Existence of positive solutions of discrete third-order three-point BVP with sign-changing Green’s function

Abstract

Consider positive solutions and multiple positive solutions for a discrete nonlinear third-order boundary value problem

$$ \textstyle\begin{cases} \Delta ^{3}u(t-1)=a(t)f(t,u(t)), \quad t\in [1,T-2]_{\mathbb{Z}},\\ \Delta u(0)=u(T)=0,\qquad \Delta ^{2} u(\eta )-\alpha \Delta u(T-1)=0, \end{cases} $$

which has the sign-changing Green’s function. Here $T>8$ is a positive integer, $[1,T-1]_{\mathbb{Z}}=\{1,2,\dots ,T-2\}$, $\alpha \in [0, \frac{1}{T-1})$, $a:[0,T-2]_{\mathbb{Z}}\to (0,+\infty )$, $f:[1,T-2]_{ \mathbb{Z}}\times [0,\infty )\to [0,\infty )$ is continuous.

1 Introduction

Multi-point boundary value problems for differential equations have a wide application in computational physics, economics, and modern biological fields [1]. In 1992, Gupta [2] studied solvability of differential equation three-point boundary value problem. Soon afterwards, there arose many results on multi-point nonlinear boundary value problems [3,4,5,6]. In 1999, Ma [7] studied the existence of positive solution for a second-order differential equation three-point boundary value problem. Thereafter, many results for the existence of positive solutions on multi-point boundary value problems have been studied [7,8,9,10,11,12,13,14,15,16,17,18,19]. With the development of the computing science and the computer simulation, multi-point boundary value problems should be discretized, so we need to study corresponding difference equation [20,21,22,23,24,25,26]. In 1998, by using Krasnosel’skii’s fixed point theorem, Agarwal and Henderson [24] studied the discrete problem

$$ \textstyle\begin{cases} \Delta ^{3} u(t-1)=\lambda a(t)f(t,u(t)),\quad t\in [2,T]_{\mathbb{Z}}, \\ u(0)= u(1)= u(T+3)=0. \end{cases} $$

They obtained the existence of positive solutions in two cases for $\lambda =1$ and $\lambda \neq 1$. Later, there were many interesting results on the positive solutions to the discrete boundary value problems, see, for instance, [23,24,25,26] and the references therein. It is noted that Green’s functions are positive in most of these results. However, when the Green’s function is sign-changing, could we also obtain the existence of positive solutions to these kinds of problems?

In 2015, by using the Guo–Krasnosel’skii fixed point theorem, Wang and Gao [25] studied the existence of positive solutions to the discrete third-order three-point boundary value problem

$$ \textstyle\begin{cases} \Delta ^{3} u(t-1)+a(t)f(t,u(t))=0,\quad t\in [1,T-1]_{\mathbb{Z}},\\ u(0)=\Delta u(T)=\Delta ^{2} u(\eta )=0, \end{cases} $$

where $\eta \in [1,[\frac{3T^{2}-3T-2}{6T+3}]]_{\mathbb{Z}}$ and the Green’s function is sign-changing. In 2016, Geng and Gao [26], by using the Guo–Krasnosel’skii fixed point theorem, studied the discrete third-order three-point boundary value problem

$$ \textstyle\begin{cases} \Delta ^{3} u(t-1)+\lambda a(t)f(t,u(t))=0,\quad t\in [1,T-1]_{ \mathbb{Z}},\\ u(0)=\Delta u(t)=\Delta ^{2} u(\eta )=0, \end{cases} $$

when f satisfies some superlinear and sublinear condition on 0 and ∞. For the continuous case, which has the sign-changing Green’s function, one can see [27,28,29] and the references therein. Inspired by the above works, in this paper, we consider the existence and multiple positive solutions to the following discrete nonlinear third-order three-point BVP:

$$ \textstyle\begin{cases} \Delta ^{3}u(t-1)=a(t)f(t,u(t)),\quad t\in [1,T-2]_{\mathbb{Z}},\\ \Delta u(0)=u(T)=0,\qquad \Delta ^{2} u(\eta )-\alpha \Delta u(T-1)=0, \end{cases} $$

(1.1)

where $T>8$ is a positive integer, $\alpha \in [0,\frac{1}{T-1})$, $a:[1,T-2]_{\mathbb{Z}}\to (0,+\infty )$, $f:[1,T-2]_{\mathbb{Z}} \times [0,\infty )\to [0,\infty )$ is continuous, η satisfies

$(H_{0})$ :

$\eta \in [\lfloor \frac{T-2}{2}\rfloor +1,T-2]_{ \mathbb{Z}}$.

Under assumption $(H_{0})$, the Green’s function of (1.1) changes its sign. The proof of our main results is based upon the following well-known Guo–Krasnoselskii fixed point theorems [30].

Theorem 1.1

Let E be a Banach space and $K\subset E$ be a cone. Assume that $\varOmega _{1}$, $\varOmega _{2}$ are open bounded subsets of E with $0\in \varOmega _{1}$, $\overline{\varOmega }_{1}\subset \varOmega _{2}$. If

$$ A:K\cap (\overline{\varOmega }_{2}\setminus \varOmega _{1}) \to K $$

is a completely continuous operator such that

• (i) $\|Au\|\leq \|u\|$, $u\in K\cap \partial \overline{\varOmega }_{1}$, and $\|Au\|\geq \|u\|$, $u\in K\cap \partial \overline{\varOmega }_{2}$,

or

• (ii) $\|Au\|\geq \|u\|$, $u\in K\cap \partial \overline{\varOmega }_{1}$, and $\|Au\|\leq \|u\|$, $u\in K\cap \partial \overline{\varOmega }_{2}$,

then A has a fixed point in $K\cap (\overline{\varOmega }_{2}\setminus \varOmega _{1})$.

Theorem 1.2

Let E be a Banach space and K be a cone in E. For some $p>0$, define $K_{p}=\{x\in K | \|x\|\leq p\}$. Assume that $A:K_{p}\to K$ is a compact operator; if $x\in \partial K_{p}=\{x \in k | \|x\|=p\}$, $Ax\neq x$, we have

• (i) For $\forall x\in \partial K_{p}$, if $\|Ax\|\geq \|x\|$, then $i(A,K_{p},K)=0$,

• (ii) For $\forall x\in \partial K_{p}$, if $\|Ax\|\leq \|x\|$, then $i(A,K_{p},K)=1$.

2 Preliminaries

Lemma 2.1

Suppose that $(H_{0})$ holds. Then the linear problem

$$ \textstyle\begin{cases} \Delta ^{3}u(t-1)=y(t), \quad t\in [1,T-2]_{\mathbb{Z}},\\ \Delta u(0)=u(T)=0,\qquad \Delta ^{2} u(\eta )-\alpha \Delta u(T-1)=0 \end{cases} $$

(2.1)

has a unique solution

$$ u(t)=\sum^{T-2}_{s=1}G(t,s)y(s), $$

(2.2)

where

$$\begin{aligned}& G(t,s)=g(t,s)+k(t,s)+\textstyle\begin{cases} \frac{T(T-1)-t(t-1)}{2-2\alpha (T-1)}, &s\leq \eta ,\\ 0,&\eta < s, \end{cases}\displaystyle \\& g(t,s)=-\frac{\alpha (T-s-1)[T(T-1)-t(t-1)]}{2-2\alpha (T-1)}- \frac{(T-s)(T-s-1)}{2}, \\& k(t,s)=\textstyle\begin{cases} \frac{(t-s)(t-s-1)}{2}, &0< s\leq t-2\leq T-2,\\ 0,&0\leq t-2< s \leq T-2, \end{cases}\displaystyle \end{aligned}$$

and

$$ G(0,s)=G(1,s)=\textstyle\begin{cases} \frac{-\alpha T(T-1)(T-s-1)+T(T-1)}{2-2\alpha (T-1)} - \frac{(T-1)(T-s-1)}{2}, & s\leq \eta ,\\ \frac{-\alpha T(T-1)(T-s-1)}{2-2 \alpha (T-1)}-\frac{(T-1)(T-s-1)}{2}, &\eta < s. \end{cases} $$

Proof

Summing both sides of equation (2.1) from $s=1$ to $s=t-1$, we get

$$ \Delta ^{2}u(t-1)=\Delta ^{2}u(0)+\sum ^{t-1}_{s=1}y(s), $$

and

$$ \Delta u(t-1)=(t-1)\Delta ^{2}u(0)+\sum^{t-2}_{s=1}(t-s-1)y(s). $$

Then summing both sides from $\tau =1$ to $\tau =t$, we get

$$ u(t)=u(0)+\frac{t(t-1)}{2}\Delta ^{2}u(0)+\sum ^{t-2}_{s=1} \frac{(t-s)(t-s-1)}{2}y(s). $$

From boundary conditions $\Delta u(0)=u(T)=0$, $\Delta ^{2} u(\eta )- \alpha \Delta u(T-1)=0$, we have

$$ \textstyle\begin{cases} u(0)+\frac{T(T-1)}{2}\Delta ^{2}u(0)+\sum^{T-2}_{s=1} \frac{(T-s)(T-s-1)}{2}y(s)=0;\\ \Delta ^{2}u(0)+\sum^{\eta }_{s=1}y(s)- \alpha (T-1)\Delta ^{2}u(0)-\alpha \sum^{T-2}_{s=1}(T-s-1)y(s)=0. \end{cases} $$

Furthermore, we get

$$ \textstyle\begin{cases} u(0)=\sum^{T-2}_{s=1}\frac{-\alpha T(T-1)(T-s-1)}{2-2\alpha (T-1)}y(s) +\sum^{\eta }_{s=1}\frac{T(T-1)}{2-2\alpha (T-1)}y(s)\\ \hphantom{u(0)=}{} -\sum^{T-2}_{s=1}\frac{(T-s)(T-s-1)}{2}y(s),\\ \Delta ^{2}u(0)=\sum^{T-2}_{s=1}\frac{\alpha (T-s-1)}{1-\alpha (T-1)}y(s)-\sum^{\eta } _{s=1}\frac{1}{1-\alpha (T-1)}y(s). \end{cases} $$

Then we have

$$\begin{aligned} u(t) = &\sum^{T-2} _{s=1} \biggl(- \frac{\alpha (T-s-1)[T(T-1)-t(t-1)]}{2-2 \alpha (T-1)} -\frac{(T-s)(T-s-1)}{2} \biggr)y(s) \\ &{}+\sum^{t-2}_{s=1} \frac{(t-s)(t-s-1)}{2}y(s) \\ &{}+\sum^{\eta }_{s=1}\frac{T(T-1)-t(t-1)}{2-2 \alpha (T-1)}y(s). \end{aligned}$$

 □

Lemma 2.2

Suppose that ($H_{0}$) holds. Then the Green’s function $G(t,s)$ changes its sign on $[0,T]_{\mathbb{Z}}\times [0,T]_{ \mathbb{Z}}$. More precisely,

• (i) If $s\in [1,\eta ]_{\mathbb{Z}}$, then $\Delta _{t} G(t,s)\leq 0$, $G(t,s)\geq 0$ for $\forall t\in [0,T]_{\mathbb{Z}}$. Furthermore,

  $$\begin{aligned}& \min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(T,s)=0, \\& \max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(0,s)\leq \frac{T(T-1)(1+\alpha \eta )}{1-\alpha (T-1)}. \end{aligned}$$

• (ii) If $s\in [\eta +1,T-2]_{\mathbb{Z}}$, then $\Delta _{t}G(t,s) \geq 0$, $G(t,s)\leq 0$ for $\forall t\in [0,T]_{\mathbb{Z}}$. Furthermore,

  $$\begin{aligned}& \max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(T,s)=0, \\& \min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(0,s)\geq \frac{-T(T-1)(1+ \alpha \eta )}{1-\alpha (T-1)}. \end{aligned}$$

Proof

(i) If $s\in [1,\eta ]_{\mathbb{Z}}$, we have

$$ \Delta _{t}G(t,s)=\textstyle\begin{cases} \frac{\alpha t(T-s-1)-2t}{1-\alpha (T-1)}, &t-2\leq s,\\ \frac{ \alpha t(T-s-1)-2t}{1-\alpha (T-1)}+t-s,&s\leq t-2. \end{cases} $$

If $s\leq t-2$, since $\alpha (T-1)-1<0$, we get

$$\begin{aligned} \Delta _{t}G(t,s) =&\frac{2\alpha t(T-s-1)-2t}{2-2\alpha (T-1)}+t-s \\ =&\frac{2 \alpha t(T-s-1)-2t+(t-s)(2-2\alpha (T-1))}{2-2\alpha (T-1)} \\ =&\frac{-2s \alpha +2s[\alpha (T-1)-1]}{2-2\alpha (T-1)} \\ < &0. \end{aligned}$$

If $t-2\leq s$, since $\alpha <\frac{1}{T-1}$, so $2-2\alpha (T-1)>0$, $\alpha (T-s-1)-1<0$, we have

$$ \Delta _{t}G(t,s)=\frac{2\alpha t(T-s-1)-2t}{2-2\alpha (T-1)}< 0. $$

For $\forall t\in [0,T-1]_{\mathbb{Z}}$, $\Delta _{t}G(t,s)\leq 0$. If $s\in [1,\eta ]_{\mathbb{Z}}$, $G(t,s)\geq 0$. So

$$\begin{aligned}& \min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(T,s)=0, \\& \max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(0,s) \\& \hphantom{\max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)}=\frac{T(T-1)[1- \alpha (T-s-1)]}{2-2\alpha (T-1)}-\frac{(T-s)(T-s-1)}{2} \\& \hphantom{\max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)}\leq \frac{T(T-1)[1- \alpha (T-\eta -1)]}{2-2\alpha (T-1)}-\frac{(T-\eta )(T-\eta -1)}{2} \\& \hphantom{\max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)}\leq \frac{T(T-1)(1+\alpha \eta )}{1-\alpha (T-1)}. \end{aligned}$$

(ii) If $s\in [\eta +1,T]_{\mathbb{Z}}$, we have

$$ \Delta _{t} G(t,s)=\textstyle\begin{cases} \frac{2\alpha t(T-s-1)}{2-2\alpha (T-1)}, &s\geq t-2,\\ \frac{2 \alpha t (T-s-1)}{2-2\alpha (T-1)}+t-s,&s\leq t-2. \end{cases} $$

If $t-2\leq s$, since $\alpha <\frac{1}{T-1}$, $2-2\alpha (T-1)>0$, we have

$$ \Delta _{t} G(t,s)=\frac{2\alpha t(T-s-1)}{2-2\alpha (T-1)}>0. $$

If $s\leq t-2$, then $t-s>0$, so

$$ \Delta _{t} G(t,s)=\frac{2\alpha t(T-s-1)}{2-2\alpha (T-1)}+t-s>0. $$

If $s\in [\eta +1,T]_{\mathbb{Z}}$, for $\forall t\in [0,T]_{ \mathbb{Z}}$, we have

$$\begin{aligned}& \max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(T,s)=0, \\& \min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=G(0,s) \\& \hphantom{\min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)}=\frac{-\alpha T(T-1)(T-s-1)}{2-2 \alpha (T-1)}-\frac{(T-s)(T-s-1)}{2} \\& \hphantom{\min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)}\geq \frac{-\alpha T(T-1)(T- \eta -1)}{2-2\alpha (T-1)}-\frac{(T-\eta )(T-\eta -1)}{2} \\& \hphantom{\min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)}\geq \frac{-T(T-1)(1+ \alpha \eta )}{1-\alpha (T-1)}. \end{aligned}$$

In conclusion, if $s\in [1,\eta ]_{\mathbb{Z}}$, $G(t,s)$ is decreasing, then $G(t,s)\geq 0$, since $\min_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=0$; if $s\in [\eta +1,T]_{ \mathbb{Z}}$, $G(t,s)$ is increasing, then $G(t,s)\leq 0$ since $\max_{t\in [0,T]_{\mathbb{Z}}}G(t,s)=0$. So, the Green’s function $G(t,s)$ is a sign-changing function on $[0,T]_{\mathbb{Z}} \times [0,T]_{\mathbb{Z}}$. □

Remark

Now, we give a brief explanation for the reason why we choose

$$ \eta \in \biggl[ \biggl\lfloor \frac{T-2}{2} \biggr\rfloor +1,T-2 \biggr] _{\mathbb{Z}}. $$

(2.3)

Consider the problem

$$ \textstyle\begin{cases} \Delta ^{3}u(t-1)=1, \quad t\in [1,T-2]_{\mathbb{Z}},\\ \Delta u(0)=u(T)=0, \qquad \Delta ^{2} u(\eta )-\alpha \Delta u(T-1)=0. \end{cases} $$

(2.4)

From Lemma 2.1, we get

$$\begin{aligned} u(t) =&\frac{1}{12-12\alpha (T-1)} \bigl\{ 3\alpha \bigl[t(t-1)-T(T-1) \bigr](T-1) (T-2) \\ &{}+6\eta \bigl[T(T-1)-t(t-1)\bigr]-2T(T-1) (T-2)\bigl[1-\alpha (T-1)\bigr] \\ &{}+2t(t-1) (t-2)\bigl[1- \alpha (T-1)\bigr] \bigr\} \\ =&\frac{\phi (t)}{12-12\alpha (T-1)}, \end{aligned}$$

where

$$\begin{aligned} \phi (t) =&3\alpha \bigl[t(t-1)-T(T-1)\bigr](T-1) (T-2)+6\eta \bigl[T(T-1)-t(t-1)\bigr] \\ &{}-2T(T-1) (T-2)\bigl[1- \alpha (T-1)\bigr]+2t(t-1) (t-2)\bigl[1-\alpha (T-1) \bigr]. \end{aligned}$$

Clearly, $u(t)\geq 0$ is equivalent to $\phi (t)\geq 0$, and

$$ \Delta \phi (t)=6t\bigl\{ (t-1)\bigl[1-\alpha (T-1)\bigr]-2\eta +\alpha (T-1) (T-2)\bigr\} . $$

Let $\Delta \phi (t)=0$. Then $t=0$ or $t=1+\frac{2\eta -\alpha (T-1)(T-2)}{1- \alpha (T-1)}$. Therefore,

$$ \Delta \phi (t)>0\quad \Leftrightarrow\quad t>\frac{2\eta +1-\alpha (T-1)^{2}}{1- \alpha (T-1)}. $$

Now, we claim that if (2.3) holds, then $\phi (t)$ is a positive solution of (2.4). In fact, if (2.3) holds, then $\frac{2\eta +1- \alpha (T-1)^{2}}{1-\alpha (T-1)}\geq T-1$ in this case, this implies that

$$ \Delta \phi (t)\leq 0, \quad t\in [0,T-1]_{\mathbb{Z}}. $$

More precisely, $\Delta \phi (0)=0$, $\Delta \phi (t)<0$ for $t\in [1,T-2]_{\mathbb{Z}}$ and $\Delta \phi (T-1)<0$ for $\frac{2 \eta +1-\alpha (T-1)^{2}}{1-\alpha (T-1)}> T-1$ and $\Delta \phi (T-1)=0$ for $\frac{2\eta +1-\alpha (T-1)^{2}}{1-\alpha (T-1)}= T-1$. Therefore, $\phi (t)$ is a positive solution of the linear problem (2.4) since $\phi (T)=0$.

Let

$$ E=\bigl\{ u:[0,T]_{\mathbb{Z}}\to \mathbb{R}|\Delta u(0)=u(T), \Delta ^{2}u( \eta )-\alpha \Delta u(T-1)=0\bigr\} . $$

Then E is a Banach space with norm $\|u\|=\max_{t\in [0,T]_{ \mathbb{Z}}}|u(t)|$.

Let

$$ K_{0}=\bigl\{ y\in E:y(t)\geq 0,t\in [0,T]_{\mathbb{Z}}; \Delta y(t) \leq 0, t\in [0,T-1]_{\mathbb{Z}}\bigr\} . $$

Then $K_{0}$ is a cone in E.

Lemma 2.3

Let ($H_{0}$) hold. If $y\in K_{0}$, then the solution $u(t)$ of problem (2.1) belongs to $K_{0}$, i.e., $u\in K _{0}$. Furthermore, $u(t)$ is concave on $[0,\eta ]_{\mathbb{Z}}$.

Proof

Firstly, if $0\leq t-2\leq \eta$, then

$$\begin{aligned} u(t) = &\sum_{s=\eta +1}^{T-2} \biggl\{ \frac{-\alpha (T-s-1)[T(T-1)-t(t-1)]}{2-2 \alpha (T-1)}-\frac{(T-s)(T-s-1)}{2} \biggr\} y(s) \\ &{}+\sum_{s=1}^{ \eta } \biggl\{ \frac{[1-\alpha (T-s-1)][T(T-1)-t(t-1)]}{2-2\alpha (T-1)}- \frac{(T-s)(T-s-1)}{2} \biggr\} y(s) \\ &{}+\sum_{s=1}^{t-2} \frac{(t-s)(t-s-1)}{2}y(s). \end{aligned}$$

Then we obtain that

$$\begin{aligned} \Delta u(t) = &u(t+1)-u(t) \\ =&-\sum_{s=1}^{\eta }\frac{t[1-\alpha (T-s-1)]}{1- \alpha (T-1)}y(s)+ \sum_{s=1}^{t-2}(t-s)y(s)+y(t-1)+\sum _{s=\eta +1} ^{T-2}\frac{\alpha t (T-s-1)}{1-\alpha (T-1)}y(s) \\ =&\sum_{s=1}^{t-2}\frac{- \alpha ts+s(\alpha (T-1)-1)}{1-\alpha (T-1)}y(s)- \sum_{s=t-1}^{\eta }\frac{t[1- \alpha (T-s-1)]}{1-\alpha (T-1)}y(s)+y(t-1) \\ &{}+\sum_{s=\eta +1}^{T-2}\frac{ \alpha t (T-s-1)}{1-\alpha (T-1)}y(s). \end{aligned}$$

(2.5)

Since $y\in {K_{0}}$, we know that y is decreasing on $[0,T]_{ \mathbb{Z}}$. That is to say, $y(t)\geq y(\eta )$ for $t\leq \eta$ and $y(t)\leq y(\eta )$ for $t\geq \eta$. Therefore, if $t-1\leq \eta$, then

$$\begin{aligned}& \Delta u(t) \\& \quad =\sum_{s=1}^{t-1}\frac{-\alpha ts+s(\alpha (T-1)-1)}{1- \alpha (T-1)}y(s)- \sum_{s=t-1}^{\eta }\frac{t[1-\alpha (T-s-1)]}{1- \alpha (T-1)}y(s) \\& \qquad {}+\sum_{s=\eta +1}^{T-2} \frac{\alpha t (T-s-1)}{1- \alpha (T-1)}y(s) \\& \quad \leq y(\eta ) \Biggl\{ \sum_{s=1}^{t-1} \frac{- \alpha ts+s(\alpha (T-1)-1)}{1-\alpha (T-1)}-\sum_{s=t-1}^{\eta } \frac{t[1- \alpha (T-s-1)]}{1-\alpha (T-1)}+\sum_{s=\eta +1}^{T-2} \frac{\alpha t (T-s-1)}{1-\alpha (T-1)} \Biggr\} \\& \quad =y(\eta ) \Biggl\{ -\sum_{s=1}^{ \eta } \frac{t[1-\alpha (T-s-1)]}{1-\alpha (T-1)}+\sum_{s=1}^{t-1}(t-s)+ \sum _{s=\eta +1}^{T-2}\frac{\alpha t (T-s-1)}{1-\alpha (T-1)} \Biggr\} . \end{aligned}$$

If $t-1>\eta$, then $y(t-1)\leq y(\eta )$. Therefore, by (2.5), no matter $t-1\leq \eta$ or $t-1>\eta$, we always have

$$\begin{aligned} \Delta u(t) \leq& y(\eta ) \Biggl\{ -\sum _{s=1}^{\eta }\frac{t[1-\alpha (T-s-1)]}{1-\alpha (T-1)}+\sum _{s=1}^{t-1}(t-s)+\sum_{s=\eta +1}^{T-2} \frac{ \alpha t (T-s-1)}{1-\alpha (T-1)} \Biggr\} \\ =&\frac{ty(\eta )}{1- \alpha (T-1)} \biggl\{ -\eta +\frac{(t-1)[1-\alpha (T-1)]}{2}+\frac{ \alpha (T-1)(T-2)}{2} \biggr\} \\ =&\frac{1}{1-\alpha (T-1)} \bigl[-2 \eta +(t-1)+\alpha (T-1) (T-t-1) \bigr]. \end{aligned}$$

Combining this with $\eta \geq \frac{T-2}{2}$, we have

$$ \Delta u(t)\leq (T-t-1) \bigl(\alpha (T-1)-1\bigr)< 0. $$

Furthermore, we have

$$\begin{aligned}& \Delta ^{2}u(t-1) \\& \quad =-\sum_{s=1}^{\eta }\frac{1-\alpha (T-s-1)}{1- \alpha (T-1)}y(s)+ \sum_{s=1}^{t}y(s)+\sum _{s=\eta +1}^{T-2}\frac{ \alpha (T-s-1)}{1-\alpha (T-1)}y(s) \\& \quad =-\sum_{s=1}^{\eta }\frac{1- \alpha (T-s-1)}{1-\alpha (T-1)}y(s)+ \sum_{s=1}^{t-2}y(s)+y(t-1)+y(t)+ \sum _{s=\eta +1}^{T-2}\frac{\alpha (T-s-1)}{1-\alpha (T-1)}y(s). \end{aligned}$$

Now, if $t-1\leq \eta$ and $t\leq \eta$, then

$$\begin{aligned}& \Delta ^{2}u(t-1) \\& \quad \leq -\sum_{s=1}^{t-2} \frac{s\alpha }{1-\alpha (T-1)}y(s)-\sum_{s=t-1}^{\eta } \frac{1- \alpha (T-s-1)}{1-\alpha (T-1)}y(s)+\sum_{s=\eta +1}^{T-2} \frac{ \alpha (T-s-1)}{1-\alpha (T-1)}y(s)+2y(\eta ). \end{aligned}$$

Combining this with the monotonicity of y, we get that

$$\begin{aligned}& \Delta ^{2}u(t-1) \\& \quad \leq y(\eta ) \Biggl\{ -\sum_{s=1}^{t-2} \frac{s \alpha }{1-\alpha (T-1)}-\sum_{s=t-1}^{\eta } \frac{1-\alpha (T-s-1)}{1- \alpha (T-1)}+\sum_{s=\eta +1}^{T-2} \frac{\alpha (T-s-1)}{1-\alpha (T-1)}+2 \Biggr\} \\& \quad =y(\eta ) \Biggl\{ - \sum_{s=1}^{\eta } \frac{1-\alpha (T-s-1)}{1-\alpha (T-1)}+t+ \sum_{s=\eta +1}^{T-2} \frac{\alpha (T-s-1)}{1-\alpha (T-1)} \Biggr\} \\& \quad =\frac{y(\eta )}{1-\alpha (T-1)} \biggl\{ -\eta + \frac{t[1-\alpha (T-1)]}{2}+\frac{\alpha (T-1)}{2} \biggr\} \\& \quad =\frac{y( \eta )}{1-\alpha (T-1)}\frac{-2\eta +2t+\alpha (T-1)(T-2)}{2} \\& \quad \leq 0. \end{aligned}$$

For other cases, $t-1\leq \eta < t$ and $t-1> \eta$, we could also obtain $\Delta ^{2}u(t-1)\leq 0$ for $0\leq t-2\leq \eta$.

Secondly, if $\eta < t-2\leq T-2$, then

$$\begin{aligned} u(t) = &\sum_{s=t-1}^{T-2} \biggl\{ - \frac{\alpha (T-s-1)[T(T-1)-t(t-1)]}{2-2 \alpha (T-1)}-\frac{(T-s)(T-s-1)}{2} \biggr\} y(s) \\ &{}+\sum_{s=1}^{t-2} \biggl\{ \frac{-\alpha (T-s-1)[T(T-1)-t(t-1)]}{2-2\alpha (T-1)}- \frac{(T-s)(T-s-1)}{2} \\ &{}+\frac{(t-s)(t-s-1)}{2} \biggr\} y(s)+\sum_{s=1} ^{\eta }\frac{T(T-1)-t(t-1)}{2-2\alpha (T-1)}y(s) \end{aligned}$$

and

$$\begin{aligned} \Delta u(t) =&\sum_{s=1}^{T-2} \frac{\alpha t(T-s-1)}{1-\alpha (T-1)}y(s)+ \sum_{s=1}^{t-1}(t-s)y(s)- \sum_{s=1}^{\eta }\frac{t}{1-\alpha (T-1)}y(s) \\ =&-\sum_{s=1}^{\eta } \frac{\alpha ts+s(1-\alpha (T-1))}{1-\alpha (T-1)}y(s)+ \sum_{s=\eta +1} ^{T-2}\frac{\alpha t(T-s-1)}{(1-\alpha (T-1))}y(s)+\sum _{s=\eta +1} ^{t-1}(t-s)y(s). \end{aligned}$$

Combining this with the fact that $y\in K_{0}$ and the monotonicity of y, we get that

$$\begin{aligned} \Delta u(t) \leq& y(\eta ) \Biggl\{ -\sum _{s=1}^{\eta }\frac{\alpha ts+s(1- \alpha (T-1))}{1-\alpha (T-1)}+\sum _{s=\eta +1}^{T-2}\frac{\alpha t(T-s-1)}{1- \alpha (T-1)}+\sum _{s=\eta +1}^{t-1}(t-s) \Biggr\} \\ =&\frac{y(\eta )}{2-2 \alpha (T-1)} \bigl[\alpha t(T-1) (T-2)+t(t-1) \bigl(1-\alpha (T-1) \bigr)-2t\eta \bigr]. \end{aligned}$$

Since $\eta \geq \frac{T-2}{2}$, then we get that

$$ \Delta u(t)\leq y(\eta ) \bigl(1-\alpha (T-2)\bigr)t(t-T)\leq 0. $$

So, for $\forall t\in [0,T-1]_{\mathbb{Z}}$, $\Delta u(t)\leq 0$, which implies that $u(t)$ is decreasing. Since $u(T)=0$, for $\forall t \in [0,T]_{\mathbb{Z}}$, we have $u(t)\geq 0$ and $u\in K_{0}$. For $\forall t\in [1,\eta +1]_{\mathbb{Z}}$, $\Delta ^{2}u(t-1)\leq 0$, we get that $u(t)$ is concave on $[0,\eta +2]_{\mathbb{Z}}$. □

Lemma 2.4

Let ($H_{0}$) hold. If $y\in K_{0}$, then the solution $u(t)$ of (2.1) satisfies

$$ \min_{t\in [T-\theta ,\theta ]_{\mathbb{Z}}}u(t)=u(\theta )\geq \frac{ \eta +2-\theta }{\eta +2} \Vert u \Vert =\theta ^{\ast } \Vert u \Vert , $$

where $\theta ^{\ast }=\frac{\eta +2-\theta }{\eta +2}$, $\theta \in [ \lfloor \frac{T}{2} \rfloor +1,\eta +2 ] _{\mathbb{Z}}$.

Proof

From Lemma 2.3, $u(t)$ is concave on $[0,\eta +2]_{ \mathbb{Z}}$. So, $u(t)$ satisfies

$$ \frac{u(t)-u(0)}{t}\geq \frac{u(\eta +2)-u(0)}{\eta +2}, \quad t\in [0,\eta +2]_{\mathbb{Z}}. $$

Meanwhile, from Lemma 2.3, $u(t)$ is non-increasing on $[0,T]_{ \mathbb{Z}}$, which implies that $u(0)=\|u\|$. Therefore,

$$\begin{aligned}& u(t)\geq \frac{\eta +2-t}{\eta +2}u(0)=\frac{\eta +2-t}{\eta +2} \Vert u \Vert , \\& \min_{t\in [T-\theta ,\theta ]_{\mathbb{Z}}}u(t)=u(\theta )\geq \frac{ \eta +2-\theta }{\eta +2} \Vert u \Vert =\theta ^{\ast } \Vert u \Vert . \end{aligned}$$

 □

3 Main results

($H_{1}$):

$f:[1,T-2]_{\mathbb{Z}}\times [0,\infty )\to [0,\infty )$ is continuous. For $u\in [0,\infty )$, $f(t,u)$ is a decreasing function with respect to t, and for $t\in [1,T-2]_{\mathbb{Z}}$, $f(t,u)$ is an increasing function with respect to u.

($H_{2}$):

$a:[1,T-2]_{\mathbb{Z}}\to [0,\infty )$ is a decreasing function.

Let

$$ K= \Bigl\{ u\in K_{0}| \min_{t\in [T-\theta ,\theta ]_{\mathbb{Z}}}u(t) \geq \theta ^{\ast } \Vert u \Vert \Bigr\} . $$

Then K is a cone in E. Define the operator $S:K\to E$ as

$$ Su(t)=\sum_{s=1}^{T-2}G(t,s)a(s)f\bigl(s,u(s) \bigr). $$

(3.1)

Lemma 3.1

$S:K\to K$ is completely continuous.

Proof

It is obvious that $S:K\to E$ is completely continuous since the Banach space E is finite dimensional. Now, let us prove that $S:K\to K$, that is to say, for any $u\in K$, $Su\in K$.

Let $u\in K$. Then $u\in K_{0}$, which implies that $\Delta u(t) \leq 0$ and u is decreasing on t. Therefore, by $(H_{1})$, $f(t,u(t))$ is a decreasing function of t. Let $y(t):=a(t)f(t,u(t))$. Then, from ($H_{1}$) and ($H_{2}$), we obtain that $y(t)\geq 0$ and y is also a decreasing function of t. Thus, $y\in K_{0}$. Furthermore, by (3.1), we know that

$$ \Delta ^{3} (Su) (t-1)=y(t), \quad t\in [1,T-2]_{\mathbb{Z}}, $$

(3.2)

and

$$ \Delta (Su) (0)=(Su) (T)=0, \qquad \Delta ^{2} (Su) (\eta )-\alpha \Delta (Su) (T-1)=0. $$

(3.3)

Therefore, Su satisfies problem (2.1). Now, similar to the proof of Lemma 2.3, and using the fact $y\in K_{0}$, we obtain that $Su\in K _{0}$ and Su is concave on $[0,\eta ]_{\mathbb{Z}}$. Furthermore, by Lemma 2.4 and the fact $Su\in K_{0}$, we know that

$$ \min_{t\in [T-\theta ,\theta ]_{\mathbb{Z}}}(Su) (t)\geq \theta ^{*} \Vert Su \Vert . $$

Therefore, $Su\in K$ and $S:K\to K$ is completely continuous. □

From (3.1) and Lemma 3.1, we know that if u is a fixed point of S in K, then u is a positive solution of (1.1). In the rest of this paper, we try to prove S has at least one or two fixed point(s) in K by using Theorem 1.1 and Theorem 1.2.

Let

$$ A=\sum_{s=1}^{T-2}\frac{T(T-1)(1+\alpha \eta )}{1-\alpha (T-1)}a(s), \qquad B=\sum_{s=T-\theta }^{\theta }\frac{\theta (T-\theta )[2-\alpha (\theta -1)]}{2-2\alpha (T-1)}a(s). $$

Theorem 3.1

Suppose that ($H_{0}$), ($H_{1}$), and ($H_{2}$) hold. If there exist two constants r and R ($r\neq R$) such that

(A₁):

$f(t,u)\leq \frac{r}{A}$, $(t,u)\in [1,T-2]_{\mathbb{Z}} \times [0,r]$;

(A₂):

$f(t,u)\geq \frac{R}{B}$, $(t,u)\in [1,T-2]_{\mathbb{Z}} \times [\theta ^{\ast }R,R]$,

then problem (1.1) has at least one positive solution $u\in K$ with $\min \{r,R\}\leq \|u\|\leq \max \{r,R\}$.

Proof

Without loss of generality, suppose that $r< R$, the other case could be treated similarly. Let $\varOmega _{1}=\{u\in E:\|u\|< r \}$. From Lemma 2.2, $G(t,s)\leq 0$ for $s\in [\eta +1,T-2]_{ \mathbb{Z}}$; $G(t,s)\geq 0$ for $s\in [1,\eta ]_{\mathbb{Z}}$. Since (A₁), we get, for $\forall u\in K\cap \partial \varOmega _{1}$,

$$\begin{aligned} \Vert Su \Vert =&\max_{t\in [0,T]_{\mathbb{Z}}} \Biggl\vert \sum_{s=1}^{T-2}G(t,s)a(s)f\bigl(s,u(s)\bigr) \Biggr\vert \\ \leq& \max_{t\in [0,T]_{\mathbb{Z}}}\sum_{s=1}^{T-2} \bigl\vert G(t,s) \bigr\vert a(s)f\bigl(s,u(s)\bigr) \\ \leq& \sum_{s=1}^{T-2} \frac{T(T-1)(1+\alpha \eta )}{1-\alpha (T-1)}a(s)f \bigl(s,u(s)\bigr) \\ \leq& \sum_{s=1}^{T-2}\frac{T(T-1)(1+\alpha \eta )}{1-\alpha (T-1)}a(s) \frac{r}{A} \\ =&r. \end{aligned}$$

So, for $u\in K\cap \partial \varOmega _{1}$,

$$ \Vert Su \Vert \leq \Vert u \Vert . $$

(3.4)

Let $\varOmega _{2}=\{u\in E:\|u\|< R\}$. Then, for $u\in K \cap \partial {\varOmega _{2}}$,

$$ Su(T-\theta )=\sum_{s=1}^{T-2}G(T-\theta ,s)a(s)f\bigl(s,u(s)\bigr)\geq \sum_{s=\theta }^{T-\theta }G(T- \theta ,s)a(s)f\bigl(s,u(s)\bigr). $$

(3.5)

In fact, by Lemma 2.2,

$$\begin{aligned}& \sum_{s=1}^{T-\theta -1}G(T-\theta ,s)a(s)f\bigl(s,u(s)\bigr)+\sum_{s=\theta +1} ^{T-2}G(T-\theta ,s)a(s)f\bigl(s,u(s)\bigr) \\& \quad \geq \sum_{s=1}^{T-\theta -1}G(T- \theta ,s)a(s)f\bigl(s,u(s)\bigr)+\sum_{s=\eta +1}^{T-2}G(T- \theta ,s)a(s)f\bigl(s,u(s)\bigr) \\& \quad \geq a(\eta )f\bigl(\eta ,u(\eta )\bigr) \Biggl[\sum _{s=1}^{T-\theta -1}G(T- \theta ,s)+\sum _{s=\eta +1}^{T-2}G(T-\theta ,s) \Biggr]. \end{aligned}$$

Furthermore,

$$\begin{aligned}& \sum_{s=1}^{T-\theta -1}G(T-\theta ,s)+ \sum_{s=\eta +1}^{T-2}G(T- \theta ,s) \\& \quad =-\sum_{s=1}^{T-\theta -1}\frac{\alpha T(T-1)(T-s-1)}{2-2 \alpha (T-1)}- \sum_{s=1}^{T-\theta -1}\frac{(T-s)(T-s-1)}{2} \\& \qquad {}- \sum_{s=\eta +1}^{T-2}\frac{\alpha T(T-1)(T-s-1)}{2-2\alpha (T-1)} - \sum_{s=\eta +1}^{T-2}\frac{(T-s)(T-s-1)}{2} +\sum _{s=1}^{T-\theta -1}\frac{T(T-1)}{2-2 \alpha (T-1)} \\& \qquad {}+\sum_{s=1}^{T-\theta -1} \frac{ \alpha (T-s-1)(T- \theta )(T-\theta -1)}{2-2\alpha (T-1)}+\sum_{s=1}^{T-\theta -1} \frac{(T- \theta -s)(T-\theta -s-1)}{2-2\alpha (T-1)} \\& \qquad {}-\sum_{s=1}^{T-\theta -1} \frac{(T-\theta )(T-\theta -1)}{2-2\alpha (T-1)}. \end{aligned}$$

Let

$$\begin{aligned} I_{1} := &-\sum_{s=1}^{T-\theta -1} \frac{\alpha T(T-1)(T-s-1)}{2-2 \alpha (T-1)}-\sum_{s=1}^{T-\theta -1} \frac{(T-s)(T-s-1)}{2} \\ &{}- \sum_{s=\eta +1}^{T-2}\frac{\alpha T(T-1)(T-s-1)}{2-2\alpha (T-1)} - \sum_{s=\eta +1}^{T-2}\frac{(T-s)(T-s-1)}{2} \\ &{}+\sum _{s=1}^{T-\theta -1}\frac{T(T-1)}{2-2 \alpha (T-1)} \end{aligned}$$

and

$$\begin{aligned} I_{2} := &\sum_{s=1}^{T-\theta -1} \frac{ \alpha (T-s-1)(T-\theta )(T- \theta -1)}{2-2\alpha (T-1)}+\sum_{s=1}^{T-\theta -1} \frac{(T-\theta -s)(T-\theta -s-1)}{2-2\alpha (T-1)} \\ &{}-\sum_{s=1}^{T-\theta -1}\frac{(T- \theta )(T-\theta -1)}{2-2\alpha (T-1)}. \end{aligned}$$

Then

$$\begin{aligned} I_{1} \geq &-\sum_{s=1}^{T-\theta -1}(T-s-1) \frac{(T-1)(1+\alpha ) }{2-2 \alpha (T-1)}-\sum_{s=\theta }^{T-2}(T-s-1) \frac{(T-1)(1+\alpha ) }{2-2 \alpha (T-1)} \\ &{}+\sum_{s=1}^{T-\theta -1} \frac{T(T-1)}{2-2\alpha (T-1)} \\ =&-\frac{(T-1)(1+\alpha )(T-1-\theta )(T-2+\theta )}{4-4\alpha (T-1)}-\frac{(T-1)(1+\alpha )(T-1-\theta )(T- \theta )}{4-4\alpha (T-1)} \\ &{}+ \frac{T(T-1)(T-\theta -1)}{2-2\alpha (T-1)} \\ =&\frac{(T-1)(T-\theta -1)}{2-2\alpha (T-1)}\bigl(T-(1+\alpha ) (T-1)\bigr) \\ \geq &\frac{(T-1)(T- \theta -1)}{2-2\alpha (T-1)}\biggl(T-\frac{T}{T-1}(T-1)\biggr) \\ =&0 \end{aligned}$$

and

$$\begin{aligned} I_{2} =&\frac{(T-\theta )(T-\theta -1)(T-\theta -1)}{2-2\alpha (T-1)}\bigl( \alpha (T-2+\theta )-1 \bigr)+\sum_{s=1}^{T-\theta -1}\frac{(T-\theta -s)(T- \theta -s-1)}{2-2\alpha (T-1)} \\ =&\frac{(T-\theta )(T-\theta -1)(T- \theta -1)}{2-2\alpha (T-1)}\bigl(\alpha (T-2+\theta )-1\bigr)+\frac{(T-\theta )(T-\theta -1)(T-\theta -2)}{3} \\ =&\frac{(T-\theta )(T-\theta -1)(T- \theta -2)}{6-6\alpha (T-1)}\bigl[\alpha (T-8)+3(\theta -1)\bigr] \\ \geq& 0. \end{aligned}$$

Therefore, (3.5) holds. This implies that

$$\begin{aligned} Su(T-\theta ) \geq& \sum_{s=T-\theta }^{\theta }G(T- \theta ,s)a(s)f\bigl(s,u(s)\bigr) \\ =&\sum_{s=T-\theta }^{\theta } \biggl\{ \frac{1-\alpha (T-s-s)[T(T-1)-(T- \theta )(T-\theta -1)]}{2-2\alpha (T-1)} \\ &{} - \frac{(T-s)(T-s-1)}{2}+\frac{(T-\theta -s)(T-\theta -s-1)}{2} \biggr\} a(s)f\bigl(s,u(s)\bigr) \\ \geq& \sum_{s=T-\theta }^{\theta }\frac{\theta (T-\theta )[2- \alpha (\theta -1)]}{2-2\alpha (T-1)}a(s)f \bigl(s,u(s)\bigr) \\ \geq& \sum_{s=T-\theta }^{\theta }\frac{\theta (T-\theta )[2-\alpha (\theta -1)]}{2-2\alpha (T-1)}a(s) \frac{R}{B} \\ \geq& R. \end{aligned}$$

So, for $\forall u\in K\cap \partial \varOmega _{2}$,

$$ \Vert Su \Vert \geq \Vert u \Vert . $$

(3.6)

Then, by Theorem 1.1, S has at least one fixed point $u\in K$ and u will be a positive solution of problem (1.1). □

Theorem 3.2

Suppose that ($H_{0}$), ($H_{1}$), and ($H_{2}$) hold. If one of the following conditions holds:

$$ (B_{1})\qquad f^{0}:=\lim_{u\to 0^{+}}\max _{t\in [1,T-2]_{\mathbb{Z}}} \frac{f(t,u)}{u}=0, \qquad f_{\infty }:=\lim _{u\to \infty } \max_{t\in [1,T-2]_{\mathbb{Z}}}\frac{f(t,u)}{u}=\infty , $$

or

$$ (B_{2})\qquad f_{0}:=\lim_{u\to 0^{+}}\max _{t\in [1,T-2]_{\mathbb{Z}}} \frac{f(t,u)}{u}=\infty ,\qquad f^{\infty }:=\lim _{u\to \infty } \max_{t\in [1,T-2]_{\mathbb{Z}}}\frac{f(t,u)}{u}=0, $$

then (1.1) has at least one positive solution.

Proof

Firstly, we prove the case that $(B_{1})$ holds. Since $f^{0}=0$, there exists a constant $R_{1}>0$ such that

$$ \frac{f(t,u)}{u}\leq \frac{R_{1}}{A}, \quad (t,u)\in [1,T-2]_{ \mathbb{Z}} \times [0,R_{1}]. $$

Since $f_{\infty }=\infty$, there exists a constant $R_{2}>R_{1}$ such that

$$ f(t,u)\geq \frac{u}{\theta ^{\ast }B}\geq \frac{\theta ^{\ast }R_{2}}{ \theta ^{\ast }B}=\frac{R_{2}}{B}, \quad (t,u)\in [1,T-2]_{\mathbb{Z}} \times \bigl[\theta ^{\ast }R_{2},R_{2} \bigr]. $$

From Theorem 3.1, problem (1.1) has at least one positive solution $u\in K$.

Secondly, suppose that $(B_{2})$ holds. Since $f_{0}=\infty$, there exists a constant $r_{1}>0$ such that

$$ f(t,u)\geq \frac{u}{\theta ^{\ast }B},\quad (t,u)\in [1,T-2]_{ \mathbb{Z}}\times [0,r_{1}]. $$

Let $\varOmega _{1}=\{u\in E:\|u\|< r_{1} \}$. If $u\in K\cap \partial \varOmega _{1}$, we have

$$ \min_{s\in [T-\theta ,\theta ]_{\mathbb{Z}}}u(s)\geq \theta ^{\ast } \Vert u \Vert = \theta ^{\ast }r_{1}. $$

Therefore, similar to the proof of (3.6), we have

$$ \begin{aligned} \Vert Su \Vert \geq \Vert u \Vert \quad \text{for } u\in K\cap \partial \varOmega _{1}. \end{aligned} $$

On the other hand, since $f^{\infty }=0$, there exists a constant $r_{2}>0$ such that

$$ f(t,u)\leq \frac{u}{A},\quad (t,u)\in [1,T-2]_{\mathbb{Z}}\times [r _{2},\infty ). $$

If f is bounded, then there exists a constant $N>0$ such that $f\leq N$. So, we choose $R'=\max \{2r_{1},NA\}$. If f is unbounded, then let $R'>\max \{2r_{1},r_{2}\}$ such that $f(t,u)\leq f(t,R_{2})$. Let $\varOmega _{2}=\{u\in K:\|u\|< R'\}$ for $\forall (t,u)\in [1,T-2]_{ \mathbb{Z}}\times [0,R_{2}]$. Similar to the proof of Theorem 3.1, we get, for $\forall u\in k\cap \partial \varOmega _{2}$,

$$ \Vert Su \Vert \leq \Vert u \Vert . $$

Thus, by Theorem 1.1, S has at least one fixed point $u\in K\cap \overline{ \varOmega _{2}}\setminus \varOmega _{1}$, which is a positive solution of problem (1.1). □

Theorem 3.3

Assume that ($H_{0}$), ($H_{1}$), and ($H_{2}$) hold. If

$(C_{1})$ :

$f_{0}:=\lim_{u\to 0^{+}}\min_{t\in [1,T-2]_{\mathbb{Z}}}= \frac{f(t,u)}{u}=+\infty$, $f_{\infty }:=\lim_{u\to \infty } \min_{t\in [1,T-2]_{\mathbb{Z}}}=\frac{f(t,u)}{u}=+\infty$,

and

($C_{2}$):

There exists a constant $p>0$ such that $f(t,u)<\gamma p$ for $0\leq u\leq p$ and $t\in [1,T-2]_{\mathbb{Z}}$, where

$$ \gamma = \Biggl( \frac{T(T-1)(1+\alpha \eta )}{1-\alpha (T-1)}\sum_{s=1} ^{T-2}a(s) \Biggr)^{-1}, $$

then problem (1.1) has at least two positive solutions $u_{1}$ and $u_{2}$ with $0\leq \|u_{1}\|\leq p\leq \|u_{2}\|$.

Proof

Choose $M>0$ such that

$$ M\theta ^{\ast }\frac{\theta (T-\theta )[2-\alpha (\theta -1)]}{2-2 \alpha (T-1)}\sum_{s=T-\theta }^{\theta }a(s) \geq 1. $$

Since $f_{0}=+\infty$, there exists a constant r with $0< r< p$ such that $f(t,u)\ge Mu$ for $0\le u\le r$. Then, for $\forall u\in \partial K_{r}$, we have

$$\begin{aligned} Su(T-\theta ) =&\sum_{s=1}^{T-2}G(T- \theta ,s)a(s)f\bigl(s,u(s)\bigr) \\ \ge& \sum_{s=T-\theta }^{\theta }G(T-\theta ,s)a(s)f \bigl(s,u(s)\bigr) \\ \ge& M \theta ^{\ast }\sum_{s=T-\theta }^{\theta }G(T- \theta ,s)a(s) \Vert u \Vert \\ \ge& M\theta ^{\ast }\frac{\theta (T-\theta )[2-\alpha (\theta -1)]}{2-2 \alpha (T-1)}\sum _{s=T-\theta }^{\theta }a(s) \Vert u \Vert \\ \ge& \Vert u \Vert . \end{aligned}$$

From Theorem 1.2, we get

$$ i(S,K_{r},K)=0. $$

Since $f_{\infty }=+\infty$, there exists a constant $R_{1}>0$ such that $f(t,u)\ge Mu$ for $\forall u\ge R_{1}$. Choose $R>\max \{p, \frac{R _{1}}{\theta ^{*}}\}$, then for $\forall u\in \partial K_{R}$, $\min_{t\in [T-\theta ,\theta ]_{\mathbb{Z}}}u(t)\ge \theta ^{*}\|u\|>R_{1}$. Similar to the above proof, we have

$$ \Vert Su \Vert \ge \Vert u \Vert \quad \text{for } u\in \partial K_{R}. $$

Therefore,

$$ i(S,K_{R}, K)=0. $$

From ($C_{2}$), for $\forall u\in \partial K_{p}$,

$$\begin{aligned} \Vert Su \Vert =&\max_{t\in [0,T]_{\mathbb{Z}}} \Biggl\vert \sum_{s=1}^{T-2}G(t,s)a(s)f\bigl(s,u(s)\bigr) \Biggr\vert \\ \le& \max_{t\in [0,T]_{\mathbb{Z}}}\sum_{s=1}^{T-2} \bigl\vert G(t,s) \bigr\vert a(s)f\bigl(s,u(s)\bigr) \\ \le& \sum_{s=1}^{T-2}\frac{T(T-1)(1+\alpha \eta )}{1-\alpha (T-1)}a(s)f \bigl(s,u(s)\bigr) \\ =& \Vert u \Vert . \end{aligned}$$

Therefore, for $\forall u\in \partial K_{p}$, $\|Tu\|\le \|u\|$. By Theorem 1.2,

$$ i(S,K_{p}, K)=1. $$

Thus,

$$ i(S, K_{R}\setminus \mathring{K}_{p},K)=-1, \qquad i(S, K_{p}\setminus \mathring{K}_{r},K)=1. $$

So, S has a fixed point $u_{1}$ in $K_{p}\setminus \mathring{K}_{r}$ and another fixed point $u_{2}$ in $K_{R}\setminus \mathring{K}_{p}$. So, problem (1.1) has at least two positive solutions $u_{1}$ and $u_{2}$ with $0\le \|u_{1}\|\le p\le \|u_{2}\|$. □

Theorem 3.4

Assume that $(H_{0})$, $(H_{1})$, and $(H_{2})$ hold. If

$(D_{1})$ :

$f^{0}:=\lim_{u\to 0^{+}}\max_{t\in [1,T-2]_{\mathbb{Z}}} \frac{f(t,u)}{u}=0$, $f^{\infty }:=\lim_{u\to \infty } \max_{t\in [1,T-2]_{\mathbb{Z}}}\frac{f(t,u)}{u}=0$,

and

$(D_{2})$ :

there exists a constant $p>0$ such that $f(t,u)>\beta p$ for $\theta ^{*}p\le u\le p$ and $t\in [1,T-2]_{\mathbb{Z}}$, where

$$ \beta = \Biggl( \theta ^{\ast }\frac{\theta (T-\theta )[2-\alpha (\theta -1)]}{2-2\alpha (T-1)}\sum _{s=T-\theta }^{\theta }a(s) \Biggr)^{-1}, $$

then (1.1) has at least two positive solutions $u_{1}$ and $u_{2}$ with $0\le \|u_{1}\|\le p\le \|u_{2}\|$.

Proof

From ($D_{1}$), for $\forall \epsilon >0$, there exists $M_{1}>0$, if $u>0$, $t\in [0,T]_{\mathbb{Z}}$, we have $f(t,u)\leq M _{1}+\epsilon u$. Then, for $\forall u\in K$,

$$ \begin{aligned} \Vert Su \Vert &=\max_{t\in [0,T]_{\mathbb{Z}}} \Biggl\vert \sum_{s=1}^{T-2}G(t,s)a(s)f \bigl(s,u(s)\bigr) \Biggr\vert \\ &\leq \frac{T(T-1)(1+ \alpha \eta )}{1-\alpha (T-1)} \sum_{s=1}^{T-2}a(s) (M_{1}+\epsilon u). \end{aligned} $$

Choose $\epsilon >0$ sufficiently small and $R>p$ sufficiently large, then for $\forall u\in \partial K_{R}$, $\|Su\|\le \|u\|$, from Theorem 1.2, we have

$$ i(S,K_{R},K)=1. $$

In a similar way, if $0< r< p$,

$$ i(S,K_{R},K)=1. $$

From ($D_{2}$), for $\forall u\in \partial K_{p}$,

$$\begin{aligned} Su(T-\theta ) =&\sum_{s=1}^{T-2}G(T- \theta ,s)a(s)f\bigl(s,u(s)\bigr) \\ \ge& \sum_{s=T-\theta }^{\theta }G(T-\theta ,s)a(s)f \bigl(s,u(s)\bigr) \\ >&\beta \theta ^{\ast }p \sum_{s=T-\theta }^{\theta }G(T- \theta ,s)a(s) \\ \geq& \beta \theta ^{\ast }p\frac{\theta (T-\theta )[2-\alpha (\theta -1)]}{2-2 \alpha (T-1)}\sum _{s=T-\theta }^{\theta }a(s) \\ =& p. \end{aligned}$$

Then, for $\forall u\in \partial K_{p}$, $\|Su\|\geq \|u\|$. From Theorem 1.2, we have

$$ i(S,K_{p}, K)=0. $$

Then, problem (1.1) has at least two solutions $u_{1}$ and $u_{2}$ with $0\le \|u_{1}\|\le p\le \|u_{2}\|$. □

4 Example

Example 4.1

Consider the discrete three-point boundary problem

$$ \textstyle\begin{cases} \Delta ^{3}u(t-1)-a(t)f(t,u(t))=0, \quad t\in [1,7]_{\mathbb{Z}},\\ \Delta u(0)=u(9)=0, \qquad \Delta ^{2}u(7)-\frac{1}{9}\Delta u(8)=0, \end{cases} $$

(4.1)

where $a(t)=\frac{9-t}{10}$, and

$$ f(t,u)=\textstyle\begin{cases} 15-t+\frac{u}{1000}, & (t,u)\in [1,7]_{\mathbb{Z}}\times [0,1000], \\ \sqrt[3]{u}+6-t,& (t,u)\in [1,7]_{\mathbb{Z}}\times [1000,\infty ]. \end{cases} $$

Since $T=9$ and $\alpha =\frac{1}{9}$, $\eta \in [4,7]_{\mathbb{Z}}$. Without loss of generality, let $\eta =4$. Then, by the direct calculation, we get $\theta \in [5,6]_{\mathbb{Z}}$. Choose $\theta =5$, then $\theta ^{\ast }=\frac{\eta +2-\theta }{\eta +2}= \frac{1}{6}$. So,

$$\begin{aligned}& A=\sum_{s=1}^{T-2}\frac{T(T-1)(1+\alpha \eta )}{1-\alpha (T-1)}a(s)=936 \sum_{s=1}^{7}\frac{9-s}{10}=3276, \\& B=\sum_{s=T-\theta }^{\theta }\frac{\theta (T-\theta )[2-\alpha ( \theta -1)]}{2-2\alpha (T-1)}a(s)=140 \sum_{s=4}^{5}\frac{9-s}{10}=136. \end{aligned}$$

If we choose $R=330$, $r=1{,}000{,}000$, from Theorem 3.1, problem (4.1) has at least one positive solution.

Example 4.2

In this example, we continue to consider problem (4.1) with

$$ f(t,u)=\textstyle\begin{cases} \frac{u^{2}(10-t)}{10}, &(t,u)\in [1,7]_{\mathbb{Z}}\times [0,1], \\ \frac{\sqrt[3]{u}+9-t}{10},&(t,u)\in [1,7]_{\mathbb{Z}}\times [1,3], \\ \frac{\sqrt[3]{u}+1890-t}{1000},&(t,u)\in [1,7]_{\mathbb{Z}} \times [3,\infty ). \end{cases} $$

Continue to take $\alpha =\frac{1}{9}$, $\eta =4$, $\theta =5$, and $\theta ^{\ast }=\frac{1}{6}$. Then

$$ \beta = \Biggl( \theta ^{\ast }\frac{\theta (T-\theta )[2-\alpha (\theta -1)]}{2-2\alpha (T-1)} \sum _{s=T-\theta }^{\theta }a(s) \Biggr)^{-1}= \frac{3}{68}. $$

Furthermore, if we choose $p=1$, then for $\theta ^{\ast }p\leq u \leq p$, $f(t,u)\geq \beta p=\frac{3}{68}$. From Theorem 3.4, problem (4.1) has at least two positive solutions $u_{1}$ and $u_{2}$ with $0\le \|u_{1}\|\le p\le \|u_{2}\|$.