# Positive solutions for a system of 2nth-order boundary value problems involving semipositone nonlinearities

## Abstract

In this paper we use the fixed point index to study the existence of positive solutions for a system of 2nth-order boundary value problems involving semipositone nonlinearities.

## Introduction

In this paper we investigate the existence of positive solutions for the following system of 2nth-order boundary value problems involving semipositone nonlinearities:

$$\textstyle\begin{cases} (-1)^{n} x^{(2 n)}=f_{1} (t, x, x^{\prime }, \ldots ,(-1)^{n-2}x ^{(2n-4)}, (-1)^{n-2}x^{(2n-3)}, (-1)^{n-1} x^{(2 n-2)}, \\ \hphantom{(-1)^{n} x^{(2 n)}={}}y, y^{\prime }, \ldots ,(-1)^{n-2}y^{(2n-4)}, (-1)^{n-2}y^{(2n-3)}, (-1)^{n-1} y^{(2 n-2)} ), \\ (-1)^{n} y^{(2 n)}=f_{2} (t, x, x^{\prime }, \ldots ,(-1)^{n-2}x ^{(2n-4)}, (-1)^{n-2}x^{(2n-3)}, (-1)^{n-1} x^{(2 n-2)}, \\ \hphantom{(-1)^{n} y^{(2 n)}={}}y, y^{\prime }, \ldots ,(-1)^{n-2}y^{(2n-4)}, (-1)^{n-2}y^{(2n-3)}, (-1)^{n-1} y^{(2 n-2)} ), \\ x^{(2 i)}(0)=x^{(2 i+1)}(1)=0,\qquad y^{(2 i)}(0)=y^{(2i+1)}(1)=0,\quad i=0,1, \ldots , n-1, \end{cases}$$
(1.1)

where $$n\in N$$ with $$n\ge 1$$, and $$f_{j}\in C ([0,1]\times R_{+} ^{4 n-2}, R )$$ ($$R_{+}:=[0, \infty )$$, $$R:=(-\infty ,+\infty )$$, $$j=1,2$$) satisfy the semipositone condition:

1. (H0)

there is a positive constant M such that

$$f_{j}(t,z_{1},z_{2},\ldots,z_{4n-2}) \ge -M,\quad t\in [0,1], z_{i} \in R_{+}, i=1,2,\ldots,4n-2, j=1,2.$$

In recent years, coupled systems of boundary value problems have been investigated by many authors since such systems appear naturally in many real-world situations. Some recent results on the topic can be found in a series of papers  and the references therein. In , Yang used nonnegative matrix theory to study the existence of positive solutions for the system of generalized Lidstone problems,

$$\textstyle\begin{cases} (-1)^{m} u^{(2m)} =f_{1} (t,u,-u^{\prime \prime }, \ldots ,(-1)^{m-1} u^{(2 m-2)}, v,-v^{\prime \prime }, \ldots ,(-1)^{n-1} v^{(2 n-2)} ), \\ (-1)^{n} v^{(2n)}= f_{2} (t, u,-u^{\prime \prime }, \ldots ,(-1)^{m-1} u^{(2 m-2)}, v,-v^{\prime \prime }, \ldots ,(-1)^{n-1} v^{(2 n-2)} ), \\ \alpha _{0} u^{(2 i)}(0)-\beta _{0} u^{(2 i+1)}(0) =\alpha _{1} u^{(2i)}(1)+ \beta _{1} u^{(2 i+1)}(1)=0,\quad i=0,1, \ldots , m-1, \\ \alpha _{0} v^{(2 j)}(0)-\beta _{0} v^{(2j+1)} (0)=\alpha _{1}v^{(2j)}(1)+ \beta _{1} v^{(2 j+1)}(1)=0,\quad j=0,1, \ldots , n-1, \end{cases}$$
(1.2)

where $$f_{1}, f_{2} \in C ([0,1] \times R_{+}^{m+n},R_{+} )$$, and in  Xu and Yang used some concave functions to depict the coupling behaviors for the nonlinearities $$f_{i}$$ ($$i=1,2$$), and they established the existence of positive solutions for (1.2). In , Wang and Yang used similar methods as in  to study the existence of positive solutions for the system of higher-order boundary value problems involving all derivatives of odd orders

$$\textstyle\begin{cases} (-1)^{m} w^{(2 m)} =f (t, w, w^{\prime },-w^{\prime \prime \prime }, \ldots ,(-1)^{m-1} w^{(2 m-1)}, z, z^{\prime },-z^{\prime \prime \prime },\ldots ,(-1)^{n-1} z^{(2n-1)} ), \\ (-1)^{n} z^{(2 n)} =g (t, w,w^{\prime },-w^{\prime \prime \prime }, \ldots ,(-1)^{m-1} w^{(2 m-1)},z, z^{\prime },-z^{\prime \prime \prime }, \ldots ,(-1)^{n-1} z^{(2n-1)} ), \\ w^{(2 i)}(0)=w^{(2 i+1)}(1)=0,\quad i=0,1, \ldots , m-1, \\ z^{(2 j)}(0)=z^{(2 j+1)}(1)=0,\quad j=0,1, \ldots , n-1, \end{cases}$$

where $$f, g \in C ([0,1] \times R_{+}^{m+n+2},R_{+} )$$. Moreover, they used a condition of Berstein–Nagumo type to obtain a priori estimates for $$w^{(2m-1)}$$ and $$z^{(2 n-1)}$$. For related papers, we refer the reader to . In  the authors used topological degree theory to study the existence of nontrivial solutions for the higher-order nonlinear fractional boundary value problem involving Riemann–Liouville fractional derivatives:

$$\textstyle\begin{cases} D_{0+}^{\alpha } u(t) =-f (t, u(t), D_{0+}^{\beta _{1}}u(t), D _{0+}^{\beta _{2}} u(t), \ldots , D_{0+}^{\beta _{n-1}}(t) ),\quad 0< t< 1, \\ u(0)=u^{\prime }(0)=\cdots =u^{(n-2)}(0)=D_{0+}^{\beta } u(1)=0, \end{cases}$$

where $$D_{0+}^{\alpha }$$, $$D_{0+}^{\beta }$$, $$D_{0+}^{\beta _{i}}$$ are the Riemann–Liouville fractional derivatives, and $$f\in C([0,1]\times R ^{n}, R)$$.

Motivated by the above work, in this paper we investigate the positive solutions for the system of 2nth-order boundary value problems (1.1) involving semipositone nonlinearities. We first use the method of order reduction to transform (1.1) into an equivalent system of integro-integral equations, and then we establish a system of nonnegative operator equations. Using the fixed point index and nonnegative matrix theory, we study the existence of positive fixed points for the operator equations, and obtain positive solutions for (1.1).

## Preliminaries

Let $$E=C[0,1]$$, $$\|z\|=\max_{t\in [0,1]}|z(t)|$$, $$P=\{t\in [0,1]: z(t) \ge 0, \forall t\in [0,1]\}$$. Then $$(E,\|\cdot \|)$$ is a Banach space, and P a cone on E. Let

$$k_{1}(t, s) :=\min \{t, s\},\qquad k_{i}(t, s) := \int _{0}^{1} k_{i-1}(t, \tau ) k_{1}(\tau , s) \,d\tau ,\quad t, s \in [0,1], i=2,3,\ldots,n,$$

and

$$(B_{i} z ) (t) := \int _{0}^{1} k_{i}(t, s) z(s) \,ds, \qquad h_{i}(t, s) :=\partial k_{i}(t, s) / \partial t,\quad i=1,2, \ldots ,n-1.$$

Note

$$\bigl( (B_{i} z ) (t) \bigr)^{\prime } := \int _{0}^{1} h _{i}(t, s) z(s)\,ds,\quad i=1,2, \ldots ,$$

and $$B_{i},B_{i}':E\to E$$ are completely continuous linear operators, $$B_{i}$$, $$B_{i}'$$ are also positive operators, i.e., they will map P into P.

### Lemma 2.1

()

Let $$\kappa _{\psi }=1-2/e$$, and $$\psi (t)=te^{t}$$, $$t\in [0,1]$$. Then we have

$$\kappa _{\psi } \psi (s) \leq \int _{0}^{1} k_{1}(t, s) \psi (t) \,dt \leq \psi (s).$$

### Lemma 2.2

()

Let $$z\in P$$. Then we have

$$\int _{0}^{1} \Biggl[ (B_{n-1} z ) (t)+2\sum_{i=0}^{n-2} \bigl( (B _{n-1-i} z ) (t) \bigr)^{\prime } \Biggr] \psi (t) \,dt= \int _{0} ^{1} z(t)\psi (t) \,dt.$$

### Lemma 2.3

()

LetEbe a real Banach space andPa cone onE. Suppose that $$\varOmega \subset E$$is a bounded open set and that $$A:\overline{ \varOmega }\cap P\to P$$is a continuous compact operator. If there exists a $$\omega _{0}\in P\setminus \{0\}$$such that

$$\omega -A\omega \neq \lambda \omega _{0},\quad \forall \lambda \geq 0, \omega \in \partial \varOmega \cap P,$$

then $$i(A,\varOmega \cap P,P)=0$$, whereidenotes the fixed point index onP.

### Lemma 2.4

()

LetEbe a real Banach space andPa cone onE. Suppose that $$\varOmega \subset E$$is a bounded open set with $$0\in \varOmega$$and that $$A:\overline{\varOmega }\cap P\to P$$is a continuous compact operator. If

$$\omega -\lambda A\omega \neq 0,\quad \forall \lambda \in [0,1], \omega \in \partial \varOmega \cap P,$$

then $$i(A,\varOmega \cap P,P)=1$$.

Now, we consider the following auxiliary problem associated with (1.1):

$$\textstyle\begin{cases} (-1)^{n} x^{(2 n)}=f (t, x, x^{\prime }, \ldots ,(-1)^{n-2}x^{(2n-4)}, (-1)^{n-2}x^{(2n-3)}, (-1)^{n-1} x^{(2 n-2)} ), \\ x^{(2 i)}(0)=x^{(2 i+1)}(1)=0,\quad i=0,1, \ldots ,n-1, \end{cases}$$

where $$f\in C ([0,1]\times R_{+}^{2 n-1}, R )$$ satisfies the condition:

(H0)′:

there is a positive constant M such that

$$f(t,z_{1},z_{2},\ldots,z_{2n-1})\ge -M, \quad t\in [0,1], z_{i}\in R _{+}, i=1,2,\ldots,2n-1.$$

Let $$(-1)^{n-1} x^{(2 n-2)}(t)=z(t)$$, $$t\in [0,1]$$. Then we have

$$\textstyle\begin{cases} -z''(t)=f (t, (B_{n-1} z )(t), ( (B_{n-1} z )(t) ) ^{\prime }, \ldots , (B_{1} z )(t), ( (B_{1} z )(t) ) ^{\prime }, z(t) ), \\ z(0)=z^{\prime }(1)=0, \end{cases}$$
(2.1)

which can be expressed in the integral form

$$z(t)= \int _{0}^{1} k_{1}(t,s) f \bigl(s, (B_{n-1}z ) (s), \bigl( (B_{n-1} z ) (s) \bigr)^{\prime }, \ldots , (B _{1} z ) (s), \bigl( (B_{1}z ) (s) \bigr)^{\prime }, z(s) \bigr) \,ds.$$
(2.2)

For convenience, let

$$(A_{i} z) (t)= \bigl( (B_{i} z ) (t) \bigr)^{\prime }, \quad t\in [0,1], i=1,2,\ldots ,n-1.$$

As a result, we can also write (2.2) in the form

$$z(t)= \int _{0}^{1} k_{1}(t,s) f \bigl(s, (B_{n-1} z ) (s), (A_{n-1} z ) (s), \ldots , (B_{1} z ) (s), (A_{1} z ) (s), z(s) \bigr) \,ds.$$

Let $$w(t)=M\int _{0}^{1}k_{1}(t,s)\,ds=M(t-t^{2}/2)$$. We need to consider the following problem:

$$\textstyle\begin{cases} -z''(t)=\widetilde{f} (t, (B_{n-1}(z-w) )(t), (A _{n-1} (z-w) )(t), \ldots , \\ \hphantom{-z''(t)={}} (B_{1}(z-w) )(t), (A_{1} (z-w) )(t), (z-w)(t) ), \\ z(0)=z^{\prime }(1)=0, \end{cases}$$
(2.3)

where

$$\widetilde{f}(t,z_{1},\ldots,z_{2n-1})= \textstyle\begin{cases} f(t,z_{1},\ldots,z_{2n-1})+M, &t\in [0,1], z_{i}\ge 0, i=1,2,\ldots,2n-1, \\ f(t,0,\ldots,0)+M,& t\in [0,1], \mbox{for else cases}. \end{cases}$$

Note that (2.3) can be expressed in the integral form

\begin{aligned} z(t) =& \int _{0}^{1} k_{1}(t,s) \widetilde{f} \bigl(s, \bigl(B_{n-1}(z-w) \bigr) (s), \bigl(A_{n-1} (z-w) \bigr) (s), \ldots , \\ &\bigl(B_{1}(z-w) \bigr) (s), \bigl(A_{1} (z-w) \bigr) (s), (z-w) (s) \bigr) \,ds. \end{aligned}

Using (H0)′, we see that $$\widetilde{f}\in C ([0,1] \times R _{+}^{2 n-1}, R_{+} )$$.

### Lemma 2.5

1. (i)

If $$z^{*}$$is a positive solution of (2.1), then $$z^{*}+w$$is a positive solution of (2.3).

2. (ii)

If $$z^{**}$$is a positive solution of (2.3), and greater thanw, then $$z^{**}-w$$is a positive solution of (2.1).

### Proof

Substituting $$z^{*}+w$$ into (2.3), we have

$$\textstyle\begin{cases} -{z^{*}}''(t)-w''(t)=\widetilde{f} (t, (B_{n-1}(z^{*}+w-w) )(t), (A_{n-1} (z^{*}+w-w) )(t), \ldots , \\ \hphantom{-{z^{*}}''(t)-w''(t)={}} (B_{1} (z^{*}+w-w) )(t), (A_{1} (z^{*}+w-w) )(t), (z^{*}+w-w)(t) ), \\ (z^{*}+w)(0)=(z^{*}+w)^{\prime }(1)=0. \end{cases}$$
(2.4)

Note that w satisfies the boundary value problem

$$\textstyle\begin{cases} -z''(t)=M, \\ z(0)=z^{\prime }(1)=0. \end{cases}$$

By virtue of (2.4), we have

$$\textstyle\begin{cases} -{z^{*}}''(t)-w''(t)=f (t, (B_{n-1} z^{*} )(t), (A _{n-1}z^{*} )(t), \ldots , (B_{1} z^{*} )(t), (A _{1} z^{*} )(t), z^{*}(t) )+M, \\ z^{*}(0)={z^{*}}^{\prime }(1)=0, \end{cases}$$

which is (2.1).

On the other hand, we substitute $$z^{**}-w$$ into (2.1), and obtain

$$\textstyle\begin{cases} -{z^{**}}''(t)+w''(t)=f (t, (B_{n-1} (z^{**}-w) )(t), (A_{n-1} (z^{**}-w) )(t), \ldots , \\ \hphantom{-{z^{**}}''(t)+w''(t)={}} (B_{1} (z^{**}-w) )(t), (A_{1} (z^{**}-w) )(t), (z^{**}-w)(t) ), \\ (z^{**}-w)(0)=(z^{**}-w)^{\prime }(1)=0. \end{cases}$$

Note that, from the definitions of w and , we have

$$\textstyle\begin{cases} -{z^{**}}''(t)=\widetilde{f} (t, (B_{n-1} (z^{**}-w) )(t), (A_{n-1} (z^{**}-w) )(t), \ldots , \\ \hphantom{-{z^{**}}''(t)={}} (B_{1} (z^{**}-w) )(t), (A_{1} (z^{**}-w) )(t), (z^{**}-w)(t) ), \\ z^{**}(0)={z^{**}}^{\prime }(1)=0, \end{cases}$$

which is (2.3). This completes the proof. □

From Lemma 2.5, if we wish to seek the positive solutions for (2.1), we only need to study the positive solutions for (2.3), which are greater than w. Consequently, we define an operator $$T:P\to E$$ as follows:

\begin{aligned} (Tz) (t) =& \int _{0}^{1} k_{1}(t,s) \widetilde{f} \bigl(s, \bigl(B_{n-1} (z-w) \bigr) (s), \bigl(A_{n-1} (z-w) \bigr) (s), \ldots , \\ &\bigl(B_{1} (z-w) \bigr) (s), \bigl(A_{1} (z-w) \bigr) (s), (z-w) (s) \bigr) \,ds. \end{aligned}

Then T is a completely continuous operator, and if there exists a $$\overline{z}\in P$$ with $$\overline{z}\ge w$$ such that $$T\overline{z}= \overline{z}$$, we see that $$\overline{z}-w$$ is a positive solution of (2.1).

Let

$$P_{0}= \bigl\{ z\in P: z(t)\ge t \Vert z \Vert , \forall t\in [0,1] \bigr\} .$$

Then $$P_{0}$$ is also a cone on E, and we have the following lemma.

### Lemma 2.6

$$T(P)\subset P_{0}$$.

Note that, for $$t,s\in [0,1]$$, $$tk_{1}(s,s)\le k_{1}(t,s)\le k_{1}(s,s)$$ and $$k_{1}(s,s)=s$$, so we can easily obtain this lemma (the details are omitted).

From Lemmas 2.5 and 2.6, we have $$\overline{z}\in P_{0}$$ if is a fixed point of T. Consequently, if $$\| \overline{z}\|\ge M$$ we have

$$\overline{z}(t)-w(t)\ge t \Vert \overline{z} \Vert - M \bigl(t-t^{2}/2 \bigr) \ge t \Vert \overline{z} \Vert - t M\ge 0.$$

Hence, we only need to seek T’s positive fixed point with $$\|\overline{z}\|\ge M$$, and then $$\overline{z}-w$$ is a positive solution of (2.1).

## Main results

In (1.1), let $$(-1)^{n-1} x^{(2 n-2)}=u$$ and $$(-1)^{n-1} y^{(2n-2)}=v$$, then we obtain the following system of boundary value problems:

$$\textstyle\begin{cases} -u''(t)= f_{1} (t, (B_{n-1}u)(t), (A_{n-1}u)(t), \ldots ,(B_{1}u)(t), (A_{1}u)(t), u(t), \\ \hphantom{-u''(t)={}} (B_{n-1}v)(t), (A_{n-1}v)(t), \ldots ,(B_{1}v)(t), (A_{1}v)(t), v(t) ), \\ -v''(t)= f_{2} (t, (B_{n-1}u)(t), (A_{n-1}u)(t), \ldots ,(B_{1}u)(t), (A_{1}u)(t), u(t), \\ \hphantom{-v''(t)={}}(B_{n-1}v)(t), (A_{n-1}v)(t), \ldots ,(B_{1}v)(t), (A_{1}v)(t), v(t) ), \\ u(0)=u'(1)=0, \qquad v(0)=v'(1)=0, \end{cases}$$
(3.1)

which has the integral form

$$\textstyle\begin{cases} u(t)=\int _{0}^{1} k_{1}(t,s) f_{1} (s, (B_{n-1}u)(s), (A_{n-1}u)(s), \ldots ,(B_{1}u)(s), (A_{1}u)(s), u(s), \\ \hphantom{u(t)={}} (B_{n-1}v)(s), (A_{n-1}v)(s), \ldots ,(B_{1}v)(s), (A _{1}v)(s), v(s) )\,ds, \\ v(t)= \int _{0}^{1} k_{1}(t,s) f_{2} (s, (B_{n-1}u)(s), (A_{n-1}u)(s), \ldots ,(B_{1}u)(s), (A_{1}u)(s), u(s), \\ \hphantom{v(t)={}} (B_{n-1}v)(s), (A_{n-1}v)(s), \ldots ,(B_{1}v)(s), (A _{1}v)(s), v(s) )\,ds. \end{cases}$$

For $$j=1,2$$, let

\begin{aligned}& F_{j}(t,z_{1},z_{2},\ldots,z_{4n-2}) \\& \quad = \textstyle\begin{cases} f_{j}(t,z_{1},z_{2},\ldots,z_{4n-2})+M, & t\in [0,1], z_{i}\ge 0, i=1,2,\ldots,4n-2, \\ f_{j}(t,0,0,\ldots,0)+M, &t\in [0,1], \mbox{for other cases}. \end{cases}\displaystyle \end{aligned}

Then we can define the operators $$T_{j}\ (j=1,2):P^{4n-2}\to P$$ and $$T:P^{2} \to P^{2}$$ as follows:

\begin{aligned} T_{j} (u,v) (t) =& \int _{0}^{1} k_{1}(t,s)F_{j} \bigl(s, \bigl(B_{n-1}(u-w) \bigr) (s), \bigl(A_{n-1}(u-w) \bigr) (s), \ldots , \\ &\bigl(B_{1}(u-w) \bigr) (s), \bigl(A_{1}(u-w) \bigr) (s), (u-w) (s), \\ &\bigl(B_{n-1}(v-w) \bigr) (s), \bigl(A_{n-1}(v-w) \bigr) (s), \ldots , \\ &\bigl(B_{1}(v-w) \bigr) (s), \bigl(A_{1}(v-w) \bigr) (s), (v-w) (s) \bigr)\,ds \end{aligned}

and

$$T(u,v) (t)=(T_{1},T_{2}) (u,v) (t),\quad t\in [0,1].$$

Then, if we find the positive fixed point $$(u^{*},v^{*})$$ of T with $$u^{*},v^{*}\ge w$$, then $$(u^{*}-w,v^{*}-w)$$ is a positive solution for (3.1). Let

$$x(t)= \int _{0}^{1}k_{n-1}(t,s) \bigl(u^{*}(s)-w(s) \bigr)\,ds,\qquad y(t)= \int _{0} ^{1}k_{n-1}(t,s) \bigl(v^{*}(s)-w(s) \bigr)\,ds,$$
(3.2)

and we will obtain the positive solution for (1.1) (note from the discussion in Sect. 2, we need the norms of $$u^{*}$$, $$v^{*}$$ to be greater than M).

Now, we list our assumptions for $$F_{j}$$ ($$j=1,2$$):

(H1) There exist $$a_{j1},b_{j1},c_{j1},d_{j1},l_{j}>0$$ ($$j=1,2$$) such that

\begin{aligned}& \kappa _{\psi } \bigl(b_{11}+a_{11}(n-1) \bigr)< 1, \qquad \kappa _{\psi } \bigl(d_{21}+c _{21}(n-1) \bigr)< 1, \\& \Delta _{11}= \det \begin{pmatrix} \kappa _{\psi }(d_{11}+c_{11}(n-1)) & \kappa _{\psi }(b_{11}+a_{11}(n-1))-1 \\ \kappa _{\psi }(d_{21}+c_{21}(n-1))-1 & \kappa _{\psi }(b_{21}+a_{21}(n-1)) \end{pmatrix}>0, \end{aligned}

and, for all $$t\in [0,1]$$, $$z_{i},\widetilde{z}_{i}\in R_{+}$$, $$i=1,2,\ldots,2n-1$$,

\begin{aligned}& F_{1}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1}) \\& \quad \ge a_{11}(z_{1}+z_{3}+\cdots +z_{2n-3})+a_{11}\bigl(2z_{2}+4z_{4}+ \cdots +2(n-1)z_{2n-2}\bigr)+b_{11} z_{2n-1} \\& \qquad {}+c_{11}(\widetilde{z}_{1}+\widetilde{z}_{3}+ \cdots + \widetilde{z}_{2n-3})+c_{11}\bigl(2 \widetilde{z}_{2}+4\widetilde{z}_{4}+ \cdots +2(n-1) \widetilde{z}_{2n-2}\bigr)+d_{11} \widetilde{z}_{2n-1}-l_{1}, \\& F_{2}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1}) \\& \quad \ge a_{21}(z_{1}+z_{3}+\cdots +z_{2n-3})+a_{21}\bigl(2z_{2}+4z_{4}+ \cdots +2(n-1)z_{2n-2}\bigr)+b_{21} z_{2n-1} \\& \qquad {}+c_{21}(\widetilde{z}_{1}+\widetilde{z}_{3}+ \cdots + \widetilde{z}_{2n-3})+c_{21}\bigl(2 \widetilde{z}_{2}+4\widetilde{z}_{4}+ \cdots +2(n-1) \widetilde{z}_{2n-2}\bigr)+d_{21} \widetilde{z}_{2n-1}-l_{2}. \end{aligned}

(H2) There exist $$Q_{j}\ (j=1,2):[0,1]\to R$$ such that

$$\int _{0}^{1} k_{1}(s,s)Q_{j}(s)\,ds< M$$

and

$$F_{j}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1}) \le Q_{j}(t),$$

for all $$t\in [0,1]$$, $$z_{i},\widetilde{z}_{i}\in [0,M]$$, $$i=1, 2,\ldots,2n-1$$, $$j=1,2$$.

(H3) There exist $$\widetilde{a}_{j1},\widetilde{b}_{j1},\widetilde{c} _{j1},\widetilde{d}_{j1},\widetilde{l}_{j}>0$$ ($$j=1,2$$) such that

\begin{aligned}& \widetilde{b}_{11}+\widetilde{a}_{11}(n-1)< 1,\qquad \widetilde{d}_{21}+ \widetilde{c}_{21}(n-1)< 1, \\& \Delta _{22}=\det \begin{pmatrix} 1- [\widetilde{b}_{11}+\widetilde{a}_{11}(n-1)] & - [\widetilde{d} _{11}+\widetilde{c}_{11}(n-1)] \\ - [\widetilde{b}_{21}+\widetilde{a}_{21}(n-1)] & 1- [\widetilde{d} _{21}+\widetilde{c}_{21}(n-1)] \end{pmatrix}>0, \end{aligned}

and, for all $$t\in [0,1]$$, $$z_{i},\widetilde{z}_{i}\in R_{+}$$, $$i=1,2,\ldots,2n-1$$,

\begin{aligned}& F_{1}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1}) \\& \quad \le \widetilde{a}_{11}(z_{1}+z_{3}+\cdots +z_{2n-3})+\widetilde{a} _{11}\bigl(2z_{2}+4z_{4}+ \cdots +2(n-1)z_{2n-2}\bigr)+\widetilde{b}_{11} z_{2n-1} \\& \qquad {}+\widetilde{c}_{11}(\widetilde{z}_{1}+ \widetilde{z}_{3}+\cdots +\widetilde{z}_{2n-3})+ \widetilde{c}_{11}\bigl(2\widetilde{z}_{2}+4 \widetilde{z}_{4}+\cdots +2(n-1)\widetilde{z}_{2n-2}\bigr)+ \widetilde{d} _{11} \widetilde{z}_{2n-1}+ \widetilde{l}_{1}, \\& F_{2}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1}) \\& \quad \le \widetilde{a}_{21}(z_{1}+z_{3}+\cdots +z_{2n-3})+\widetilde{a} _{21}\bigl(2z_{2}+4z_{4}+ \cdots +2(n-1)z_{2n-2}\bigr)+\widetilde{b}_{21} z_{2n-1} \\& \qquad {}+\widetilde{c}_{21}(\widetilde{z}_{1}+ \widetilde{z}_{3}+\cdots +\widetilde{z}_{2n-3})+ \widetilde{c}_{21}\bigl(2\widetilde{z}_{2}+4 \widetilde{z}_{4}+\cdots +2(n-1)\widetilde{z}_{2n-2}\bigr)+ \widetilde{d} _{21} \widetilde{z}_{2n-1}+ \widetilde{l}_{2}. \end{aligned}

(H4) There exist $$\widetilde{Q}_{j}\ (j=1,2):[0,1]\to R$$ and $$t_{1},t _{2}\in (0,1]$$ such that

$$\int _{0}^{1} k_{1}(t_{j},s) \widetilde{Q}_{j}(s)\,ds> M$$

and

$$F_{j}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1}) \ge \widetilde{Q}_{j}(t),$$

for all $$t\in [0,1]$$, $$z_{i},\widetilde{z}_{i}\in [0,M]$$, $$i=1, 2,\ldots,2n-1$$, $$j=1,2$$.

Let $$B_{\rho }=\{u\in P: \|u\|<\rho \}$$ for $$\rho >0$$ in the sequel. Then we easily have $$\partial B_{\rho }=\{u\in P: \|u\|=\rho \}$$, $$\overline{B}_{\rho }=\{u\in P: \|u\|\le \rho \}$$.

### Theorem 3.1

Suppose that (H0)(H2) hold. Then (1.1) has at least one positive solution.

### Proof

We first prove that there exists $$R_{1}>M$$ such that

$$(u,v)\neq T(u,v)+\lambda (\phi _{1},\phi _{2}),\quad \mbox{for } (u,v) \in \partial B_{R_{1}}\cap (P\times P), \lambda \ge 0,$$
(3.3)

where $$\phi _{i}$$ ($$i=1,2$$) are given elements in the cone $$P_{0}$$. We argue by contradiction. Suppose there exist $$(u,v)\in \partial B_{R_{1}} \cap (P\times P)$$ and $$\lambda _{0}\ge 0$$ with

$$(u,v) = T(u,v)+\lambda _{0} (\phi _{1},\phi _{2}).$$
(3.4)

This, together with Lemma 2.6, implies that $$u,v\in P_{0}$$. Moreover, from (H1) we have

\begin{aligned}& \begin{pmatrix} u(t) \\ v(t) \end{pmatrix} \\& \quad = \begin{pmatrix} T_{1}(u,v)(t)+\lambda _{0} \phi _{1}(t) \\ T_{2}(u,v)(t)+\lambda _{0} \phi _{2}(t) \end{pmatrix} \ge \begin{pmatrix} T_{1}(u,v)(t) \\ T_{2}(u,v)(t) \end{pmatrix} \\& \quad \ge \left(\textstyle\begin{array}{@{}l@{}} \int _{0}^{1} k_{1}(t,s) (a_{11}\sum_{i=1}^{n-1}[(B_{i}(u-w))(s)+2(n-i)(A _{i}(u-w))(s)]+b_{11}(u-w)(s) ) \,ds \\ \quad {}+ \int _{0}^{1} k_{1}(t,s) (c_{11}\sum_{i=1}^{n-1}[(B_{i}(v-w))(s)+2(n-i)(A _{i}(v-w))(s)]+d_{11}(v-w)(s) ) \,ds \\ \quad {}- l_{1} \int _{0}^{1} k_{1}(t,s) \,ds \\ \int _{0}^{1} k_{1}(t,s) (a_{21}\sum_{i=1}^{n-1}[(B_{i}(u-w))(s)+2(n-i)(A _{i}(u-w))(s)]+b_{21}(u-w)(s) ) \,ds \\ \quad {}+ \int _{0}^{1} k_{1}(t,s) (c_{21}\sum_{i=1}^{n-1}[(B_{i}(v-w))(s)+2(n-i)(A _{i}(v-w))(s)]+d_{21}(v-w)(s) ) \,ds \\ \quad {}- l_{2} \int _{0}^{1} k_{1}(t,s) \,ds \end{array}\displaystyle \right). \end{aligned}

Multiply by $$\psi (t)$$ on both sides, integrate over $$[0,1]$$, and use Lemma 2.1, and we have

\begin{aligned}& \begin{pmatrix} \int _{0}^{1} u(t) \psi (t)\,dt \\ \int _{0}^{1} v(t) \psi (t)\,dt \end{pmatrix} \\& \quad \ge \left(\textstyle\begin{array}{@{}l@{}} \kappa _{\psi } \int _{0}^{1} \psi (t) (a_{11}\sum_{i=1}^{n-1}[(B _{i}(u-w))(t)+2(n-i)(A_{i}(u-w))(t)]+b_{11}(u-w)(t) ) \,dt \\ \quad {}+ \kappa _{\psi } \int _{0}^{1} \psi (t) (c_{11}\sum_{i=1}^{n-1}[(B _{i}(v-w))(t)+2(n-i)(A_{i}(v-w))(t)]+d_{11}(v-w)(t) ) \,dt \\ \quad {}- l_{1} \int _{0}^{1} \psi (t) \,dt \\ \kappa _{\psi } \int _{0}^{1} \psi (t) (a_{21}\sum_{i=1}^{n-1}[(B _{i}(u-w))(t)+2(n-i)(A_{i}(u-w))(t)]+b_{21}(u-w)(t) ) \,dt \\ \quad {}+ \kappa _{\psi } \int _{0}^{1} \psi (t) (c_{21}\sum_{i=1}^{n-1}[(B _{i}(v-w))(t)+2(n-i)(A_{i}(v-w))(t)]+d_{21}(v-w)(t) ) \,dt \\ \quad {}- l_{2} \int _{0}^{1} \psi (t) \,dt \end{array}\displaystyle \right) \\& \quad = \left(\textstyle\begin{array}{@{}l@{}} \kappa _{\psi } \int _{0}^{1} \psi (t) (a_{11}\sum_{i=1}^{n-1}[(B _{i}(u-w))(t)+2(n-i)(B_{i}(u-w))'(t)]+b_{11}(u-w)(t) ) \,dt \\ \quad {}+ \kappa _{\psi } \int _{0}^{1} \psi (t) (c_{11}\sum_{i=1}^{n-1}[(B _{i}(v-w))(t)+2(n-i)(B_{i}(v-w))'(t)]+d_{11}(v-w)(t) ) \,dt- l _{1} \\ \kappa _{\psi } \int _{0}^{1} \psi (t) (a_{21}\sum_{i=1}^{n-1}[(B _{i}(u-w))(t)+2(n-i)(B_{i}(u-w))'(t)]+b_{21}(u-w)(t) ) \,dt \\ \quad {}+ \kappa _{\psi } \int _{0}^{1} \psi (t) (c_{21}\sum_{i=1}^{n-1}[(B _{i}(v-w))(t)+2(n-i)(B_{i}(v-w))'(t)]+d_{21}(v-w)(t) ) \,dt- l _{2} \end{array}\displaystyle \right). \end{aligned}

Using Lemma 2.2 we obtain

\begin{aligned}& \begin{pmatrix} \int _{0}^{1} u(t) \psi (t)\,dt \\ \int _{0}^{1} v(t) \psi (t)\,dt \end{pmatrix} \\& \quad \ge \begin{pmatrix} \kappa _{\psi }(b_{11}+a_{11}(n-1)) \int _{0}^{1} (u-w)(t) \psi (t)\,dt+ \kappa _{\psi }(d_{11}+c_{11}(n-1)) \int _{0}^{1} (v-w)(t) \psi (t)\,dt -l _{1} \\ \kappa _{\psi }(b_{21}+a_{21}(n-1)) \int _{0}^{1} (u-w)(t) \psi (t)\,dt+ \kappa _{\psi }(d_{21}+c_{21}(n-1)) \int _{0}^{1} (v-w)(t) \psi (t)\,dt -l _{2} \end{pmatrix}. \end{aligned}

Let

\begin{aligned}& \begin{aligned} \mathcal{N}_{1}&=\kappa _{\psi }\bigl[\bigl(b_{11}+a_{11}(n-1) \bigr)+\bigl(d_{11}+c_{11}(n-1)\bigr)\bigr] \int _{0}^{1} w(t) \psi (t)\,dt+l_{1} \\ &=\kappa _{\psi }\bigl[\bigl(b_{11}+a_{11}(n-1) \bigr)+\bigl(d_{11}+c_{11}(n-1)\bigr)\bigr] (2e-5)M+l _{1}, \end{aligned} \\& \begin{aligned} \mathcal{N}_{2}&=\kappa _{\psi }\bigl[\bigl(b_{21}+a_{21}(n-1) \bigr)+\bigl(d_{21}+c_{21}(n-1)\bigr)\bigr] \int _{0}^{1} w(t) \psi (t)\,dt+l_{2} \\ &=\kappa _{\psi }\bigl[\bigl(b_{21}+a_{21}(n-1) \bigr)+\bigl(d_{21}+c_{21}(n-1)\bigr)\bigr] (2e-5)M+l _{2}. \end{aligned} \end{aligned}

Therefore, we have

\begin{aligned}& \begin{pmatrix}[\kappa _{\psi }(b_{11}+a_{11}(n-1))-1] \int _{0}^{1} u(t) \psi (t)\,dt+ \kappa _{\psi }(d_{11}+c_{11}(n-1)) \int _{0}^{1} v(t) \psi (t)\,dt \\ \kappa _{\psi }(b_{21}+a_{21}(n-1)) \int _{0}^{1} u(t) \psi (t)\,dt+ [ \kappa _{\psi }(d_{21}+c_{21}(n-1))-1] \int _{0}^{1} v(t) \psi (t)\,dt \end{pmatrix} \\& \quad \le \begin{pmatrix} \mathcal{N}_{1} \\ \mathcal{N}_{2} \end{pmatrix} \end{aligned}

and

$$\begin{pmatrix} \kappa _{\psi }(d_{11}+c_{11}(n-1)) & \kappa _{\psi }(b_{11}+a_{11}(n-1))-1 \\ \kappa _{\psi }(d_{21}+c_{21}(n-1))-1 & \kappa _{\psi }(b_{21}+a_{21}(n-1)) \end{pmatrix} \begin{pmatrix} \int _{0}^{1} v(t) \psi (t)\,dt \\ \int _{0}^{1} u(t) \psi (t)\,dt \end{pmatrix} \le \begin{pmatrix} \mathcal{N}_{1} \\ \mathcal{N}_{2} \end{pmatrix}.$$

Solving this matrix inequality, we obtain

$$\begin{pmatrix} \int _{0}^{1} v(t) \psi (t)\,dt \\ \int _{0}^{1} u(t) \psi (t)\,dt \end{pmatrix} \le \frac{1}{\Delta _{11}} \begin{pmatrix} \kappa _{\psi }(b_{21}+a_{21}(n-1)) & 1-\kappa _{\psi }(b_{11}+a_{11}(n-1)) \\ 1-\kappa _{\psi }(d_{21}+c_{21}(n-1)) & \kappa _{\psi }(d_{11}+c_{11}(n-1)) \end{pmatrix} \begin{pmatrix} \mathcal{N}_{1} \\ \mathcal{N}_{2} \end{pmatrix}.$$

Consequently, there exist $$\widetilde{\mathcal{N}}_{1}, \widetilde{\mathcal{N}}_{2}>0$$ such that

$$\begin{pmatrix} \int _{0}^{1} v(t) \psi (t)\,dt \\ \int _{0}^{1} u(t) \psi (t)\,dt \end{pmatrix} \le \begin{pmatrix} \widetilde{\mathcal{N}}_{1} \\ \widetilde{\mathcal{N}}_{2} \end{pmatrix}.$$

Note that $$u,v\in P_{0}$$, and we have

$$\begin{pmatrix} \Vert v \Vert \\ \Vert u \Vert \end{pmatrix} \le \begin{pmatrix} \frac{\widetilde{\mathcal{N}}_{1}}{e-2} \\ \frac{\widetilde{\mathcal{N}}_{2}}{e-2} \end{pmatrix}.$$

Therefore, we can choose $$R_{1}>\max \{ M,\frac{ \widetilde{\mathcal{N}}_{1}}{e-2}, \frac{\widetilde{\mathcal{N}}_{2}}{e-2} \}$$ such that (3.4) is false, and thus (3.3) holds. From Lemma 2.3 we have

$$i \bigl(T, B_{R_{1}}\cap (P \times P),P \times P \bigr) = 0.$$
(3.5)

Next we prove that

$$(u, v) \neq \lambda T(u, v),\quad \mbox{for }(u, v) \in \partial B_{{M}} \cap (P \times P), \forall \lambda \in [0,1].$$
(3.6)

If not, there exist $$(u, v) \in \partial B_{{M}} \cap (P \times P)$$ and $$\lambda _{1} \in [0,1]$$ such that

$$(u, v) = \lambda _{1} T(u, v).$$

This, combining with (H2), implies that

$$\begin{pmatrix} M \\ M \end{pmatrix}= \begin{pmatrix} \Vert u \Vert \\ \Vert v \Vert \end{pmatrix} \le \begin{pmatrix} \Vert T_{1} (u,v) \Vert \\ \Vert T_{2}(u, v) \Vert \end{pmatrix}\le \begin{pmatrix} \int _{0}^{1} k_{1}(s,s)Q_{1}(s)\,ds \\ \int _{0}^{1} k_{1}(s,s)Q_{2}(s)\,ds \end{pmatrix}< \begin{pmatrix} M \\ M \end{pmatrix}.$$

This is a contradiction, and thus (3.6) is true. From Lemma 2.4 we have

$$\bigl(T, B_{M}\cap (P \times P),P \times P \bigr) = 1.$$
(3.7)

From (3.5) and (3.7) we have

\begin{aligned}& i \bigl(T, (B_{R_{1}} \setminus \overline{B}_{{M}} ) \cap (P \times P), P \times P \bigr) \\& \quad =i \bigl(T, B_{R_{1}} \cap (P \times P),P \times P \bigr)-i \bigl(T, B_{{M}} \cap (P \times P), P \times P \bigr)=0-1=-1. \end{aligned}

Therefore the operator T has at least one fixed point $$(u^{*},v^{*})$$ on $$(B_{R_{1}} \setminus \overline{B}_{{M}} ) \cap (P \times P)$$ with $$\|u^{*}\|\ge M$$, $$\|v^{*}\|\ge M$$, and note from (3.2) we see that (1.1) has at least one positive solution. This completes the proof. □

### Theorem 3.2

Suppose that (H0), and (H3)(H4) hold. Then (1.1) has at least one positive solution.

### Proof

We first prove that

$$(u,v)\neq T(u,v)+\lambda (\widetilde{\phi }_{1},\widetilde{\phi } _{2}), \quad \mbox{for } (u,v)\in \partial B_{M}\cap (P \times P), \lambda \ge 0,$$
(3.8)

where $$\widetilde{\phi }_{i}\ (i=1,2)\in P$$ are fixed elements. If this claim is false, there exist $$(u,v)\in \partial B_{M}\cap (P\times P)$$ and $$\lambda _{2}\ge 0$$ such that

$$(u,v) = T(u,v)+\lambda _{2} (\widetilde{\phi }_{1}, \widetilde{\phi } _{2}).$$

This, together with (H4), gives

$$\begin{pmatrix} \Vert u \Vert \\ \Vert v \Vert \end{pmatrix}\ge \begin{pmatrix} u(t_{1}) \\ v(t_{2}) \end{pmatrix} \ge \begin{pmatrix} T_{1} (u,v)(t_{1}) \\ T_{2} (u,v)(t_{2}) \end{pmatrix}\ge \begin{pmatrix} \int _{0}^{1} k_{1}(t_{1},s)\widetilde{Q}_{1}(s)\,ds \\ \int _{0}^{1} k_{1}(t_{2},s)\widetilde{Q}_{2}(s)\,ds \end{pmatrix}> \begin{pmatrix} M \\ M \end{pmatrix}.$$

This is a contradiction, and thus (3.8) holds. From Lemma 2.3 we have

$$i \bigl(T, B_{M}\cap (P \times P),P \times P \bigr) = 0.$$
(3.9)

Next we show that there is a large number $$R_{2}>M$$ such that

$$(u, v) \neq \lambda T(u, v), \quad \mbox{for }(u, v) \in \partial B_{R_{2}} \cap (P \times P), \forall \lambda \in [0,1].$$
(3.10)

We argue by contradiction, so we assume there exist $$(u, v) \in \partial B_{R_{2}} \cap (P \times P)$$ and $$\lambda _{3}\in [0,1]$$ such that

$$(u, v) = \lambda _{3} T(u, v).$$

Lemma 2.6 implies that $$u,v\in P_{0}$$, and from (H3) we obtain

\begin{aligned}& \begin{pmatrix} u(t) \\ v(t) \end{pmatrix} \\& \quad \le \begin{pmatrix} T_{1} (u,v)(t) \\ T_{2} (u,v)(t) \end{pmatrix} \\& \quad \le \left(\textstyle\begin{array}{@{}l@{}} \int _{0}^{1} k_{1}(t,s) (\widetilde{a}_{11}\sum_{i=1}^{n-1}[(B _{i}(u-w))(s)+2(n-i)(A_{i}(u-w))(s)]+\widetilde{b}_{11}(u-w)(s) ) \,ds \\ \quad {}+ \int _{0}^{1} k_{1}(t,s) (\widetilde{c}_{11}\sum_{i=1}^{n-1}[(B _{i}(v-w))(s)+2(n-i)(A_{i}(v-w))(s)]+\widetilde{d}_{11}(v-w)(s) ) \,ds \\ \quad {}+ \widetilde{l}_{1} \int _{0}^{1} k_{1}(t,s) \,ds \\ \int _{0}^{1} k_{1}(t,s) (\widetilde{a}_{21}\sum_{i=1}^{n-1}[(B _{i}(u-w))(s)+2(n-i)(A_{i}(u-w))(s)]+\widetilde{b}_{21}(u-w)(s) ) \,ds \\ \quad {}+ \int _{0}^{1} k_{1}(t,s) (\widetilde{c}_{21}\sum_{i=1}^{n-1}[(B _{i}(v-w))(s)+2(n-i)(A_{i}(v-w))(s)]+\widetilde{d}_{21}(v-w)(s) ) \,ds \\ \quad {}+ \widetilde{l}_{2} \int _{0}^{1} k_{1}(t,s) \,ds \end{array}\displaystyle \right). \end{aligned}

Multiply by $$\psi (t)$$ on both sides, integrate over $$[0,1]$$, and use Lemma 2.1, we have

\begin{aligned}& \begin{pmatrix} \int _{0}^{1} u(t) \psi (t)\,dt \\ \int _{0}^{1} v(t)\psi (t)\,dt \end{pmatrix} \\& \quad \le \left(\textstyle\begin{array}{@{}l@{}} \int _{0}^{1} \psi (t) (\widetilde{a}_{11}\sum_{i=1}^{n-1}[(B_{i}(u-w))(t)+2(n-i)(A _{i}(u-w))(t)]+\widetilde{b}_{11}(u-w)(t) ) \,dt \\ \quad {}+ \int _{0}^{1} \psi (t) (\widetilde{c}_{11}\sum_{i=1}^{n-1}[(B _{i}(v-w))(t)+2(n-i)(A_{i}(v-w))(t)]+\widetilde{d}_{11}(v-w)(t) ) \,dt + \widetilde{l}_{1} \\ \int _{0}^{1} \psi (t) (\widetilde{a}_{21}\sum_{i=1}^{n-1}[(B _{i}(u-w))(t)+2(n-i)(A_{i}(u-w))(t)]+\widetilde{b}_{21}(u-w)(t) ) \,dt \\ \quad {}+ \int _{0}^{1}\psi (t) (\widetilde{c}_{21}\sum_{i=1}^{n-1}[(B _{i}(v-w))(t)+2(n-i)(A_{i}(v-w))(t)]+\widetilde{d}_{21}(v-w)(t) ) \,dt + \widetilde{l}_{2} \end{array}\displaystyle \right). \end{aligned}

This, combining with Lemma 2.2, implies that

\begin{aligned}& \begin{pmatrix} \int _{0}^{1} u(t) \psi (t)\,dt \\ \int _{0}^{1} v(t)\psi (t)\,dt \end{pmatrix} \\& \quad \le \left(\textstyle\begin{array}{@{}l@{}} \int _{0}^{1} \psi (t) (\widetilde{a}_{11}\sum_{i=1}^{n-1}[(B_{i}u)(t)+2(n-i)(B _{i}u)'(t)]+\widetilde{b}_{11}u(t) ) \,dt \\ \quad {}+ \int _{0}^{1} \psi (t) (\widetilde{c}_{11}\sum_{i=1}^{n-1}[(B _{i}v)(t)+2(n-i)(B_{i}v)'(t)]+\widetilde{d}_{11}v(t) ) \,dt+ \widetilde{l}_{1} \\ \int _{0}^{1} \psi (t) (\widetilde{a}_{21}\sum_{i=1}^{n-1}[(B _{i}u)(t)+2(n-i)(B_{i}u)'(t)]+\widetilde{b}_{21}u(t) ) \,dt \\ \quad {}+ \int _{0}^{1}\psi (t) (\widetilde{c}_{21}\sum_{i=1}^{n-1}[(B _{i}v)(t)+2(n-i)(B_{i}v)'(t)]+\widetilde{d}_{21}v(t) ) \,dt + \widetilde{l}_{2} \end{array}\displaystyle \right) \\& \quad = \begin{pmatrix}[\widetilde{b}_{11}+\widetilde{a}_{11}(n-1)] \int _{0}^{1} u(t) \psi (t)\,dt +[\widetilde{d}_{11}+\widetilde{c}_{11}(n-1)]\int _{0}^{1}v(t) \psi (t)\,dt+\widetilde{l}_{1} \\ [\widetilde{b}_{21}+\widetilde{a}_{21}(n-1)] \int _{0}^{1} u(t) \psi (t)\,dt +[\widetilde{d}_{21}+\widetilde{c}_{21}(n-1)]\int _{0}^{1}v(t) \psi (t)\,dt+ \widetilde{l}_{2} \end{pmatrix}. \end{aligned}

Consequently, we have

$$\begin{pmatrix} 1- [\widetilde{b}_{11}+\widetilde{a}_{11}(n-1)] & - [\widetilde{d} _{11}+\widetilde{c}_{11}(n-1)] \\ - [\widetilde{b}_{21}+\widetilde{a}_{21}(n-1)] &1- [\widetilde{d}_{21}+ \widetilde{c}_{21}(n-1)] \end{pmatrix} \begin{pmatrix} \int _{0}^{1} u(t) \psi (t)\,dt \\ \int _{0}^{1} v(t)\psi (t)\,dt \end{pmatrix} \le \begin{pmatrix} \widetilde{l}_{1} \\ \widetilde{l}_{2} \end{pmatrix}.$$

Solving this matrix inequality, we obtain

$$\begin{pmatrix} \int _{0}^{1} u(t) \psi (t)\,dt \\ \int _{0}^{1} v(t)\psi (t)\,dt \end{pmatrix}\le \frac{1}{\Delta _{22}} \begin{pmatrix} 1- [\widetilde{d}_{21}+\widetilde{c}_{21}(n-1)] & \widetilde{d}_{11}+ \widetilde{c}_{11}(n-1) \\ \widetilde{b}_{21}+\widetilde{a}_{21}(n-1) & 1- [\widetilde{b}_{11}+ \widetilde{a}_{11}(n-1)] \end{pmatrix} \begin{pmatrix} \widetilde{l}_{1} \\ \widetilde{l}_{2} \end{pmatrix}.$$

Therefore, there exist $$\widetilde{\mathcal{N}}_{3}, \widetilde{\mathcal{N}}_{4}>0$$ such that

$$\begin{pmatrix} \int _{0}^{1} u(t) \psi (t)\,dt \\ \int _{0}^{1} v(t)\psi (t)\,dt \end{pmatrix}\le \begin{pmatrix} \widetilde{\mathcal{N}}_{3} \\ \widetilde{\mathcal{N}}_{4} \end{pmatrix}.$$

Note that $$u,v\in P_{0}$$, and then we obtain

$$\begin{pmatrix} \Vert u \Vert \\ \Vert v \Vert \end{pmatrix}\le \begin{pmatrix} \frac{\widetilde{\mathcal{N}}_{3}}{e-2} \\ \frac{\widetilde{\mathcal{N}}_{4}}{e-2} \end{pmatrix}.$$

If we choose $$R_{2}>\max \{ M,\frac{\widetilde{\mathcal{N}} _{3}}{e-2},\frac{\widetilde{\mathcal{N}}_{4}}{e-2} \}$$ then (3.10) holds. From Lemma 2.4 we have

$$i \bigl(T, B_{R_{2}}\cap (P \times P),P \times P \bigr) = 1.$$
(3.11)

From (3.9) and (3.11) we have

\begin{aligned}& i \bigl(T, (B_{R_{2}} \setminus \overline{B}_{{M}} ) \cap (P \times P), P \times P \bigr) \\& \quad =i \bigl(T, B_{R_{2}} \cap (P \times P), P \times P \bigr)-i \bigl(T, B_{{M}} \cap (P \times P), P \times P \bigr)=1-0=1. \end{aligned}

Therefore the operator T has at least one fixed point $$(u^{*},v^{*})$$ on $$(B_{R_{2}} \setminus \overline{B}_{{M}} ) \cap (P \times P)$$ with $$\|u^{*}\|\ge M$$, $$\|v^{*}\|\ge M$$, and note from (3.2) we see that (1.1) has at least one positive solution. This completes the proof. □

### Example 3.3

Let

\begin{aligned}& a_{11}=\frac{e}{4(e-2)(n-1)},\qquad b_{11}= \frac{e}{2(e-2)}, \\& c_{11}= \frac{e}{(e-2)(n-1)},\qquad d_{11}=\frac{e}{e-2}, \\& a_{21}=\frac{e}{2(e-2)(n-1)},\qquad b_{21}= \frac{e}{5(e-2)}, \\& c_{21}= \frac{e}{3(e-2)(n-1)},\qquad d_{21}=\frac{e}{2(e-2)}. \end{aligned}

Then

\begin{aligned}& \kappa _{\psi }\bigl(b_{11}+a_{11}(n-1)\bigr)= \frac{e-2}{e} \biggl(\frac{e}{2(e-2)}+ \frac{e}{4(e-2)(n-1)}(n-1) \biggr)=\frac{3}{4}< 1, \\& \kappa _{\psi }\bigl(d_{21}+c_{21}(n-1)\bigr)= \frac{e-2}{e} \biggl(\frac{e}{2(e-2)}+ \frac{e}{3(e-2)(n-1)}(n-1) \biggr)=\frac{5}{6}< 1, \\& \Delta _{11}= \begin{vmatrix} \frac{e-2}{e} ( \frac{e}{e-2}+\frac{e}{(e-2)(n-1)}(n-1) ) & -\frac{1}{4} \\ -\frac{1 }{6} & \frac{e-2}{e} ( \frac{e}{5(e-2)} + \frac{e}{2(e-2)(n-1)}(n-1) ) \end{vmatrix}= \frac{163}{120}>0. \end{aligned}

Consider

\begin{aligned}& F_{1}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1}) \\& \quad =\frac{9}{5}M \biggl[\frac{eM}{e-2} \biggl( \frac{3}{2}+\frac{5}{4}(1+n) \biggr) \biggr] ^{-\delta _{1}} \\& \qquad{}\times \bigl[a_{11}(z_{1}+z_{3}+\cdots +z_{2n-3})+a_{11}\bigl(2z_{2}+4z _{4}+\cdots +2(n-1)z_{2n-2}\bigr) \\& \qquad {}+b_{11}z_{2n-1}+c_{11}(\widetilde{z}_{1}+ \widetilde{z}_{3}+\cdots + \widetilde{z}_{2n-3}) \\& \qquad {}+c_{11} \bigl(2\widetilde{z}_{2}+4\widetilde{z}_{4}+ \cdots +2(n-1)\widetilde{z}_{2n-2}\bigr)+d_{11} \widetilde{z}_{2n-1} \bigr]^{ \delta _{1}}, \\& F_{2}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1}) \\& \quad = \frac{19}{10}M \biggl[\frac{eM}{e-2} \biggl( \frac{7}{10}+\frac{5}{6}(1+n) \biggr) \biggr] ^{-\delta _{2}} \\& \qquad {}\times \bigl[a_{21}(z_{1}+z_{3}+\cdots +z_{2n-3})+a_{21}\bigl(2z_{2}+4z _{4}+\cdots +2(n-1)z_{2n-2}\bigr) \\& \qquad {}+b_{21}z_{2n-1}+c_{21}(\widetilde{z}_{1}+ \widetilde{z}_{3}+\cdots + \widetilde{z}_{2n-3}) \\& \qquad {}+c_{21} \bigl(2\widetilde{z}_{2}+4\widetilde{z}_{4}+ \cdots +2(n-1)\widetilde{z}_{2n-2}\bigr)+d_{21} \widetilde{z}_{2n-1} \bigr]^{ \delta _{2}}, \end{aligned}

for all $$t\in [0,1]$$, $$z_{i},\widetilde{z}_{i}\in R_{+}$$, $$i=1,2,\ldots,2n-1$$, and $$\delta _{1},\delta _{2}>1$$.

For all $$t\in [0,1]$$, $$z_{i},\widetilde{z}_{i}\in [0,M]$$, $$i=1, 2,\ldots,2n-1$$, $$j=1,2$$, we have

\begin{aligned}& F_{1}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1})\le \frac{9}{5}M, \\& F_{2}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1})\le \frac{19}{10}M. \end{aligned}

Consequently, if let $$Q_{1}(t)\equiv \frac{9}{5}M$$, $$Q_{2}(t)\equiv \frac{19}{10}M$$ for $$t\in [0,1]$$, then (H2) holds.

On the other hand, for all $$t\in [0,1]$$ we note that

\begin{aligned}& \liminf_{a_{11}\sum _{i=1}^{n-1}(z_{2i-1}+2iz_{2i})+b_{11}z_{2n-1}+c _{11}\sum _{i=1}^{n-1}(\widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+d _{11}\widetilde{z}_{2n-1}\to + \infty } \\& \qquad \frac{F_{1}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1},\widetilde{z}_{2},\ldots,\widetilde{z}_{2n-3}, \widetilde{z}_{2n-2},\widetilde{z}_{2n-1})}{a_{11}\sum_{i=1} ^{n-1}(z_{2i-1} +2iz_{2i})+b_{11}z_{2n-1}+c_{11}\sum_{i=1}^{n-1}( \widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+d_{11}\widetilde{z}_{2n-1}} \\& \quad =\liminf_{a_{11}\sum _{i=1}^{n-1}(z_{2i-1}+2iz_{2i})+b_{11}z_{2n-1}+c _{11}\sum _{i=1}^{n-1}(\widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+d _{11}\widetilde{z}_{2n-1}\to + \infty } \\& \qquad \frac{\frac{9}{5}M [\frac{eM}{e-2} (\frac{3}{2}+ \frac{5}{4}(1+n) ) ]^{-\delta _{1}} [a_{11}\sum_{i=1}^{n-1}(z_{2i-1} +2iz_{2i})+b_{11}z_{2n-1}+c_{11}\sum_{i=1}^{n-1}(\widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+d_{11} \widetilde{z}_{2n-1} ]^{\delta _{1}}}{a_{11}\sum_{i=1} ^{n-1}(z_{2i-1} +2iz_{2i})+b_{11}z_{2n-1}+c_{11}\sum_{i=1}^{n-1}( \widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+d_{11}\widetilde{z}_{2n-1}} \\& \quad = +\infty \end{aligned}

and

\begin{aligned}& \liminf_{a_{21}\sum _{i=1}^{n-1}(z_{2i-1}+2iz_{2i})+b_{21}z_{2n-1}+c _{21}\sum _{i=1}^{n-1}(\widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+d _{21}\widetilde{z}_{2n-1}\to + \infty } \\& \qquad \frac{F_{2}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1},\widetilde{z}_{2},\ldots,\widetilde{z}_{2n-3}, \widetilde{z}_{2n-2},\widetilde{z}_{2n-1})}{a_{21}\sum_{i=1} ^{n-1}(z_{2i-1} +2iz_{2i})+b_{21}z_{2n-1}+c_{21}\sum_{i=1}^{n-1}( \widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+d_{21}\widetilde{z}_{2n-1}} \\& \quad =\liminf_{a_{21}\sum _{i=1}^{n-1}(z_{2i-1}+2iz_{2i})+b_{21}z_{2n-1}+c _{21}\sum _{i=1}^{n-1}(\widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+d _{21}\widetilde{z}_{2n-1}\to + \infty } \\& \qquad \frac{\frac{19}{10}M [\frac{eM}{e-2} (\frac{7}{10}+ \frac{5}{6}(1+n) ) ]^{-\delta _{2}} [a_{21}\sum_{i=1}^{n-1}(z_{2i-1} +2iz_{2i})+b_{21}z_{2n-1}+c_{21}\sum_{i=1}^{n-1}(\widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+d_{21} \widetilde{z}_{2n-1} ]^{\delta _{2}}}{a_{21}\sum_{i=1} ^{n-1}(z_{2i-1} +2iz_{2i})+b_{21}z_{2n-1}+c_{21}\sum_{i=1}^{n-1}( \widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+d_{21}\widetilde{z}_{2n-1}} \\& \quad = +\infty . \end{aligned}

Therefore, (H1) holds.

### Example 3.4

Let $$t_{1}=\frac{1}{2}$$, $$t_{2}=1$$, and note that $$\int _{0}^{1} k_{1}(t,s)\,ds=t- \frac{1}{2}t^{2}$$ for $$t\in [0,1]$$, and if we consider the case $$\widetilde{Q}_{j}\equiv \mbox{constant}$$, we have

$$\int _{0}^{1} k_{1}(t_{1},s) \widetilde{Q}_{1}(s)\,ds=\frac{3}{8} \widetilde{Q}_{1},\qquad \int _{0}^{1} k_{1}(t_{2},s) \widetilde{Q}_{2}(s)\,ds= \frac{1}{2}\widetilde{Q}_{2}.$$

To obtain the first inequality in (H4), we can take $$\widetilde{Q} _{1}=3M$$, $$\widetilde{Q}_{2}=\frac{5}{2}M$$.

Let

$$\begin{pmatrix} \widetilde{a}_{11} & \widetilde{b}_{11} & \widetilde{c}_{11} & \widetilde{d}_{11} \\ \widetilde{a}_{21} & \widetilde{b}_{21} & \widetilde{c}_{21} & \widetilde{d}_{21} \end{pmatrix}= \begin{pmatrix} \frac{1}{20(n-1)} & \frac{1}{10}& \frac{1}{80(n-1)} & \frac{1}{40} \\ \frac{1}{40(n-1)} & \frac{1}{20} & \frac{1}{100(n-1)} & \frac{1}{50} \end{pmatrix}.$$

Then we have

\begin{aligned}& \widetilde{b}_{11}+\widetilde{a}_{11}(n-1)= \frac{1}{10}+\frac{1}{20}< 1,\qquad \widetilde{d}_{21}+ \widetilde{c}_{21}(n-1)=\frac{1}{50}+ \frac{1}{100}< 1, \\& \Delta _{22}= \begin{vmatrix} 0.85 & - [\frac{1}{40}+\frac{1}{80}] \\ - [\frac{1}{20}+\frac{1}{40}] & 0.97 \end{vmatrix}\approx 0.82 >0. \end{aligned}

Let

\begin{aligned}& F_{1}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1}) \\& \quad = 3M \exp \bigl\{ M\bigl(0.125+0.0625(1+n)\bigr)\bigr\} \\& \qquad {}\times\exp \bigl\{ - \widetilde{a}_{11}(z_{1}+z _{3}+\cdots +z_{2n-3})-\widetilde{a}_{11}\bigl(2z_{2}+4z_{4}+ \cdots +2(n-1)z _{2n-2}\bigr) \\& \qquad {}-\widetilde{b}_{11} z_{2n-1} -\widetilde{c}_{11}( \widetilde{z}_{1}+ \widetilde{z}_{3}+\cdots + \widetilde{z}_{2n-3}) \\& \qquad {}-\widetilde{c}_{11}\bigl(2 \widetilde{z}_{2}+4\widetilde{z}_{4}+\cdots +2(n-1) \widetilde{z}_{2n-2}\bigr)- \widetilde{d}_{11} \widetilde{z}_{2n-1}\bigr\} , \\& F_{2}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1}) \\& \quad = 2.5 M \exp \bigl\{ M\bigl(0.07+0.035(1+n)\bigr)\bigr\} \\& \qquad {}\times\exp \bigl\{ - \widetilde{a}_{21}(z_{1}+z _{3}+\cdots +z_{2n-3})-\widetilde{a}_{21}\bigl(2z_{2}+4z_{4}+ \cdots +2(n-1)z _{2n-2}\bigr) \\& \qquad {}-\widetilde{b}_{21} z_{2n-1} -\widetilde{c}_{21}( \widetilde{z}_{1}+ \widetilde{z}_{3}+\cdots + \widetilde{z}_{2n-3}) \\& \qquad {}-\widetilde{c}_{21}\bigl(2 \widetilde{z}_{2}+4\widetilde{z}_{4}+\cdots +2(n-1) \widetilde{z}_{2n-2}\bigr)- \widetilde{d}_{21} \widetilde{z}_{2n-1}\bigr\} , \end{aligned}

for all $$t\in [0,1]$$, $$z_{i},\widetilde{z}_{i}\in R_{+}$$, $$i=1,2,\ldots,2n-1$$.

For all $$t\in [0,1]$$, $$z_{i},\widetilde{z}_{i}\in [0,M]$$, $$i=1, 2,\ldots,2n-1$$, $$j=1,2$$, we have

\begin{aligned}& F_{1}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1})\ge 3M, \\& F_{2}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1}, \widetilde{z}_{2},\ldots, \widetilde{z}_{2n-3},\widetilde{z}_{2n-2}, \widetilde{z}_{2n-1})\ge 2.5 M, \end{aligned}

and thus (H4) holds. On the other hand, for all $$t\in [0,1]$$ we also have

\begin{aligned}& \limsup_{\widetilde{a}_{11}\sum _{i=1}^{n-1}(z_{2i-1}+2iz_{2i})+ \widetilde{b}_{11}z_{2n-1}+\widetilde{c}_{11}\sum _{i=1}^{n-1}( \widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+\widetilde{d}_{11} \widetilde{z}_{2n-1}\to + \infty } \\& \qquad \frac{F_{1}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1},\widetilde{z}_{2},\ldots,\widetilde{z}_{2n-3}, \widetilde{z}_{2n-2},\widetilde{z}_{2n-1})}{\widetilde{a}_{11}\sum_{i=1}^{n-1}(z_{2i-1} +2iz_{2i})+\widetilde{b}_{11}z_{2n-1}+ \widetilde{c}_{11}\sum_{i=1}^{n-1}(\widetilde{z}_{2i-1}+2i \widetilde{z}_{2i})+\widetilde{d}_{11}\widetilde{z}_{2n-1}} \\& \quad =\limsup_{\widetilde{a}_{11}\sum _{i=1}^{n-1}(z_{2i-1}+2iz_{2i})+ \widetilde{b}_{11}z_{2n-1}+\widetilde{c}_{11}\sum _{i=1}^{n-1}( \widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+\widetilde{d}_{11} \widetilde{z}_{2n-1}\to + \infty } \\& \qquad \Biggl(3M \exp \bigl\{ M\bigl(0.125+0.0625(1+n)\bigr)\bigr\} \\& \qquad {}\times\exp \Biggl\{ - \Biggl[\widetilde{a} _{11}\sum_{i=1}^{n-1}(z_{2i-1} +2iz_{2i})+\widetilde{b}_{11}z _{2n-1}+\widetilde{c}_{11}\sum_{i=1}^{n-1}(\widetilde{z}_{2i-1}+2i \widetilde{z}_{2i})+\widetilde{d}_{11}\widetilde{z}_{2n-1} \Biggr] \Biggr\} \\& \qquad{} \Big/ \Biggl(\widetilde{a}_{11}\sum_{i=1}^{n-1}(z_{2i-1} +2iz_{2i})+ \widetilde{b}_{11}z_{2n-1}+\widetilde{c}_{11}\sum_{i=1}^{n-1}( \widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+\widetilde{d}_{11} \widetilde{z}_{2n-1}\Biggr)\Biggr) \\& \quad = 0 \end{aligned}

and

\begin{aligned}& \limsup_{\widetilde{a}_{21}\sum _{i=1}^{n-1}(z_{2i-1}+2iz_{2i})+ \widetilde{b}_{21}z_{2n-1}+\widetilde{c}_{21}\sum _{i=1}^{n-1}( \widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+\widetilde{d}_{21} \widetilde{z}_{2n-1}\to + \infty } \\& \qquad \frac{F_{2}(t,z_{1},z_{2},\ldots,z_{2n-3},z_{2n-2},z_{2n-1}, \widetilde{z}_{1},\widetilde{z}_{2},\ldots,\widetilde{z}_{2n-3}, \widetilde{z}_{2n-2},\widetilde{z}_{2n-1})}{\widetilde{a}_{21}\sum_{i=1}^{n-1}(z_{2i-1} +2iz_{2i})+\widetilde{b}_{21}z_{2n-1}+ \widetilde{c}_{21}\sum_{i=1}^{n-1}(\widetilde{z}_{2i-1}+2i \widetilde{z}_{2i})+\widetilde{d}_{21}\widetilde{z}_{2n-1}} \\& \quad =\limsup_{\widetilde{a}_{21}\sum _{i=1}^{n-1}(z_{2i-1}+2iz_{2i})+ \widetilde{b}_{21}z_{2n-1}+\widetilde{c}_{21}\sum _{i=1}^{n-1}( \widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+\widetilde{d}_{21} \widetilde{z}_{2n-1}\to + \infty } \\& \qquad \Biggl(2.5 M \exp \bigl\{ M\bigl(0.07+0.035(1+n)\bigr)\bigr\} \\& \qquad {}\times\exp \Biggl\{ - \Biggl[\widetilde{a} _{21}\sum_{i=1}^{n-1}(z_{2i-1} +2iz_{2i})+\widetilde{b}_{21}z _{2n-1}+\widetilde{c}_{21}\sum_{i=1}^{n-1}(\widetilde{z}_{2i-1}+2i \widetilde{z}_{2i})+\widetilde{d}_{21}\widetilde{z}_{2n-1} \Biggr] \Biggr\} \\& \qquad{} \Big/ \Biggl(\widetilde{a}_{21}\sum_{i=1}^{n-1}(z_{2i-1} +2iz_{2i})+ \widetilde{b}_{21}z_{2n-1}+\widetilde{c}_{21}\sum_{i=1}^{n-1}( \widetilde{z}_{2i-1}+2i\widetilde{z}_{2i})+\widetilde{d}_{21} \widetilde{z}_{2n-1}\Biggr)\Biggr) \\& \quad = 0. \end{aligned}

Therefore, (H3) holds.

## Conclusion

In this paper we use the fixed point index to study the existence of positive solutions for the system of 2nth-order boundary value problems (1.1) involving semipositone nonlinearities. Our nonlinearities not only depend on all derivatives of unknown functions, but they also grow superlinearly and sublinearly at infinity.

## References

1. 1.

Yang, Z.: Existence of positive solutions for a system of generalized Lidstone problems. Comput. Math. Appl. 60, 501–510 (2010)

2. 2.

Xu, J., Yang, Z.: Positive solutions for a system of generalized Lidstone problems. J. Appl. Math. Comput. 37, 13–35 (2011)

3. 3.

Wang, K., Yang, Z.: Positive solutions for a system of higher order boundary-value problems involving all derivatives of odd orders. Electron. J. Differ. Equ. 2012, 52 (2012)

4. 4.

Cheng, W., Xu, J., Cui, Y.: Positive solutions for a system of nonlinear semipositone fractional q-difference equations with q-integral boundary conditions. J. Nonlinear Sci. Appl. 10, 4430–4440 (2017)

5. 5.

Xu, J., Goodrich, C., Cui, Y.: Positive solutions for a system of first-order discrete fractional boundary value problems with semipositone nonlinearities. Rev. R. Acad. Cienc. Exactas Fís. Nat., Ser. A Mat. 113, 1343–1358 (2019)

6. 6.

Qiu, X., Xu, J., O’Regan, D., Cui, Y.: Positive solutions for a system of nonlinear semipositone boundary value problems with Riemann–Liouville fractional derivatives. J. Funct. Spaces 2018, Article ID 7351653 (2018)

7. 7.

Chen, C., Xu, J., O’Regan, D., Fu, Z.: Positive solutions for a system of semipositone fractional difference boundary value problems. J. Funct. Spaces 2018, Article ID 6835028 (2018)

8. 8.

Jiang, J., O’Regan, D., Xu, J., Fu, Z.: Positive solutions for a system of nonlinear Hadamard fractional differential equations involving coupled integral boundary conditions. J. Inequal. Appl. 2019, 204 (2019)

9. 9.

Zhang, X., Liu, L., Wu, Y., Zou, Y.: Existence and uniqueness of solutions for systems of fractional differential equations with Riemann–Stieltjes integral boundary condition. Adv. Differ. Equ. 2018, 204 (2018)

10. 10.

Zhang, X., Liu, L., Zou, Y.: Fixed-point theorems for systems of operator equations and their applications to the fractional differential equations. J. Funct. Spaces 2018, Article ID 7469868 (2018)

11. 11.

Hao, X., Wang, H., Liu, L., Cui, Y.: Positive solutions for a system of nonlinear fractional nonlocal boundary value problems with parameters and p-Laplacian operator. Bound. Value Probl. 2017, 182 (2017)

12. 12.

Hao, X., Wang, H.: Positive solutions of semipositone singular fractional differential systems with a parameter and integral boundary conditions. Open Math. 16, 581–596 (2018)

13. 13.

Qi, T., Liu, Y., Cui, Y.: Existence of solutions for a class of coupled fractional differential systems with nonlocal boundary conditions. J. Funct. Spaces 2017, Article ID 6703860 (2017)

14. 14.

Hao, X., Zhang, L., Liu, L.: Positive solutions of higher order fractional integral boundary value problem with a parameter. Nonlinear Anal., Model. Control 24, 210–223 (2019)

15. 15.

Li, H., Zhang, J.: Positive solutions for a system of fractional differential equations with two parameters. J. Funct. Spaces 2018, Article ID 1462505 (2018)

16. 16.

Zhai, C., Wang, W., Li, H.: A uniqueness method to a new Hadamard fractional differential system with four-point boundary conditions. J. Inequal. Appl. 2018, 207 (2018)

17. 17.

Cheng, W., Xu, J., Cui, Y., Ge, Q.: Positive solutions for a class of fractional difference systems with coupled boundary conditions. Adv. Differ. Equ. 2019, 249 (2019)

18. 18.

Wang, F., Cui, Y., Zhou, H.: Solvability for an infinite system of fractional order boundary value problems. Ann. Funct. Anal. 10, 395–411 (2019)

19. 19.

Wang, F., Cui, Y.: Positive solutions for an infinite system of fractional order boundary value problems. Adv. Differ. Equ. 2019, 169 (2019)

20. 20.

Hao, X., Zuo, M., Liu, L.: Multiple positive solutions for a system of impulsive integral boundary value problems with sign-changing nonlinearities. Appl. Math. Lett. 82, 24–31 (2018)

21. 21.

Riaz, U., Zada, A., Ali, Z., Ahmad, M., Xu, J., Fu, Z.: Analysis of nonlinear coupled systems of impulsive fractional differential equations with Hadamard derivatives. Math. Probl. Eng. 2019, Article ID 5093572 (2019)

22. 22.

Riaz, U., Zada, A., Ali, Z., Cui, Y., Xu, J.: Analysis of coupled systems of implicit impulsive fractional differential equations involving Hadamard derivatives. Adv. Differ. Equ. 2019, 226 (2019)

23. 23.

Zada, A., Waheed, H., Alzabut, J., Wang, X.: Existence and stability of impulsive coupled system of fractional integrodifferential equations. Demonstr. Math. 52, 296–335 (2019)

24. 24.

Zada, A., Fatima, S., Ali, Z., Xu, J., Cui, Y.: Stability results for a coupled system of impulsive fractional differential equations. Mathematics 7, 927 (2019)

25. 25.

Wang, J., Zada, A., Waheed, H.: Stability analysis of a coupled system of nonlinear implicit fractional anti-periodic boundary value problem. Math. Methods Appl. Sci. 42, 6706–6732 (2019)

26. 26.

Ali, Z., Zada, A., Shah, K.: On Ulam’s stability for a coupled systems of nonlinear implicit fractional differential equations. Bull. Malays. Math. Sci. Soc. 42, 2681–2699 (2019)

27. 27.

Zhang, K., O’Regan, D., Xu, J., Fu, Z.: Nontrivial solutions for a higher order nonlinear fractional boundary value problem involving Riemann–Liouville fractional derivatives. J. Funct. Spaces 2019, Article ID 2381530 (2019)

28. 28.

Ding, Y., Xu, J., Zhang, X.: Positive solutions for a 2nth-order p-Laplacian boundary value problem involving all derivatives. Electron. J. Differ. Equ. 2013, 36 (2013)

29. 29.

Xu, J., Wei, Z., Ding, Y.: Positive solutions for a 2nth-order p-Laplacian boundary value problem involving all even derivatives. Topol. Methods Nonlinear Anal. 39, 23–36 (2012)

30. 30.

Hao, X., Sun, H., Liu, L., Wang, D.: Positive solutions for semipositone fractional integral boundary value problem on the half-line. Rev. R. Acad. Cienc. Exactas Fís. Nat., Ser. A Mat. 113, 3055–3067 (2019)

31. 31.

Hao, X., Zhang, L.: Positive solutions of a fractional thermostat model with a parameter. Symmetry 11, 122 (2019)

32. 32.

Yang, Z.: Positive solutions of a 2nth-order boundary value problem involving all derivatives via the order reduction. Comput. Math. Appl. 61, 822–831 (2011)

33. 33.

Yang, Z., O’Regan, D.: Positive solutions for a 2nth-order boundary value problem involving all derivatives of odd orders. Topol. Methods Nonlinear Anal. 37, 87–101 (2011)

34. 34.

Guo, D., Lakshmikantham, V.: Nonlinear Problems in Abstract Cones. Academic Press, New York (1988)

### Acknowledgements

The authors would like to thank the referees for their pertinent comments and valuable suggestions.

Not applicable.

## Funding

This work is supported by the China Postdoctoral Science Foundation (Grant Nos. 2017M612230, 2019M652348), and the Natural Science Foundation of Chongqing Normal University (Grant No. 16XYY24).

## Author information

Authors

### Contributions

All authors contributed equally to the manuscript. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Xinan Hao.

## Ethics declarations

### Competing interests

The authors declare that they have no competing interests. 