The existence of symmetric positive solutions for a seconder-order difference equation with sum form boundary conditions

Abstract

In this paper, we consider the existence of positive solutions for a second-order discrete boundary value problem $\mathrm{\Delta }(g(k-1)\mathrm{\Delta }u(k-1))+w(k)f(k,u(k))=0$ subject to the boundary conditions: $au(0)-bg(0)\mathrm{\Delta }u(0)={\sum }_{i=1}^{n-1}h(i)u(i)$, $au(n)+bg(n-1)\mathrm{\Delta }u(n-1)={\sum }_{i=1}^{n-1}h(i)u(i)$, where $a,b>0$, $\mathrm{\Delta }u(k)=u(k+1)-u(k)$ for $k\in \{0,1,\dots ,n-1\}$, $g(k)>0$ is symmetric on $\{0,1,\dots ,n-1\}$, $w(k)$ is symmetric on $\{0,1,\dots ,n\}$, $f:\{0,1,\dots ,n\}\times [0,+\mathrm{\infty })$ is continuous, $f(k,u)=f(n-k,u)$ for all $(k,u)\in \{0,1,\dots ,n\}\times [0,+\mathrm{\infty })$, and $h(i)$ is nonnegative and symmetric on $\{0,1,\dots ,n\}$. By the fixed point theorem and the Hölder inequality, we study the existence of symmetric positive solutions for the above difference equation with sum form boundary conditions.

1 Introduction

A class of boundary value problems (BVPs) with integral boundary conditions arise in thermal conduction problems, semiconductor problems, and hydrodynamic problems [1–3]. Recently, such problems have been investigated by many authors [4–10]. The equation ${(g(t)u^{\prime }(t))}^{\prime }+w(t)f(t,u(t))=0$, $0<t<1$, describes many phenomena in the fields of gas dynamics, nuclear physics, chemically reacting systems and atomic structures [11–15]. In [10], Feng considered the following differential equation BVP with integral boundary conditions:

$${(g(t)u^{\prime }(t))}^{\prime }+w(t)f(t,u(t))=0,0<t<1,$$

(1.1)

$$au(0)-b\underset{t\to 0^{+}}{lim}g(t)u^{\prime }(t)={\int }_0^{1}h(s)u(s)\mathrm{d}s,$$

(1.2)

$$au(1)+b\underset{t\to 1^{-}}{lim}g(t)u^{\prime }(t)={\int }_0^{1}h(s)u(s)\mathrm{d}s.$$

(1.3)

Applying the fixed point index theorem and the Hölder inequality, the author studied the existence of symmetric positive solutions for BVP (1.1)-(1.3).

Motivated by the above works, we will study the following BVP with sum form boundary conditions:

$$\mathrm{\Delta }(g(k-1)\mathrm{\Delta }u(k-1))+w(k)f(k,u(k))=0,k\in \{1,\dots ,n-1\},$$

(1.4)

$$au(0)-bg(0)\mathrm{\Delta }u(0)=\sum _{i=1}^{n-1}h(i)u(i),$$

(1.5)

$$au(n)+bg(n-1)\mathrm{\Delta }u(n-1)=\sum _{i=1}^{n-1}h(i)u(i).$$

(1.6)

Throughout this paper, the following conditions are assumed:

(A₁) $a,b>0$, $w(k)$ is symmetric on $\{0,1,\dots ,n\}$, and there exists $m>0$ such that $w(k)\ge \frac{m}{n-1}$ on $\{0,1,\dots ,n\}$, $g(k)>0$ for $k\in \{0,1,\dots ,n\}$, and $g(k)$ is symmetric on $\{0,1,\dots ,n-1\}$, h is nonnegative, symmetric on $\{0,1,\dots ,n\}$, and $0\le s<a$, where $s={\sum }_{i=1}^{n-1}h(i)$, $f:\{0,1,\dots ,n\}\times [0,+\mathrm{\infty })$ is continuous and $f(\cdot ,u)$ is symmetric on $\{0,1,\dots ,n\}$ for all $u\ge 0$.

Remark 1 The conditions that g and h are symmetric on the different sets, which can guarantee the symmetry of associated kernel function for BVP (1.4)-(1.6). The kernel functions are then used to obtain the existence of symmetric positive solutions for BVP (1.4)-(1.6) by constructing a suitable operator.

In order to study the existence of symmetric positive solutions of problem (1.4)-(1.6), we need the following lemmas.

Lemma 1.1 [16]

Let P be a cone of the real Banach space E and Ω be a bounded open subset of E and $\theta \in \mathrm{\Omega }$. Assume $A:P\cap \overline{\mathrm{\Omega }}\to P$ is a completely continuous operator and satisfies $Au=\mu u$, $u\in P\cap \partial \mathrm{\Omega }$, $\mu <1$. Then $i(A,P\cap \mathrm{\Omega },P)=1$.

Lemma 1.2 [16]

Suppose $A:P\cap \overline{\mathrm{\Omega }}\to P$ is a completely continuous operator, and satisfies

• (1) ${inf}_{u\in P\cap \partial \mathrm{\Omega }}\parallel Au\parallel >0$;

• (2) $Au=\mu u$, $u\in P\cap \partial \mathrm{\Omega }$, $\mu \notin (0,1]$.

Then $i(A,P\cap \mathrm{\Omega },P)=0$.

Lemma 1.3 (Hölder)

Suppose $u=\{u_1,u_2,\dots ,u_n\}$ is a real-valued column, let

$${\parallel u\parallel }_p=\{\begin{array}{ll}{({\sum }_{k=1}^n{|u_k|}^p)}^{1/p},& 0<p<\mathrm{\infty },\\ {sup}_{k\in \{1,2,\dots ,n\}}|u_k|,& p=\mathrm{\infty },\end{array}$$

where p, q satisfy the condition $\frac{1}{p}+\frac{1}{q}=1$, which are called conjugate exponents, and $q=\mathrm{\infty }$ for $p=1$. If $1\le p\le \mathrm{\infty }$, then

$${\parallel uv\parallel }_1\le {\parallel u\parallel }_p{\parallel v\parallel }_q,$$

which can be denoted as

$$\sum _{k=1}^n|u_k{v}_k|\le \{\begin{array}{ll}{({\sum }_{k=1}^n{|u_k|}^p)}^{1/p}{({\sum }_{k=1}^n{|v_k|}^q)}^{1/q},& 1<p<\mathrm{\infty },\\ ({\sum }_{k=1}^n|u_k|)({sup}_{k\in \{1,2,\dots ,n\}}|v_k|),& p=1,\\ ({sup}_{k\in \{1,2,\dots ,n\}}|u_k|)({\sum }_{k=1}^n|v_k|),& p=\mathrm{\infty }.\end{array}$$

2 Preliminaries

Let $E=\{u(k):\{0,1,\dots ,n\}\to \mathbb{R}\}$. It is well known that E is a real Banach space with the norm $\parallel \cdot \parallel$ defined by $\parallel u\parallel ={max}_{k\in \{0,1,\dots ,n\}}|u(k)|$. Let K be a cone of E,

$$K_r=\{u\in K:\parallel u\parallel \le r\},\partial K_r=\{u\in K:\parallel u\parallel =r\},$$

where $r>0$.

In our main results, we will use the following lemmas.

Lemma 2.1 Assume that (A₁) holds. Then for any $y\in E$, the BVP

$$-\mathrm{\Delta }(g(k-1)\mathrm{\Delta }u(k-1))=y(k),k\in \{1,\dots ,n-1\},$$

(2.1)

$$au(0)-bg(0)\mathrm{\Delta }u(0)=\sum _{i=1}^{n-1}h(i)u(i),$$

(2.2)

$$au(n)+bg(n-1)\mathrm{\Delta }u(n-1)=\sum _{i=1}^{n-1}h(i)u(i)$$

(2.3)

has a unique solution u given by

$$u(k)=\sum _{i=1}^{n-1}H(k,i)y(i),$$

where

$$H(k,i)=G(k,i)+\frac{1}{a-s}\sum _{\tau =1}^{n-1}G(\tau ,i)h(\tau ),$$

(2.4)

$$G(k,i)=\frac{1}{\mathrm{\Delta }}\{\begin{array}{ll}(b+a{\sum }_{j=k}^{n-1}\frac{1}{g(j)})(b+a{\sum }_{j=0}^{i-1}\frac{1}{g(j)}),& 0\le i<k,\\ (b+a{\sum }_{j=i}^{n-1}\frac{1}{g(j)})(b+a{\sum }_{j=0}^{k-1}\frac{1}{g(j)}),& k\le i\le n,\end{array}$$

(2.5)

and $\mathrm{\Delta }=2ab+a^2{\sum }_{j=0}^{n-1}\frac{1}{g(j)}$, $s={\sum }_{i=1}^{n-1}h(i)$.

Proof From the properties of the difference operator, it is easy to see that

$$-g(k)\mathrm{\Delta }u(k)+g(k-1)\mathrm{\Delta }u(k-1)=y(k),$$

then we have

$$\begin{array}{c}-g(1)\mathrm{\Delta }u(1)+g(0)\mathrm{\Delta }u(0)=y(1),\hfill \\ -g(2)\mathrm{\Delta }u(2)+g(1)\mathrm{\Delta }u(1)=y(2),\hfill \\ \cdots \hfill \\ -g(k)\mathrm{\Delta }u(k)+g(k-1)\mathrm{\Delta }u(k-1)=y(k).\hfill \end{array}$$

From the above equalities, we can obtain

$$-g(k)\mathrm{\Delta }u(k)+g(0)\mathrm{\Delta }u(0)=\sum _{i=1}^{k}y(i).$$

Let $g(0)\mathrm{\Delta }u(0)=A$, then

$$\mathrm{\Delta }u(k)=\frac{1}{g(k)}A-\frac{1}{g(k)}\sum _{i=1}^{k}y(i),$$

that is,

$$u(k+1)-u(k)=\frac{1}{g(k)}A-\frac{1}{g(k)}\sum _{i=1}^{k}y(i).$$

So,

$$\begin{array}{c}u(1)-u(0)=\frac{1}{g(0)}A,\hfill \\ u(2)-u(1)=\frac{1}{g(1)}A-\frac{1}{g(1)}\sum _{i=1}^{1}y(i),\hfill \\ u(3)-u(2)=\frac{1}{g(2)}A-\frac{1}{g(2)}\sum _{i=1}^{2}y(i),\hfill \\ \cdots \hfill \\ u(k)-u(k-1)=\frac{1}{g(k-1)}A-\frac{1}{g(k-1)}\sum _{i=1}^{k-1}y(i).\hfill \end{array}$$

It follows that

$$u(k)=u(0)+A\sum _{j=0}^{k-1}\frac{1}{g(j)}-\sum _{j=1}^{k-1}\frac{1}{g(j)}\sum _{i=1}^{j}y(i).$$

By the boundary conditions, we get

$$\begin{array}{c}au(0)-bA=\sum _{i=1}^{n-1}h(i)u(i),\hfill \\ au(0)+(b+a\sum _{j=0}^{n-1}\frac{1}{g(j)})A=\sum _{i=1}^{n-1}h(i)u(i)+a\sum _{j=1}^{n-1}\frac{1}{g(j)}\sum _{i=1}^{j}y(i)+b\sum _{i=1}^{n-1}y(i).\hfill \end{array}$$

Then

$$\begin{array}{c}A=\frac{1}{2b+a{\sum }_{j=0}^{n-1}\frac{1}{g(j)}}(a\sum _{j=1}^{n-1}\frac{1}{g(j)}\sum _{i=1}^{j}y(i)+b\sum _{i=1}^{n-1}y(i)),\hfill \\ u(0)=\frac{b}{2ab+a^2{\sum }_{j=0}^{n-1}\frac{1}{g(j)}}(a\sum _{j=1}^{n-1}\frac{1}{g(j)}\sum _{i=1}^{j}y(i)+b\sum _{i=1}^{n-1}y(i))+\frac{1}{a}\sum _{i=1}^{n-1}h(i)u(i).\hfill \end{array}$$

Thus,

$$\begin{array}{rcl}u(k)& =& \frac{1}{a}\sum _{i=1}^{n-1}h(i)u(i)+\frac{b}{2ab+a^2{\sum }_{j=0}^{n-1}\frac{1}{g(j)}}(a\sum _{j=1}^{n-1}\frac{1}{g(j)}\sum _{i=1}^{j}y(i)+b\sum _{i=1}^{n-1}y(i))\\ +\sum _{j=0}^{k-1}\frac{1}{g(j)}\cdot \frac{1}{2b+a{\sum }_{j=0}^{n-1}\frac{1}{g(j)}}(a\sum _{j=1}^{n-1}\frac{1}{g(j)}\sum _{i=1}^{j}y(i)+b\sum _{i=1}^{n-1}y(i))\\ -\sum _{j=1}^{k-1}\frac{1}{g(j)}\sum _{i=1}^{j}y(i)\\ =& \frac{1}{a}\sum _{i=1}^{n-1}h(i)u(i)+\sum _{i=1}^{n-1}G(k,i)y(i),\end{array}$$

where $G(k,i)$ is defined by (2.5). Multiplying the above equation with $h(k)$, and summing from 1 to $n-1$, we can get

$$\sum _{i=1}^{n-1}h(i)u(i)=\frac{a}{a-{\sum }_{k=2}^{n-1}h(k)}\sum _{k=1}^{n-1}h(k)\sum _{i=1}^{n-1}G(k,i)y(i).$$

One deduces that

$$\begin{array}{rcl}u(k)& =& \sum _{i=1}^{n-1}G(k,i)y(i)+\frac{1}{a-{\sum }_{k=1}^{n-1}h(k)}\sum _{k=1}^{n-1}h(k)\sum _{i=1}^{n-1}G(k,i)y(i)\\ =& \sum _{i=1}^{n-1}H(k,i)y(i),\end{array}$$

where $H(k,i)$ is defined by (2.4). The proof is complete. □

From the above work, we can prove that $H(k,i)$ and $G(k,i)$ have the following properties.

Proposition 2.1 If (A₁) holds, then we have

$$H(k,i)>0,G(k,i)>0,\mathit{\text{for }}k,i\in \{0,1,\dots ,n\};$$

(2.6)

$$G(n-k,n-i)=G(k,i),H(n-k,n-i)=H(k,i),\mathit{\text{for }}k,i\in \{0,1,\dots ,n\};$$

(2.7)

$$\frac{1}{\mathrm{\Delta }}{b}^2\le G(k,i)\le G(i,i)\le \frac{1}{\mathrm{\Delta }}D,\frac{1}{\mathrm{\Delta }}a{b}^2\gamma \le H(k,i)\le H(i,i)\le \frac{1}{\mathrm{\Delta }}a\gamma D,$$

(2.8)

where $D={(b+a{\sum }_{j=0}^n\frac{1}{g(j)})}^2$, $\gamma =\frac{1}{a-s}$, $k,i\in \{0,1,\dots ,n\}$.

Proof It is clear that (2.6) holds. Now we prove (2.7) holds.

If $i\in \{0,1,\dots ,k-1\}$, then $n-i\ge n-k$, from (2.5) and (A₁) we get

$$\begin{array}{rcl}G(n-k,n-i)& =& \frac{1}{\mathrm{\Delta }}(b+a\sum _{j=n-i}^{n-1}\frac{1}{g(j)})(b+a\sum _{j=0}^{n-k-1}\frac{1}{g(j)})\\ =& \frac{1}{\mathrm{\Delta }}(b+a\sum _{j=n-i}^{n-1}\frac{1}{g(n-1-j)})(b+a\sum _{j=0}^{n-k-1}\frac{1}{g(n-1-j)})\\ =& \frac{1}{\mathrm{\Delta }}(b+a\sum _{j=0}^{i-1}\frac{1}{g(j)})(b+a\sum _{j=k}^{n-1}\frac{1}{g(j)})\\ =& G(k,i),i\in \{0,1,\dots ,k-1\}.\end{array}$$

Similarly, we can prove that $G(n-k,n-i)=G(k,i)$, $i\in \{k,\dots ,n\}$. So we have $G(n-k,n-i)=G(k,i)$, for $k,i\in \{0,1,\dots ,n\}$. From (2.4) and (A₁), we have

$$\begin{array}{rcl}H(n-k,n-i)& =& G(n-k,n-i)+\frac{1}{a-s}\sum _{\tau =1}^{n-1}G(\tau ,n-i)h(\tau )\\ =& G(k,i)+\frac{1}{a-s}\sum _{\tau =1}^{n-1}G(n-\tau ,i)h(n-\tau )\\ =& G(k,i)+\frac{1}{a-s}\sum _{\tau =1}^{n-1}G(\tau ,i)h(\tau )\\ =& H(k,i).\end{array}$$

So, (2.7) is established. Next we prove (2.8) holds. In fact, for $k,i\in \{0,1,\dots ,n\}$, if $i\in \{0,1,\dots ,k-1\}$, then

$$\begin{array}{rcl}G(k,i)& =& \frac{1}{\mathrm{\Delta }}(b+a\sum _{j=k}^{n-1}\frac{1}{g(j)})(b+a\sum _{j=0}^{i-1}\frac{1}{g(j)})\\ \le & \frac{1}{\mathrm{\Delta }}(b+a\sum _{j=i}^{n-1}\frac{1}{g(j)})(b+a\sum _{j=0}^{i-1}\frac{1}{g(j)})\\ =& G(i,i)\\ \le & \frac{1}{\mathrm{\Delta }}(b+a\sum _{j=0}^n\frac{1}{g(j)})(b+a\sum _{j=0}^n\frac{1}{g(j)})\\ \le & \frac{1}{\mathrm{\Delta }}{(b+a\sum _{j=0}^n\frac{1}{g(j)})}^2\\ =& \frac{1}{\mathrm{\Delta }}D.\end{array}$$

Similarly, we can prove that $G(k,i)\le G(i,i)\le \frac{1}{\mathrm{\Delta }}D$, for $i\in \{k,k+1,\dots ,n\}$. Therefore $G(k,i)\le G(i,i)\le \frac{1}{\mathrm{\Delta }}D$. For $k,i\in \{0,1,\dots ,n\}$, we can get

$$\begin{array}{rcl}H(k,i)& =& G(k,i)+\frac{1}{a-s}\sum _{\tau =1}^{n-1}G(\tau ,i)h(\tau )\\ \le & G(i,i)+\frac{1}{a-s}\sum _{\tau =1}^{n-1}G(\tau ,i)h(\tau )\\ =& H(i,i)\\ \le & G(i,i)+\frac{1}{a-s}\sum _{\tau =1}^{n-1}G(\tau ,\tau )h(\tau )\\ \le & \frac{1}{\mathrm{\Delta }}D+\frac{1}{\mathrm{\Delta }}D\frac{1}{a-s}\sum _{\tau =1}^{n-1}h(\tau )\\ =& \frac{1}{\mathrm{\Delta }}(1+\frac{1}{a-s}\sum _{\tau =1}^{n-1}h(\tau ))D\\ =& \frac{a}{\mathrm{\Delta }(a-s)}D=\frac{1}{\mathrm{\Delta }}a\gamma D.\end{array}$$

On the other hand, from (2.5), we have

$$G(k,i)\ge \frac{1}{\mathrm{\Delta }}(b+a\sum _{j=n}^{n-1}\frac{1}{g(j)})(b+a\sum _{j=0}^{-1}\frac{1}{g(j)})=\frac{1}{\mathrm{\Delta }}{b}^2.$$

So, by (2.4), for $k,i\in \{0,1,\dots ,n\}$, we can obtain

$$\begin{array}{rcl}H(k,i)& =& G(k,i)+\frac{1}{a-s}\sum _{\tau =1}^{n-1}G(\tau ,i)h(\tau )\\ \ge & \frac{1}{\mathrm{\Delta }}{b}^2+\frac{b^2}{(a-s)\mathrm{\Delta }}\sum _{\tau =1}^{n-1}h(\tau )\\ \ge & \frac{1}{\mathrm{\Delta }}{b}^2(1+\frac{1}{a-s}\sum _{\tau =1}^{n-1}h(\tau ))\\ =& \frac{1}{\mathrm{\Delta }}{b}^2\frac{a}{a-s}=\frac{1}{\mathrm{\Delta }}{b}^{2}a\gamma .\end{array}$$

Thus,

$$\begin{array}{c}\frac{1}{\mathrm{\Delta }}{b}^2\le G(k,i)\le G(i,i)\le \frac{1}{\mathrm{\Delta }}D,\hfill \\ \frac{1}{\mathrm{\Delta }}{b}^{2}a\gamma \le H(k,i)\le H(i,i)\le \frac{1}{\mathrm{\Delta }}Da\gamma .\hfill \end{array}$$

The proof is completed. □

Remark 2 The symmetry of $g(k)$ on $\{0,1,\dots ,n-1\}$ can guarantee that $G(k,i)$ is symmetric for $k,i\in \{0,1,\dots ,n\}$, and the symmetry of $h(k)$ on $\{0,1,\dots ,n\}$ can guarantee that $H(k,i)$ is symmetric for $k,i\in \{0,1,\dots ,n\}$.

Next, we can construct a cone in E by

$$\begin{array}{rcl}K& =& \{u\in E:u\ge 0,u(k)\text{ is symmetric on }\{0,1,\dots ,n\},\mathrm{\Delta }(g(k)\mathrm{\Delta }u(k))\le 0,\\ k\in \{0,1,\dots ,n-2\},\text{ and }\underset{k\in \{0,1,\dots ,n\}}{min}u(k)\ge {\delta }_{\ast }\parallel u\parallel \},\end{array}$$

where ${\delta }_{\ast }=\frac{1}{D}{b}^2$. Then we define an operator

$$(Tu)(k)=\sum _{i=1}^{n-1}H(k,i)w(i)f(i,u(i)).$$

(2.9)

It can be observed that u is a solution of problem (1.4)-(1.6) if and only if u is a fixed point of operator T.

We can get the following lemma from Lemma 2.1.

Lemma 2.2 Suppose (A₁) holds. If u is a solution of the equation

$$u(k)=Tu(k)=\sum _{i=1}^{n-1}H(k,i)w(i)f(i,u(i)),$$

then u is a solution of BVP (1.4)-(1.6).

Lemma 2.3 Assume (A₁) holds. Then $T(K)\subset K$ and $T:K\to K$ is completely continuous.

Proof For $u\in K$, from (2.9), we obtain $\mathrm{\Delta }(g(k-1)\mathrm{\Delta }Tu(k-1))=-w(k)f(k,u(k))\le 0$. By Proposition 2.1, it is to see that $(Tu)(k)\ge 0$, for $k\in \{0,1,\dots ,n\}$. Using the fact that w, u, $f(k,u)$ are symmetric on $\{0,1,\dots ,n\}$, we have

$$\begin{array}{rcl}(Tu)(n-k)& =& \sum _{i=1}^{n-1}H(n-k,i)w(i)f(i,u(i))\\ =& \sum _{i=1}^{n-1}H(k,n-i)w(n-i)f(n-i,u(n-i))\\ =& \sum _{i=1}^{n-1}H(k,i)w(i)f(i,u(i))\\ =& (Tu)(k),\end{array}$$

then Tu is symmetric on $\{0,1,\dots ,n\}$ for $k\in \{0,1,\dots ,n\}$. And from (2.8) we can see

$$(Tu)(k)=\sum _{i=1}^{n-1}H(k,i)w(i)f(i,u(i))\le \frac{1}{\mathrm{\Delta }}a\gamma D\sum _{i=1}^{n-1}w(i)f(i,u(i)).$$

Thus,

$$\parallel Tu\parallel \le \frac{1}{\mathrm{\Delta }}a\gamma D\sum _{i=1}^{n-1}w(i)f(i,u(i)).$$

Similarly, by (2.8) we obtain

$$\begin{array}{rcl}(Tu)(k)& =& \sum _{i=1}^{n-1}H(k,i)w(i)f(i,u(i))\\ \ge & \frac{1}{\mathrm{\Delta }}a{b}^2\gamma \sum _{i=1}^{n-1}w(i)f(i,u(i))\\ =& \frac{1}{\mathrm{\Delta }}a{\delta }_{\ast }D\gamma \sum _{i=1}^{n-1}w(i)f(i,u(i))\\ \ge & {\delta }_{\ast }\parallel Tu\parallel .\end{array}$$

Thus, $Tu\in K$ and $T(K)\subset K$. It is clear that $T:K\to K$ is completely continuous. □

Remark 3 The symmetry of the kernel function $H(k,i)$ for $k,i\in \{0,1,\dots ,n\}$ can guarantee that Tu is symmetric on $\{0,1,\dots ,n\}$ for $u\in K$.

3 Main results

In this section, we will establish that problem (1.4)-(1.6) has at least one positive solution with Lemma 1.1 and Lemma 1.2. We need consider the following situations: $p>1$, $p=1$, $p=\mathrm{\infty }$. Next, we will prove a theorem for $p>1$. At first, we define

$$\parallel H\parallel =\underset{i\in \{1,2,\dots ,n-1\}}{sup}|H(i,i)|,{\parallel H\parallel }_p={\left(\sum _{i=1}^{n-1}{|H(i,i)|}^p\right)}^{1/p}.$$

Let

$$F^{\beta }=\underset{u\to \beta }{lim}sup\underset{k\in \{0,1,\dots ,n\}}{max}\frac{f(k,u)}{u},F_{\beta }=\underset{u\to \beta }{lim}inf\underset{k\in \{0,1,\dots ,n\}}{min}\frac{f(k,u)}{u},$$

where β denotes 0 or ∞, and

$$\begin{array}{c}\begin{array}{rl}{N}^{-1}=& max\{{\parallel H\parallel }_p{\left(\sum _{i=1}^{n-1}{|w(i)|}^q\right)}^{1/q},\\ \left(\sum _{i=1}^{n-1}|H(i,i)|\right)\left(\underset{i\in \{1,2,\dots ,n-1\}}{sup}|w(i)|\right),\parallel H\parallel \left(\sum _{i=1}^{n-1}|w(i)|\right)\},\end{array}\hfill \\ L^{-1}=\frac{1}{\mathrm{\Delta }}{\delta }_{\ast }a\gamma m{b}^2.\hfill \end{array}$$

Theorem 3.1 Assume that conditions (A₁) hold. In addition, suppose that

(A₂) $0<F^0<N$, and $L<F_{\mathrm{\infty }}<\mathrm{\infty }$, or

(A₃) $0<F^{\mathrm{\infty }}<N$, and $L<F_0<\mathrm{\infty }$

are satisfied. Then problem (1.4)-(1.6) has at least one symmetric positive solution.

Proof We only consider (A₂) case, (A₃) is similar to (A₂). If $0<F^0<N$, then there exist $r>0$, ${\epsilon }_0>0$ such that $N-{\epsilon }_0>0$ and for all $0<u\le r$, we have

$$f(k,u)\le (N-{\epsilon }_0)u\le (N-{\epsilon }_0)r,k\in \{0,1,\dots ,n\}.$$

(3.1)

For all $u\in \partial K_r$, from Lemma 1.3 we obtain

$$\begin{array}{rcl}(Tu)(k)& =& \sum _{i=1}^{n-1}H(k,i)w(i)f(i,u(i))\\ \le & \sum _{i=1}^{n-1}H(k,i)w(i)(N-{\epsilon }_0)r\\ \le & \sum _{i=1}^{n-1}H(i,i)w(i)(N-{\epsilon }_0)r\\ \le & {\left(\sum _{i=1}^{n-1}{|H(i,i)|}^p\right)}^{1/p}{\left(\sum _{i=1}^{n-1}{|w(i)|}^q\right)}^{1/q}(N-{\epsilon }_0)r\\ \le & N^{-1}(N-{\epsilon }_0)r\\ \le & r.\end{array}$$

So $Tu\ne \lambda u$, for $\mathrm{\forall }u\in \partial K_r$, $\lambda \ge 1$. From Lemma 1.1, we can get $i(T,K_r,K)=1$. Next, we prove it satisfies Lemma 1.2. Because $L<F_{\mathrm{\infty }}<\mathrm{\infty }$, there exist $R>{\delta }_{\ast }r>0$, ${\epsilon }_1>0$ such that

$$f(k,u)\ge (L+{\epsilon }_1)u,\mathrm{\forall }u\ge R,k\in \{0,1,\dots ,n\}.$$

Let $r^{\ast }={\delta }_{\ast }^{-1}R$, then $r^{\ast }>r$, and

$$\underset{k\in \{0,1,\dots ,n\}}{min}u(k)\ge {\delta }_{\ast }\parallel u\parallel =R,\mathrm{\forall }u\in \partial K_r.$$

Now we prove that $Tu\ne \lambda u$, $\mathrm{\forall }u\in \partial K_r$, $0<\lambda \le 1$. If not, then there exist $u_0\in \partial K_{r^{\ast }}$ and $0<{\lambda }_0\le 1$ such that $T{u}_0={\lambda }_0{u}_0$; thus we have

$$\begin{array}{rcl}{r}^{\ast }& \ge & u_0(k)={\lambda }_0^{-1}(T{u}_0)(k)\\ =& {\lambda }_0^{-1}\sum _{i=1}^{n-1}H(k,i)w(i)f(i,u(i))\\ \ge & \frac{1}{\mathrm{\Delta }}a{b}^2\gamma (L+{\epsilon }_1)\sum _{i=1}^{n-1}w(i)u(i)\\ \ge & \frac{1}{\mathrm{\Delta }}a{b}^2\gamma (L+{\epsilon }_1)R\sum _{i=1}^{n-1}w(i)\\ =& \frac{1}{\mathrm{\Delta }}a{b}^2\gamma (L+{\epsilon }_1){\delta }_{\ast }{r}^{\ast }\sum _{i=1}^{n-1}w(i)\\ \ge & \frac{1}{\mathrm{\Delta }}a{b}^2\gamma (L+{\epsilon }_1){\delta }_{\ast }{r}^{\ast }m\\ =& L^{-1}(L+{\epsilon }_1)r^{\ast }\\ =& r^{\ast }(1+\frac{{\epsilon }_1}{L})>r^{\ast },\end{array}$$

i.e., $r^{\ast }>r^{\ast }$, which is a contradiction. In addition, because $(Tu)(k)\ge r^{\ast }(1+\frac{{\epsilon }_1}{L})>r^{\ast }$, so ${inf}_{u\in \partial K_{r^{\ast }}}\parallel Tu\parallel \ge r^{\ast }>0$, from Lemma 1.2 we have $i(T,K_{r^{\ast }},K)=0$. On the other hand, from the above work with the additivity of the fixed point index, we get

$$i(T,K_{r^{\ast }}-\overline{K_r},K)=i(T,K_{r^{\ast }},K)-i(T,K_r,K)=0-1=-1.$$

So, T has at least one fixed point $u^{\ast }$ on $K_{r^{\ast }}-\overline{K_r}$. Then it follows that problem (1.4)-(1.6) has a symmetric positive solution $u^{\ast }$. The proof is complete. □

Remark 4 From the proof of Theorem 3.1, we can establish that problem (1.4)-(1.6) has another nonnegative solution $u^{\ast \ast }$, $u^{\ast \ast }\in K_r$.

The following corollary deals with the case $p=1$.

Corollary 3.1 Suppose that (A₁), (A₂) hold. Then problem (1.4)-(1.6) has at least one symmetric positive solution.

Proof It is similar to the proof of Theorem 3.1. Let $({\sum }_{i=1}^{n-1}|H(i,i)|)({sup}_{i\in \{1,\dots ,n-1\}}|w(i)|)$ replace ${({\sum }_{i=1}^{n-1}{|H(i,i)|}^p)}^{1/p}{({\sum }_{i=1}^{n-1}{|w(i)|}^q)}^{1/q}$ and repeat the argument of Theorem 3.1. □

Finally, we consider the case of $p=\mathrm{\infty }$.

Corollary 3.2 Assume that (A₁), (A₂) hold. Then problem (1.4)-(1.6) has at least one symmetric positive solution.

Proof It is similar to the proof of Theorem 3.1. For all $u\in \partial K_r$, we have

$$\begin{array}{rcl}(Tu)(k)& =& \sum _{i=1}^{n-1}H(k,i)w(i)f(i,u(i))\\ \le & \sum _{i=1}^{n-1}H(i,i)w(i)(N-{\epsilon }_0)r\\ \le & \left(\underset{i\in \{1,2,\dots ,n-1\}}{sup}|H(i,i)|\right)\left(\sum _{i=1}^{n-1}|w(i)|\right)(N-{\epsilon }_0)r\\ \le & N^{-1}(N-{\epsilon }_0)r\\ <& r.\end{array}$$

So $Tu\ne \lambda u$, $u\in \partial K_r$, $\lambda \ge 1$. By Lemma 1.1, we can get $i(T,K_r,K)=1$. This together with $i(T,K_{r^{\ast }},K)=0$ in the proof of Theorem 3.1 completes the proof. □