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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.

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

  1. (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

  1. (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

  1. (i)

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

  2. (ii)

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

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,

  1. (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,

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

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

 □

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 \(0\le \|u_{1}\|\le p\le \|u_{2}\|\). □

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 \(0\le \|u_{1}\|\le p\le \|u_{2}\|\).

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Acknowledgements

The authors express their gratitude to the referee for reading the paper carefully and making several corrections and remarks.

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The datasets used and analysed during the current study are available from the corresponding author on reasonable request.

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This work was supported by the National Natural Science Foundation of China (No. 71561024). The key projects of Natural Science Foundation of Gansu province of China (No. 18JR3RA084).

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Xu, Y., Tian, W. & Gao, C. Existence of positive solutions of discrete third-order three-point BVP with sign-changing Green’s function. Adv Differ Equ 2019, 206 (2019). https://doi.org/10.1186/s13662-019-2145-x

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Keywords

  • Discrete third-order three-point boundary value problem
  • Positive solutions
  • Cone
  • Fixed point
  • Sign-changing Green’s function