{P_{2}}v)$$, where $$K_{P_{1}}u=-\frac{bt+a}{a^{2}+b^{2}}\Gamma_{2} (\int_{0}^{t}(t-s)u(s)\,ds )+\int_{0}^{t}(t-s)u(s)\,ds, K_{P_{2}}v=\int_{0}^{t}(t-s)v(s)\,ds$$, then

1. (1)

$$\Vert K_{P_{1}}u \Vert \leq \Vert K_{P_{1}} \Vert \Vert u \Vert _{1}$$,

2. (2)

$$\Vert K_{P_{2}}v \Vert \leq \Vert v \Vert _{1}$$,

where $$\Vert K_{P_{1}} \Vert =(\frac{ \Vert bt+a \Vert \beta_{2}}{a^{2}+b^{2}}+1)$$.

### Proof

Observe that due to $$|\Gamma_{2}(x)|\leq\beta_{2} \Vert x \Vert$$,

\begin{aligned} \vert K_{P_{1}}u \vert =& \biggl\vert -\frac{bt+a}{a^{2}+b^{2}} \Gamma_{2}\biggl( \int_{0}^{t}(t-s)u(s)\,ds\biggr)+ \int _{0}^{t}(t-s)u(s)\,ds \biggr\vert \\ \leq & \frac{|bt+a|}{a^{2}+b^{2}}\beta_{2} \biggl\Vert \biggl( \int_{0}^{t}(t-s)u(s)\,ds\biggr) \biggr\Vert + \biggl\Vert \biggl( \int _{0}^{t}(t-s)u(s)\,ds\biggr) \biggr\Vert \\ \leq & \biggl(\frac{|bt+a|}{a^{2}+b^{2}}\beta_{2}+1\biggr) \Vert u \Vert _{1} \leq \biggl(\frac{ \Vert bt+a \Vert }{a^{2}+b^{2}}\beta_{2}+1\biggr) \Vert u \Vert _{1} \end{aligned}

and $$|(K_{P_{1}}u)'(t)|\leq(\frac{|b|}{a^{2}+b^{2}}\beta_{2}+1) \Vert u \Vert _{1}$$; (1) follows from the above two inequalities. Similarly, we can obtain (2). □

### Theorem 3.3

Assume $$(A_{1})$$ with $$a\neq0$$, $$(H)$$, $$(D_{3})$$ (of Theorem  3.1) and the following conditions hold:

$$(D_{5})$$. There exist constants $$M_{1},M_{5},M_{6}>0$$ such that, for $$(x,y)\in \operatorname{dom}L$$, if $$|x'(t)|>M_{1}$$, for $$t\in[0,1]$$, then

$$(\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int _{0}^{t}(t-s)f\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0,$$

if $$|y'(t)|>M_{5}$$,

$$\Gamma_{3} \biggl( \int_{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0,$$

or if $$|y(t)|>M_{6}$$,

$$\Gamma_{4} \biggl( \int_{0}^{t}(t-s)g\bigl(s,x(s),y(s),x'(s),y'(s) \bigr)\,ds \biggr)\neq0.$$

$$(D_{6})$$. There exist nonnegative functions $$a_{i},b_{i},e_{i},d_{i},\rho_{i}\in L^{1}[0,1],i=1,2$$ such that

\begin{aligned}& \bigl\vert f(t,x_{1},x_{2},y_{1},y_{2}) \bigr\vert < \rho _{1}(t)+a_{1}(t) \vert x_{1} \vert +b_{1}(t) \vert x_{2} \vert +e_{1}(t) \vert y_{1} \vert +d_{1}(t) \vert y_{2} \vert , \\& \bigl\vert g(t,x_{1},x_{2},y_{1},y_{2}) \bigr\vert < \rho _{2}(t)+a_{2}(t) \vert x_{1} \vert +b_{2}(t) \vert x_{2} \vert \\& \phantom{\bigl\vert g(t,x_{1},x_{2},y_{1},y_{2}) \bigr\vert < }{}+e_{2}(t) \vert y_{1} \vert +d_{2}(t) \vert y_{2} \vert , \quad t\in [0,1],x_{i},y_{i}\in\mathbb{R},i=1,2, \end{aligned}

where

\begin{aligned}& \bigl( \Vert t-b/a \Vert +\bigl( \Vert t-b/a \Vert +1\bigr) \Vert K_{P_{1}} \Vert \bigr) \bigl( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1}\bigr)+\bigl( \Vert t-b/a \Vert \\& \quad {}+\bigl( \Vert t-b/a \Vert +1\bigr) \Vert K_{P_{1}} \Vert \bigr)\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})}{1-6( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1})}< 1, \\& 6\bigl( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1}\bigr)+\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})}{1-( \Vert t-b/a \Vert +( \Vert t-b/a \Vert +1) \Vert K_{P_{1}} \Vert )( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1})}< 1. \end{aligned}

Then FBVP (1.1) has at least one solution in $$C^{1}[0,1]\times C^{1}[0,1]$$.

### Proof

As in the proof of Lemma 3.1, by $$(D_{5})$$, there exist constants $$M_{i}>0, t_{i}\in[0,1],i=5,6,7$$ such that $$|x'(t_{5})|\leq M_{1},|y'(t_{6})|\leq M_{5},|y(t_{7})|\leq M_{6}$$. Since $$x'(t)=x'(t_{5})+ \int _{t_{5}}^{t}x''(s)\,ds$$, $$y'(t)=y'(t_{6})+ \int_{t_{6}}^{t}y''(s)\,ds$$, we get

$$\bigl\vert x'(t) \bigr\vert \leq \Vert N_{1}x \Vert _{1}+M_{1},\qquad \bigl\vert y'(t) \bigr\vert \leq \Vert N_{2}y \Vert _{1}+M_{5},\quad t\in [0,1],$$
(3.22)

where $$N(x,y)=(N_{1}x,N_{2}y)$$, $$N_{1}x= f (s,x(s),y(s),x'(s),y'(s) )$$, and $$N_{2}y= g (s,x(s),y(s),x'(s), y'(s) )$$. Write $$(x,y)=(x_{1},y_{1})+(x_{2},y_{2})$$, where $$(x_{1},y_{1})=(I-P)(x,y)\in {\operatorname{dom}L\cap \operatorname{Ker}P}$$ and $$(x_{2},y_{2})=P(x,y)\in \operatorname{Im} P$$.

Then since $$(x_{1},y_{1})=(I-P)(x,y)\in{\operatorname{dom}L\cap \operatorname{Ker}P}$$, $$(x_{1},y_{1})=K_{P}L(x_{1},y_{1})=K_{P}L(I-P)(x,y)=\lambda K_{P}N(x,y)$$.

As in the proof of Lemma 3.5,

$$\Vert x_{1} \Vert \leq \Vert K_{P_{1}} \Vert \Vert N_{1}x \Vert _{1}, \qquad \Vert y_{1} \Vert \leq \Vert N_{2}y \Vert _{1}.$$
(3.23)

Now, $$(x_{2},y_{2})=(x,y)-(x_{1},y_{1})$$, so $$x_{2}'=x'-x'_{1},y_{2}'=y'-y'_{1}$$ and $$|x'_{2}(t)|\leq|x'(t)|+|x'_{1}(t)|< M_{1}+( \Vert K_{P_{1}} \Vert +1) \Vert N_{1}x \Vert _{1}$$, $$|y'_{2}(t)|\leq|y'(t)|+|y'_{1}(t)|< M_{5}+2 \Vert N_{2}y \Vert _{1}$$ by (3.23). Recall that $$(x_{2},y_{2})(t)=P(x,y)(t)=(c(x)(at-b),y'(0)t+y(0))$$, where

$$c(x)=\frac{1}{a^{2}+b^{2}}\bigl(ax'(0)-bx(0)\bigr)$$

is introduced for the sake of brevity. Hence

$$\bigl\vert x'_{2}(t) \bigr\vert = \bigl\vert c(x)a \bigr\vert < M_{1}+\bigl( \Vert K_{P_{1}} \Vert +1\bigr) \Vert N_{1}x \Vert _{1}.$$

That is,

$$\bigl\vert c(x) \bigr\vert \leq\frac{1}{|a|}\bigl(M_{1}+ \bigl( \Vert K_{P_{1}} \Vert +1\bigr)' \Vert N_{1}x \Vert _{1}\bigr).$$

Thus,

$$\Vert x_{2} \Vert = \bigl\vert c(x) \bigr\vert \Vert at-b \Vert < \Vert t-b/a \Vert \bigl(M_{1}+\bigl( \Vert K_{P_{1}} \Vert +1\bigr) \Vert N_{1}x \Vert _{1} \bigr).$$
(3.24)

Similarly, it is easy to obtain $$|y'(0)|< M_{5}+2 \Vert N_{2}y \Vert _{1}$$. In addition, $$|y_{2}(t_{7})|\leq|y(t_{7})|+|y_{1}(t_{7})|\leq M_{6}+ \Vert N_{2}y \Vert _{1}$$, so, $$|y_{2}(t_{7})|=|y'(0)t_{7}+y(0)|\leq M_{6}+ \Vert N_{2}y \Vert _{1}$$ and $$|y(0)|\leq M_{5}+M_{6}+3 \Vert N_{2}y \Vert _{1}$$, thus

$$\Vert y_{2} \Vert \leq \bigl\Vert y'(0)t \bigr\Vert + \bigl\vert y(0) \bigr\vert \leq2M_{5}+M_{6}+5 \Vert N_{2}y \Vert _{1}.$$
(3.25)

By (3.23) and (3.24), $$\Vert x \Vert \leq \Vert x_{1} \Vert + \Vert x_{2} \Vert \leq C_{3}+C_{4} \Vert N_{1}x \Vert _{1}$$, where

$$C_{3}= \Vert t-b/a \Vert M_{1},\qquad C_{4}= \Vert t-b/a \Vert +\bigl( \Vert t-b/a \Vert +1\bigr) \Vert K_{P_{1}} \Vert .$$

$$\Vert y \Vert \leq \Vert y_{1} \Vert + \Vert y_{2} \Vert \leq2M_{5}+M_{6}+6 \Vert N_{2}y \Vert _{1}$$ by (3.23) and (3.25). Finally, it follows from $$(D_{6})$$ that

\begin{aligned}& \Vert x \Vert \leq\frac{C_{3}+C_{4}( \Vert \rho_{1} \Vert +( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})\frac {2M_{5}+M_{6}+6 \Vert \rho_{2} \Vert _{1}}{1-6( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1})})}{1-C_{4}( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1})-C_{4}\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})}{1-6( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1})}}, \\& \Vert y \Vert \leq\frac{2M_{5}+M_{6}+6 \Vert \rho_{2} \Vert _{1}+\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})(C_{3}+C_{4} \Vert \rho_{1} \Vert _{1})}{1-C_{4}( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1})}}{1-6( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1})-\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})}{1-C_{4}( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1})}}. \end{aligned}

Therefore $$\Omega_{1}$$ is bounded. The rest of the proof repeats that of Theorem 3.1. □

We now provide an example that satisfies the assumptions of Theorem 3.3. Consider the kind of equation system

$$\textstyle\begin{cases} x''(t)=t-1+\frac{1}{32}\sin x(t)+\frac{1}{32}\sin y(t)+\frac {1}{32}x'(t)+\frac{1}{32}\sin y'(t),\\ y''(t)=g(t,x(t),y(t),x'(t),y'(t)),\\ \Gamma_{1}(x)=x'(0)+2x(\frac{1}{2})=0, \qquad \Gamma_{2}(x)=x(0)-2 \int_{0}^{1}x(s)\,ds=0,\\ \Gamma_{3}(y)= 2\int_{0}^{\frac{1}{2}}y(s)\,ds-y(\frac {1}{2})+\frac{1}{4}y'(\frac{1}{2})=0,\\ \Gamma_{4}(y)=y'(1)-y'(\frac{1}{2})=0, \end{cases}$$

where

\begin{aligned}& g\bigl(t,x(t),y(t),x'(t),y'(t)\bigr)\\& \quad= \textstyle\begin{cases} t+\frac{1}{32}\sin x(t)+\frac{1}{32} y(0)+\frac{1}{32}\sin x'(t)+\frac {1}{32}\sin y'(t), & t\in[0,\frac{1}{2}],\\ t+\frac{1}{32}\sin x(t)+\frac{1}{32}\sin y(t)+\frac{1}{32}\sin x'(t)+\frac{1}{32}y'(t), & t\in[\frac{1}{2},1]. \end{cases}\displaystyle \end{aligned}

It is easy to see that $$\Gamma_{1}(t)=2,\Gamma_{1}(1)=2,\Gamma _{2}(t)=-1,\Gamma_{2}(1)=-1,\Gamma_{3}(t)=\Gamma_{3}(1)=\Gamma_{4}(t)=\Gamma _{1}(1)=0$$, so that $$\alpha_{1}=-2,a=b=-1$$ and $${\operatorname{Ker}L}=\{ (c_{1}(t-1),b_{3}t+b_{4})|c_{1},b_{3},b_{4}\in\mathbb{R}\}$$. It is not difficult to verify that $$h_{1}\equiv-\frac{12}{5}$$ satisfies Lemma 1.1.

Also,

$$\bigl\vert \Gamma_{2}(x) \bigr\vert \leq \bigl\vert x(0) \bigr\vert +2 \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds \leq3 \Vert x \Vert ,$$

that is, $$\beta_{2}=3,\rho_{1}=0,\rho_{2}=1, \Vert a_{1} \Vert _{1}=\frac{1}{32}, \Vert b_{1} \Vert _{1}=\frac{1}{32}, \Vert e_{1} \Vert _{1}=\frac{1}{32}, \Vert d_{1} \Vert _{1}=\frac{1}{32}, \Vert a_{2} \Vert _{1}= \Vert b_{2} \Vert _{1}= \Vert e_{2} \Vert _{1}= \Vert d_{2} \Vert _{1}=\frac{1}{32}, \Vert K_{P_{1}} \Vert =4, \Vert K_{P_{2}} \Vert =1, \Vert t-b/a \Vert =1$$,

\begin{aligned}& \bigl( \Vert t-b/a \Vert +\bigl( \Vert t-b/a \Vert +1\bigr) \Vert K_{P_{1}} \Vert \bigr) \bigl( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1}\bigr)+\bigl( \Vert t-b/a \Vert \\& \quad{}+\bigl( \Vert t-b/a \Vert +1\bigr) \Vert K_{P_{1}} \Vert \bigr)\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})}{1-6( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1})}=\frac{144}{160}< 1, \end{aligned}

and

$$6\bigl( \Vert e_{2} \Vert _{1}+ \Vert d_{2} \Vert _{1}\bigr)+\frac{6( \Vert a_{2} \Vert _{1}+ \Vert b_{2} \Vert _{1})( \Vert e_{1} \Vert _{1}+ \Vert d_{1} \Vert _{1})}{1-( \Vert t-b/a \Vert +( \Vert t-b/a \Vert +1) \Vert K_{P_{1}} \Vert )( \Vert a_{1} \Vert _{1}+ \Vert b_{1} \Vert _{1})}=\frac {3}{7}< 1.$$

Let $$M_{1}=36$$. Since $$N(x,y)=(N_{1}x,N_{2}y)$$, if $$x'(t)>36$$, then $$N_{1}x(t)>-1-\frac{3}{32}+\frac{1}{32}M_{1}>0$$, and if $$x'(t)<-36$$, then $$N_{1}x(t)<\frac{3}{32}-\frac{1}{32}M_{1}<0$$. Taking $$M_{5}=36, M_{6}=36$$, if $$y'(t)>36$$, then $$N_{2}y(t)>0$$, and if $$y'(t)<-36$$, then $$N_{2}y(t)<0$$ for $$t\in[\frac{1}{2},1]$$. And if $$y(t)>36$$, then $$N_{2}y(t)>0$$, and $$y(t)<-36$$, then $$N_{2}y<0$$ for $$t\in[0,\frac{1}{2}]$$.

Observe that

\begin{aligned}& (\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int_{0}^{t}(t-s)N_{1}x(s)\,ds\biggr)= \int _{0}^{1}\kappa(s)N_{1}(s)\,ds, \\& \Gamma_{3} \biggl( \int_{0}^{t}(t-s)N_{2}y(s)\,ds\biggr)= \int_{\frac{1}{2}}^{1} N_{2}y(s)\,ds,\\& \Gamma_{4} \biggl( \int_{0}^{t}(t-s)N_{2}y(s)\,ds\biggr)= \int _{0}^{\frac{1}{2}}s^{2}N_{2}y(s) \,ds, \end{aligned}

where

$$\kappa(s)= \textstyle\begin{cases} -1+2s-2s^{2}, & s\in[0,\frac{1}{2}],\\ -2+4s-2s^{2}, & s\in[\frac{1}{2},1]. \end{cases}$$

Obviously, $$\kappa(s)<0,s^{2}\geq0$$ in$$[0,1]$$, therefore,

\begin{aligned}& (\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int_{0}^{t}(t-s)N_{1}x(s)\,ds\biggr)\neq 0,\qquad\Gamma_{3} \biggl( \int_{0}^{t}(t-s)N_{2}y(s)\,ds\biggr) \neq0,\\& \Gamma _{4} \biggl( \int_{0}^{t}(t-s)N_{2}y(s)\,ds\biggr)\neq0 \end{aligned}

provided $$(x,y)\in \operatorname{dom} L \backslash \operatorname{Ker} L$$ satisfies $$|x'(t)|>M_{1}=36,|y'(t)|>M_{5}=36,|y(t)|>M_{6}=36$$. Hence $$(D_{5})$$ holds.

Finally, for$$(x,y)\in{\operatorname{Ker} L},x_{c_{1}}(t)=c_{1}(t-1),y_{b}(t)=b_{3}t+b_{4}$$.

Consequently,

$$c_{1}(\Gamma_{1}-\alpha_{1}\Gamma_{2}) \biggl( \int_{0}^{t}(t-s)N_{1}x(s)\,ds\biggr)= \int _{0}^{1}\kappa(s)c_{1}N_{1}x_{c_{1}}(s) \,ds>0,$$

since $$\kappa(s)<0$$ in $$[0,1]$$ and

$$c_{1}N_{1}x_{c_{1}}(t)\leq\frac{1}{32}|c_{1}|+ \frac{1}{32}|c_{1}|+\frac {1}{32}|c_{1}|- \frac{1}{32}c_{1}^{2}< 0$$

provided $$|c_{1}|>E_{1}=3$$. When $$|b_{3}|>E_{2}=35,|b_{4}|>E_{3}=35$$,

\begin{aligned}& b_{3}\Gamma_{3} \biggl( \int_{0}^{t}(t-s)N_{2}y(s)\,ds\biggr)= \int_{\frac{1}{2}}^{1}s^{2}b_{3}N_{2}y_{b}(s) \,ds>0, \\& b_{4}\Gamma_{4} \biggl( \int_{0}^{t}(t-s)N_{2}y_{b}(s) \,ds\biggr)= \int_{0}^{\frac{1}{2}}b_{4}N_{2}y_{b}(s) \,ds>0, \end{aligned}

since $$s^{2}>0$$ in $$[\frac{1}{2},1]$$, and

\begin{aligned}& b_{3}N_{2}y_{b}(t)>- \vert b_{3} \vert -\frac{3}{32} \vert b_{3} \vert +\frac{1}{32}b_{3}^{2}>0, \quad t\in\biggl[\frac {1}{2},1\biggr], \\& b_{4}N_{2}y_{b}(t)>- \vert b_{4} \vert -\frac{3}{32} \vert b_{4} \vert + \frac{1}{32}b_{4}^{2}>0,\quad t\in\biggl[0, \frac{1}{2}\biggr], \end{aligned}

then condition $$(D_{3})$$ is satisfied. It follows from Theorem 3.3 that there must be at least one solution in $$C^{1}[0,1]\times C^{1}[0,1]$$.

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## Acknowledgements

We thank both reviewers for their valuable comments. This work was supported by The Graduate Student Innovation Project Fund of Hebei Province (No. CXZZSS2017093).

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Correspondence to Weihua Jiang.

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The main idea of this paper was proposed by SB and JW. SB prepared the manuscript initially and performed all the steps of the proofs in this research. All authors read and approved the final manuscript.

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