Existence of Positive Solutions for Singular Difference Equations with Nonlinear Boundary Conditions

Abstract

In this paper, we delve into a discrete nonlinear singular semipositone problem, characterized by a nonlinear boundary condition. The nonlinearity, given by $f\left(u\right)-{\displaystyle \frac{a}{u^{\alpha }}}$ with $\alpha >0$, exhibits a singularity at $u=0$ and tends towards $-\infty$ as u approaches $0^{+}$. By constructing some suitable auxiliary problems, the difficulty that arises from the singularity and semipositone of nonlinearity and the lack of a maximum principle is overcome. Subsequently, employing the Krasnosel’skii fixed-point theorem, we determine the parameter range that ensures the existence of at least one positive solution and the emergence of at least two positive solutions. Furthermore, based on our existence results, one can obtain the symmetry of the solutions after adding some symmetric conditions on the given functions by using a standard argument.

1. Introduction

The study of boundary value problems has a long and distinguished history within the mathematical sciences. A significant breakthrough in this area occurred in 1994 when Erbe and Wang [1] introduced the use of Guo–Krasnosel’skii’s fixed-point theorem in a conical framework. Their work played a crucial role in proving the existence of positive solutions for a particular class of second-order boundary problem:

$$\left\{\begin{array}{c}{u}^{″}+a\left(t\right)f\left(u\right)=0,t\in (0,1),\hfill \\ \alpha u\left(0\right)-\beta u^{\prime }\left(0\right)=0,\hfill \\ \gamma u\left(1\right)+\delta u^{\prime }\left(1\right)=0.\hfill \end{array}\right.$$

(1)

Since then, several papers have addressed the existence of positive solutions to boundary value problems. More recently, Hai and Wang [2] considered the one-dimensional p-Laplacian problem when f is semipositone:

$$\left\{\begin{array}{c}-{\left(r\left(t\right)\varphi \left(u^{\prime }\right)\right)}^{\prime }=\lambda g\left(t\right)f\left(u\right),t\in (0,1),\hfill \\ a_{1}u\left(0\right)-H_1\left(u^{\prime }\left(0\right)\right)=0,\hfill \\ a_{2}u\left(1\right)+H_2\left(u^{\prime }\left(1\right)\right)=0,\hfill \end{array}\right.$$

(2)

where $\varphi \left(x\right):={\left|x\right|}^{p-2}x,p>1$. Under the assumption that the nonlinearity $\underset{u\to 0^{+}}{lim}f\left(u\right)=-\infty$, they demonstrated the existence of positive solutions. For further results on nonsingular semipositone problems, we refer the reader to [3,4,5,6,7].

In the discrete case, various fixed-point theorems have been employed to investigate the existence of solutions to discrete boundary value problems, see [8,9,10,11,12] and the references therein. Notably, Bai and Xu [13] examined the existence of positive solutions for a kind of discrete semipositone boundary value problem. They proved that the semipositone problem has at least one positive solution when f is superlinear as u approaches ∞. Subsequently, Bai and Henderson [14] studied discrete Neumann problems under more general conditions, where the nonlinearity f may be unbounded below. For further related research, see [9,15,16,17,18,19,20] and the references therein.

For the discrete problems with nonlinear boundary conditions, there have only been a few results. In 2022, Mohamed et al. [21] studied a class of discrete problems with nonlinear boundary conditions of the following form:

$$\left\{\begin{array}{c}-\Delta [p(t-1)\Delta u(t-1)]=\lambda g\left(t\right)f\left(u\right),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ a_{1}u\left(0\right)-H_1(\Delta u\left(0\right))=0,\hfill \\ a_{2}u(T+1)+H_2(\Delta u\left(T\right))=0.\hfill \end{array}\right.$$

(3)

The authors established the existence and asymptotic properties of positive solutions when the nonlinear term f satisfies one of the following conditions: (i) $f_{\infty }={lim}_{z\to \infty }{\displaystyle \frac{f\left(z\right)}{z}}$; (ii) $f_{\infty }=0$ and ${lim}_{z\to \infty }f\left(z\right)=\infty$; (iii) $f_0={lim}_{z\to 0^{+}}{\displaystyle \frac{f\left(z\right)}{z}}=\infty$ and $f_0=0$. In 2023, Li et al. [22] extended the study to another type of discrete problems with nonlinear boundary conditions and obtained the existence of positive solutions for these problems.

Building on this work, we aim to explore the existence and multiplicity of positive solutions for the following discrete boundary value problem:

$$\left\{\begin{array}{c}-\Delta [p(t-1)\Delta u(t-1)]=\lambda g\left(t\right)(f\left(u\right)-{\displaystyle \frac{a}{u^{\alpha }}}),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ u\left(0\right)=0,\Delta u\left(T\right)+H\left(u\right(T+1\left)\right)=0,\hfill \end{array}\right.$$

(4)

where $T>1$ is an integer, ${[1,T]}_{\mathbb{Z}}=\left\{1,2,···,T\right\}$, $\alpha >0$, $\lambda$ is a positive parameter, and $a>0$. We assume the following conditions:

$\left(F1\right)$ $g:{[1,T]}_{\mathbb{Z}}\to (0,\infty ).$

$\left(F2\right)$ $H:[0,\infty )\to (0,\infty )$ is nondecreasing with $H\left(0\right)=0.$

$\left(F3\right)$ $p:{[0,T]}_{\mathbb{Z}}\to (0,\infty )$.

$\left(F4\right)$ $f:[0,\infty )\to R$ is continuous, nondecreasing and $\underset{z\to \infty }{lim}{\displaystyle \frac{f\left(z\right)}{z}}=\infty .$

(F5) There exist constants $m>0$ and $\beta \in (0,1)$ such that $f\left(z\right)\ge {\left({\displaystyle \frac{z}{m}}\right)}^{\beta }f\left(m\right)$ for $z\in (0,m).$

A positive solution of (4) refers to a function $u\in E$ with $u>0$ on $t\in {[1,T]}_{\mathbb{Z}}$ that satisfies (4). Notably, $\underset{u\to 0^{+}}{lim}\left(f\left(u\right)-{\displaystyle \frac{a}{u^{\alpha }}}\right)=-\infty$. This implies that $f\left(u\right)-{\displaystyle \frac{a}{u^{\alpha }}}$ is singular at $u=0$. It is worth noting that the term ${\displaystyle \frac{1}{u^{\alpha }}}$ appears in the nonlinear expression, which introduces several challenges in solving this type of problem. For example, ensuring the continuity of the summation operator is not straightforward. To address these difficulties, we first develop a new maximum principle and establish some auxiliary function and problems (see Lemma 4 and the proof of Theorem 1).

The rest of this article is structured as follows. In Section 2, we focus on the linear problem, where we specifically construct a new maximum principle for the discrete problem with a nonlinear boundary condition. In Section 3, we prove our main result, Theorem 1, by using the Krasnosel’skii’s type fixed-point theorem.

Remark 1.

It is worth noting that after adding some symmetric conditions to the functions $g\left(t\right)$, $p\left(t\right)$, the nonlinearity f and H, we can further consider the symmetric solutions of this kind of problem by using our method. For the study of symmetric solutions of discrete boundary value problems, see [23,24].

2. Preliminaries

For the remainder of this paper, let $p_0=\underset{t\in {[1,T]}_{\mathbb{Z}}}{min}p\left(t\right)$, $p_1=\underset{t\in {[1,T]}_{\mathbb{Z}}}{max}p\left(t\right)$ and $H\left(z\right)=H\left(0\right)$ for $z<0.$

Now, we present the following Krasnosel’skii’s-type fixed-point theorem in a Banach space, which serves as the primary tool for this paper.

Lemma 1

([25]). Let E be a Banach space and $K:E\to E$ be a completely continuous operator. Assume there exists $h\in E$ (with $h\ne 0$) and positive constants a and b such that $a\ne b$, satisfying the following conditions:

$\left(a\right)$ If $y\in E$ satisfies $y=\mu Ky$ for some $\mu \in (0,1]$, then $\parallel y\parallel \ne a$;

$\left(b\right)$ If $y\in E$ satisfies $y=Ky+\xi h$ for some $\xi \ge 0$, then $\parallel y\parallel \ne b$.

• Under these conditions, K has a fixed point $y\in E$ such that $min(a,b)<\parallel y\parallel <max(a,b)$.

For the rest of this paper, let $\mathbb{R}$ be the set of real numbers. Let $E=\left\{{u|u:[1,T]}_{\mathbb{Z}}\to \mathbb{R}\right\}$. Then, E is a Banach space under the norm ${\parallel u\parallel }_E=\underset{t\in {[1,T]}_{\mathbb{Z}}}{max}\left|u\left(t\right)\right|.$ Applying a similar method in [21], we can prove the following results.

Lemma 2.

Assume (F1)–(F3) hold. Let $h\in E$. Then, the following boundary value problem

$$\left\{\begin{array}{c}-\Delta [p(t-1)\Delta u(t-1)]=h\left(t\right),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ u\left(0\right)=0,\Delta u\left(T\right)+H\left(u\right(T+1\left)\right)=0\hfill \end{array}\right.$$

(5)

has a unique solution $u\equiv Sh$, where $S:E\to E$ is a completely continuous operator and

$${\parallel Sh\parallel }_E\le 2\sum _{s=1}^{T+1}{\displaystyle \frac{1}{p(s-1)}}\sum _{s=1}^T\left|h\left(s\right)\right|.$$

(6)

Lemma 3.

Let $h\in E$ and $u\in E$ with

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\Delta [p(t-1)\Delta u(t-1)]\le h\left(t\right),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ u\left(0\right)=0,\Delta u\left(T\right)+H\left(u\right(T+1\left)\right)=0.\hfill \end{array}\right.\end{array}$$

(7)

Then, $u(T+1)\ge 0$, if ${\parallel u\parallel }_E>{\displaystyle \frac{T}{p_0}}{\sum }_{s=1}^T\left|h\left(s\right)\right|$.

Proof. 

Suppose, on the contrary, that $u(T+1)<0.$ Then, it follows from $u\left(0\right)=0$ that $\Delta u\left(T\right)\ge 0.$ Let $\tau \in {[1,T]}_{\mathbb{Z}}$ be such that ${\parallel u\parallel }_E=\left|u\left(\tau \right)\right|.$

By summing the inequality in (7) on $s\in {[t+1,T]}_{\mathbb{Z}},$ we obtain

$$p\left(t\right)\Delta u\left(t\right)\ge p\left(T\right)\Delta u\left(T\right)-\sum _{s=t+1}^{T}h\left(s\right).$$

It follows that

$$\begin{array}{c}\hfill \Delta u\left(t\right)\ge -\left({\displaystyle \frac{1}{p\left(t\right)}}\sum _{s=t+1}^{T}h\left(s\right)\right)\ge -\left({\displaystyle \frac{1}{p_0}}\sum _{s=t+1}^{T}h\left(s\right)\right)t\in {[1,T]}_{\mathbb{Z}}.\end{array}$$

(8)

Next, by summing (8) from $s=0$ to $s=\tau -1,$ one has

$$\begin{array}{cc}\hfill u\left(\tau \right)& \ge u\left(0\right)-\sum _{s=0}^{\tau -1}\left({\displaystyle \frac{1}{p\left(s\right)}}\sum _{\ell =s+1}^{T}h(\ell )\right)\hfill \\ \hfill & \ge -\sum _{s=0}^{\tau -1}\left({\displaystyle \frac{1}{p_0}}\sum _{\ell =s+1}^{T}h(\ell )\right)=-{\displaystyle \frac{1}{p_0}}\left(\sum _{s=1}^{\tau -1}sh\left(s\right)+\tau \sum _{s=\tau }^{T}h\left(s\right)\right)\ge -{\displaystyle \frac{T}{p_0}}\left(\sum _{s=1}^T\left|h\left(s\right)\right|\right).\hfill \end{array}$$

(9)

Similarly from $s=\tau$ to $s=T,$ we obtain

$$\begin{array}{cc}\hfill u(T+1)-u\left(\tau \right)& \ge -\sum _{s=\tau }^T\left({\displaystyle \frac{1}{p\left(s\right)}}\sum _{\ell =s+1}^{T}h(\ell )\right)\ge -\sum _{s=\tau }^T\left({\displaystyle \frac{1}{p_0}}\sum _{\ell =s+1}^{T}h(\ell )\right)=-{\displaystyle \frac{1}{p_0}}\sum _{s=\tau }^T\left(\sum _{\ell =s+1}^{T}h(\ell )\right),\hfill \\ \hfill -u\left(\tau \right)& \ge -{\displaystyle \frac{1}{p_0}}\left(\sum _{s=\tau +1}^T(s-\tau )h\left(s\right)\right)-u(T+1)\ge -{\displaystyle \frac{T}{p_0}}\sum _{s=1}^T\left|h\left(s\right)\right|.\hfill \end{array}$$

(10)

Now, by (9) and (10), we deduce

$${\parallel u\parallel }_E\le \left({\displaystyle \frac{T}{p_0}}\sum _{s=1}^T\left|h\left(s\right)\right|\right).$$

We obtain a contradiction. So, $u(T+1)\ge 0.$ □

Let $q\left(t\right)=min\left\{{\displaystyle \frac{t}{T+1}},{\displaystyle \frac{T+1-t}{T+1}}\right\}.$ Then we get the following maximum principle.

Lemma 4.

Let $h\in E$ with $h\ge 0$ and u satisfy

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\Delta [p(t-1)\Delta u(t-1)]\le h\left(t\right),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ u\left(0\right)\ge 0,u(T+1)\ge 0.\hfill \end{array}\right.\end{array}$$

(11)

Then,

$$\begin{array}{l}\left(i\right)u\left(t\right)\ge \left({\displaystyle \frac{u(T+1)}{T+1}}-{\displaystyle \frac{1}{p_0}}\sum _{s=1}^{T}h\left(s\right)\right)t,t\in {[0,T+1]}_{\mathbb{Z}}.\\ \left(ii\right)u\left(t\right)\ge \left({\parallel u\parallel }_E-{\displaystyle \frac{T+1}{p_0}}\sum _{s=1}^{T}h\left(s\right)\right)q\left(t\right),t\in {[0,T+1]}_{\mathbb{Z}}.\end{array}$$

Furthermore, if

$${\parallel u\parallel }_E\ge {\displaystyle \frac{2(T+2)}{p_0}}\sum _{s=1}^{T}h\left(s\right),$$

then,

$$u\left(t\right)\ge {\displaystyle \frac{1}{T+2}}{\parallel u\parallel }_{E}q\left(t\right),t\in {[0,T+1]}_{\mathbb{Z}}.$$

Proof. 

Let $w\left(t\right)={\displaystyle \frac{1}{p\left(t\right)}}{\displaystyle \sum _{s=t+1}^T}h\left(s\right)$ and $z\left(t\right)={\displaystyle \sum _{s=0}^{t-1}}w\left(s\right),t\in {[1,T+1]}_{\mathbb{Z}}.$ Then, $w\left(t\right)\ge 0,$ $z\left(t\right)\ge 0$ on ${[0,T+1]}_{\mathbb{Z}}.$ Furthermore, $w\left(t\right)\le {\displaystyle \frac{1}{p_0}}{\displaystyle \sum _{s=1}^T}h\left(s\right),$ which implies that $z\left(t\right)\le {\displaystyle \frac{t}{p_0}}{\displaystyle \sum _{s=1}^T}h\left(s\right)$ for $t\in {[0,T+1]}_{\mathbb{Z}}$.

Since

$$\begin{array}{l}\Delta \left[p(t-1)(\Delta u(t-1)+\Delta z(t-1\left)\right)\right]\\ =\Delta \left[p(t-1)(\Delta u(t-1)+w(t-1\left)\right)\right]\\ =\Delta \left[p(t-1)\Delta u(t-1)]+{\displaystyle \sum _{s=t}^T}h\left(s\right)\right]\\ =\Delta [p(t-1)\Delta u(t-1)]-h\left(t\right)\le 0,t\in {[1,T]}_{\mathbb{Z}},\end{array}$$

we deduce that

$$p(t-1)\left(\Delta u(t-1)+\Delta z(t-1)\right)$$

is nonincreasing on ${[1,T]}_{\mathbb{Z}}.$ This implies the concavity of $u+z$ on ${[0,T+1]}_{\mathbb{Z}}.$ Furthermore, combining the fact that $(u+z)\left(0\right)\ge 0,$ we obtain

$${\displaystyle \frac{(u+z)\left(t\right)-(u+z)\left(0\right)}{t-0}}\ge {\displaystyle \frac{(u+z)(T+1)-(u+z)\left(0\right)}{T+1-0}}.$$

So,

$$(u+z)\left(t\right)\ge (u+z)\left(0\right)+{\displaystyle \frac{t}{T+1}}\left[(u+z)(T+1)-(u+z)\left(0\right)\right]\ge {\displaystyle \frac{t}{T+1}}(u+z)(T+1).$$

That is,

$$(u+z)\left(t\right)\ge {\displaystyle \frac{t}{T+1}}(u+z)(T+1)\ge {\displaystyle \frac{t}{T+1}}u(T+1).$$

Then, we get

$$u\left(t\right)\ge {\displaystyle \frac{t}{T+1}}u(T+1)-z\left(t\right)\ge {\displaystyle \frac{t}{T+1}}\left(u(T+1)-{\displaystyle \frac{1}{p_0}}\sum _{s=1}^{T}h\left(s\right)\right).$$

In other words, $\left(i\right)$ holds.

Note that $(u+z)(T+1)\ge 0.$ This, together with the concavity of $u+z$ on ${[0,T+1]}_{\mathbb{Z}}$, implies that

$$u\left(t\right)+z\left(t\right)\ge {\parallel u+z\parallel }_{E}q\left(t\right)\mathrm{for}t\in {[0,T+1]}_{\mathbb{Z}}.$$

Therefore,

$$\begin{array}{c}\hfill u\left(t\right)\ge \left({\parallel u\parallel }_E-{\parallel z\parallel }_E\right)q\left(t\right)-z\left(t\right)\mathrm{for}t\in {[0,T+1]}_{\mathbb{Z}}.\end{array}$$

(12)

Subsequently,

$$\begin{array}{cc}\hfill u\left(t\right)& \ge \left({\parallel u\parallel }_E-{\displaystyle \frac{t}{p_0}}\sum _{s=1}^{T}h\left(s\right)\right){\displaystyle \frac{t}{T+1}}-{\displaystyle \frac{t}{p_0}}\sum _{s=1}^{T}h\left(s\right)\hfill \\ & ={\displaystyle \frac{t}{T+1}}{\parallel u\parallel }_E-{\displaystyle \frac{t}{T+1}}\left({\displaystyle \frac{t}{p_0}}\sum _{s=1}^{T}h\left(s\right)+{\displaystyle \frac{t}{p_0}}.{\displaystyle \frac{T+1}{t}}\sum _{s=1}^{T}h\left(s\right)\right)\hfill \\ \hfill & \ge {\displaystyle \frac{t}{T+1}}\left({\parallel u\parallel }_E-{\displaystyle \frac{2(T+1)}{p_0}}\sum _{s=1}^{T}h\left(s\right)\right)\hfill \\ \hfill & \ge \left({\parallel u\parallel }_E-{\displaystyle \frac{2(T+1)}{p_0}}\sum _{s=1}^{T}h\left(s\right)\right)q\left(t\right),t\in {\left[0,\left[{\displaystyle \frac{T+1}{2}}\right]\right]}_{\mathbb{Z}}.\hfill \end{array}$$

On the other hand, let $w_0\left(t\right)=-{\displaystyle \frac{1}{p\left(t\right)}}{\displaystyle \sum _{s=1}^t}h\left(s\right)$ and $z_0\left(t\right)={\displaystyle \sum _{s=t}^T}{w}_0\left(s\right),t\in {[0,T+1]}_{\mathbb{Z}}.$ Then, by the fact that $w_0\left(t\right)\le 0,$ $z_0\left(t\right)\le 0$ on ${[0,T+1]}_{\mathbb{Z}},$ we obtain

$$|w_0\left(t\right)|\le {\displaystyle \frac{1}{p\left(t\right)}}\sum _{s=1}^{T}h\left(s\right)\le {\displaystyle \frac{1}{p_0}}\sum _{s=1}^{T}h\left(s\right)$$

and

$$|z_0\left(t\right)|\le \sum _{s=t}^T\left|w_0\left(t\right)\right|\le \sum _{s=t}^T\left({\displaystyle \frac{1}{p_0}}\sum _{s=1}^{T}h\left(s\right)\right)={\displaystyle \frac{T+1-t}{p_0}}\sum _{s=1}^{T}h\left(s\right),t\in {[0,T+1]}_{\mathbb{Z}}.$$

Furthermore,

$$\begin{array}{l}\Delta \left[p(t-1)(\Delta u(t-1)-\Delta z_0(t-1))\right]\\ =\Delta \left[p(t-1)(\Delta u(t-1)+w_0(t-1))\right]\\ =\Delta [p(t-1)\Delta u(t-1)]-\Delta {\displaystyle \sum _{s=1}^{t-1}}h\left(s\right)\\ =\Delta [p(t-1)\Delta u(t-1)]-h\left(t\right)\le 0,t\in {[1,T]}_{\mathbb{Z}},\end{array}$$

and

$$(u-z_0)\left(0\right)\ge 0,(u-z_0)(T+1)\ge 0.$$

This implies that $u+z$ is concave on ${[0,T+1]}_{\mathbb{Z}},$ and then from (12), we obtain

$$\begin{array}{cc}\hfill u\left(t\right)& \ge \left({\parallel u\parallel }_E-{\displaystyle \frac{T+1-t}{p_0}}\sum _{s=1}^{T}h\left(s\right)\right){\displaystyle \frac{T+1-t}{T+1}}-{\displaystyle \frac{T+1-t}{p_0}}\sum _{s=1}^{T}h\left(s\right)\hfill \\ \hfill & \ge \left({\parallel u\parallel }_E-{\displaystyle \frac{2(T+1)}{p_0}}\sum _{s=1}^{T}h\left(s\right)\right){\displaystyle \frac{T+1-t}{T+1}},t\in {\left[\left[{\displaystyle \frac{T+1}{2}}\right]+1,T+1\right]}_{\mathbb{Z}}.\hfill \end{array}$$

(13)

Therefore, $u\left(t\right)\ge \left({\parallel u\parallel }_E-{\displaystyle \frac{2(T+1)}{p_0}}{\displaystyle \sum _{s=1}^T}h\left(s\right)\right)q\left(t\right)$ for $t\in {[0,T+1]}_{\mathbb{Z}}.$ In particular, if ${\parallel u\parallel }_E\ge {\displaystyle \frac{2(T+2)}{p_0}}{\displaystyle \sum _{s=1}^T}h\left(s\right),$ then it is not difficult to see that $u\left(t\right)\ge {\displaystyle \frac{1}{T+2}}{\parallel u\parallel }_{E}q\left(t\right),$ which completes the proof. □

3. Main Result

Theorem 1.

(i) Assume that conditions (F1)–(F5) are satisfied. Then, there exists a positive constant L and an interval $I\subseteq (0,\infty )$ such that if $f\left(m\right)\ge L$, the problem (4) admits at least two positive solutions for $\lambda \in I.$

(ii) Assume that conditions (F1)–(F4) are satisfied. Then, there exists a positive constant ${\lambda }_0>0$ such that problem (4) has a positive solution $u_{\lambda }$ for $\lambda <{\lambda }_0$ and $u_{\lambda }\to \infty$ as $\lambda \to 0^{+}$ on ${[1,T+1]}_{\mathbb{Z}}.$

Proof. 

(i) Since $\beta \in (0,1),$ it can be seen that $\underset{x\to 0^{+}}{lim}{\displaystyle \frac{x}{x^{\beta }}}=0.$ So, it can be seen that there is a constant $\gamma \in (0,m)$ satisfying

$$\begin{array}{c}\hfill {\displaystyle \frac{\gamma }{{\left(\frac{\rho \gamma }{T+2}\right)}^{\beta }}}<{\displaystyle \frac{g_0{p}_{0}m}{128p_1{m}^{\beta }}}{\left(\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}\right)}^{-1},\end{array}$$

(14)

where $g_0=\underset{{[\rho ,T+1-\rho ]}_{\mathbb{Z}}}{min}g,$ $\rho ={\left[{\displaystyle \frac{T+1}{4}}\right]}_{\mathbb{Z}}.$ Assume that $f\left(m\right)>L$ with

$$L=max\left\{{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }}},{\displaystyle \frac{am}{4{\left(\frac{\gamma }{T+2}\right)}^{1+\alpha }}},{\displaystyle \frac{16a{p}_1{m}^{\beta }}{p_0{g}_0{\rho }^{\beta }{\left(\frac{\gamma }{T+2}\right)}^{\alpha +\beta }}}\right\}.$$

Let

$$I=\left\{{\displaystyle \frac{32\gamma p_1}{(T+1)g_{0}f\left(\frac{\rho \gamma }{T+2}\right)}},{\displaystyle \frac{p_{0}m}{4(T+1)f\left(m\right)}}{\left(\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}\right)}^{-1}\right\}.$$

Then, $I\ne \varnothing .$ In fact, by $\left(F5\right)$, we obtain

$$f\left({\displaystyle \frac{\rho \gamma }{T+2}}\right)\ge {\left({\displaystyle \frac{\rho \gamma }{m(T+2)}}\right)}^{\beta }f\left(m\right).$$

Combining this with (14), one has

$${\displaystyle \frac{32\gamma p_1}{g_0(T+1)f\left(\frac{\rho \gamma }{T+2}\right)}}\le {\displaystyle \frac{16p_1{m}^{\beta }}{g_0(T+1)f\left(m\right)}}{\displaystyle \frac{8\gamma }{{\left(\frac{\rho \gamma }{T+2}\right)}^{\beta }}}<{\displaystyle \frac{p_{0}m}{4(T+1)f\left(m\right)}}{\left(\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}\right)}^{-1}.$$

Now, let us prove that (4) has at least two positive solutions for $\lambda \in I.$ For $\lambda \in I$ and $v\in E,$ $K_{\lambda }v=u,$ where u is the solution of the linear problem

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-\Delta [p(t-1)\Delta u(t-1)]=\lambda g\left(t\right)(f\left(\tilde{v}\right)-{\displaystyle \frac{a}{{\tilde{v}}^{\alpha }}}),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ u\left(0\right)=0,\Delta u\left(T\right)+H\left(u\right(T+1\left)\right)=0.\hfill \end{array}\right.\end{array}$$

(15)

Here, $\tilde{v}\left(t\right)=max(v\left(t\right),{\displaystyle \frac{\gamma }{T+2}}q\left(t\right)).$ From Lemma 2, we know that

$$\begin{array}{c}\hfill u\left(t\right)=\sum _{s=1}^t{\displaystyle \frac{1}{p(s-1)}}\left(p\left(0\right)C-\lambda \sum _{\ell =1}^{s-1}g(\ell )\left(f\left(\tilde{v}(\ell )\right)-{\displaystyle \frac{a}{{\tilde{v}}^{\alpha }(\ell )}}\right)\right),t\in {[1,T]}_{\mathbb{Z}},\end{array}$$

(16)

where C is the unique number such that $\Delta u\left(T\right)+H\left(u\right(T+1\left)\right)=0,$ i.e., the solution of

$$\begin{array}{l}\tilde{H}\left(C\right):={\displaystyle \frac{1}{p\left(T\right)}}\left(p\left(0\right)C-\lambda {\displaystyle \sum _{\ell =1}^T}g(\ell )(f\left(\tilde{v}\left(l\right)\right)-{\displaystyle \frac{a}{{\tilde{v}}^{\alpha }}}\left(l\right))\right)\\ +H\left({\displaystyle \sum _{s=1}^{T+1}}{\displaystyle \frac{1}{p(s-1)}}\left(p\left(0\right)C-\lambda {\displaystyle \sum _{\ell =1}^{s-1}}g(\ell )(f\left(\tilde{v}\left(l\right)\right)-{\displaystyle \frac{a}{{\tilde{v}}^{\alpha }}}\left(l\right))\right)\right)=0.\end{array}$$

Therefore, by Lemma 2,

$$\left|C\right|\le \lambda \sum _{\ell =1}^{T}g(\ell )\left|f\left(\tilde{v}\left(l\right)\right)-{\displaystyle \frac{a}{{\tilde{v}}^{\alpha }\left(l\right)}}\right|.$$

Furthermore, combine this with (16) such that

$$\begin{array}{cc}\hfill {\parallel u\parallel }_E& \le {\displaystyle \frac{2(T+1)\lambda }{p_0}}\sum _{\ell =1}^{T}g(\ell )\left|f\left(\tilde{v}\left(l\right)\right)-{\displaystyle \frac{a}{{\tilde{v}}^{\alpha }\left(l\right)}}\right|\hfill \\ & \le {\displaystyle \frac{2(T+1)\lambda }{p_0}}\sum _{s=1}^{T}g\left(s\right)\left|f\left(\tilde{v}\left(l\right)\right)\right|+\left|{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(s\right)}}\right|.\hfill \end{array}$$

(17)

From this and the standard arguments, it follows that $K_{\lambda }:E\to E$ is a completely continuous operator. Next, we verify the two conditions of Lemma 1.

$\left(a\right)$ Let $u\in E$ satisfy $u=\mu K_{\lambda }u$ for some $\mu \in (0,1].$ Then, ${\parallel u\parallel }_E\ne m.$

Let u be a solution $u/\mu =K_{\lambda }u$ and ${\parallel u\parallel }_E=m.$ Then, $\parallel \tilde{u}{\parallel }_E\le m.$ This, together with (16) and the assumption $f\left(m\right)>{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }}}$ gives

$$\begin{array}{cc}\hfill m\le {∥{\displaystyle \frac{u}{\mu }}∥}_E& \le {\displaystyle \frac{2(T+1)\lambda }{p_0}}\sum _{s=1}^{T}g\left(s\right)\left(f\left(m\right)+{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(s\right)}}\right)\hfill \\ \hfill & \le {\displaystyle \frac{2(T+1)\lambda }{p_0}}\sum _{s=1}^{T}g\left(s\right)\left(f\left(m\right)+{\displaystyle \frac{f\left(m\right)}{q^{\alpha }\left(s\right)}}\right)\le {\displaystyle \frac{4(T+1)\lambda f\left(m\right)}{p_0}}\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}.\hfill \end{array}$$

Since $m\le {∥{\displaystyle \frac{u}{\mu }}∥}_E,$ we know that $m\le {\displaystyle \frac{4(T+1)\lambda f\left(m\right)}{p_0}}{\displaystyle \sum _{s=1}^T}{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}.$ Furthermore, for

$$\lambda \ge {\displaystyle \frac{p_{0}m}{4(T+1)f\left(m\right)}}{\left(\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}\right)}^{-1},$$

we get a contradiction with $\lambda \in I.$ Thus, ${\parallel u\parallel }_E\ne m.$

$\left(b\right)$ Let $u\in E$ satisfy $u=K_{\lambda }u+\xi$ for some $\xi \ge 0.$ Then, ${\parallel u\parallel }_E\notin \left\{\gamma ,R\right\}$ for $R\gg 1.$

Let u be a solution of $u-\xi =K_{\lambda }u.$ Then, u satisfies

$$\left\{\begin{array}{c}-\Delta [p(t-1)\Delta u(t-1)]=\lambda g\left(t\right)(f\left(\tilde{u}\right)-{\displaystyle \frac{a}{{\tilde{u}}^{\alpha }}}),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ u\left(0\right)=\xi \ge 0,\Delta u\left(T\right)+H\left(u\right(T+1\left)\right)=H\left(u\right(T+1\left)\right)-H\left(u\right(T+1)-\xi )\ge 0.\hfill \end{array}\right.$$

We observe that

$$\lambda g\left(t\right)\left(f\left(\tilde{u}\right)-{\displaystyle \frac{a}{{\tilde{u}}^{\alpha }}}\right)\ge -{\displaystyle \frac{a\lambda g\left(t\right)}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(t\right)}}:=-h_{\lambda }\left(t\right).$$

It follows from $f\left(m\right)>{\displaystyle \frac{am}{4{\left(\frac{\gamma }{T+2}\right)}^{1+\alpha }}}$ and $\lambda \in I$ that

$$\lambda <{\displaystyle \frac{p_{0}m}{4(T+1)f\left(m\right)}}{\left(\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}\right)}^{-1}\le {\displaystyle \frac{p_0{\left(\frac{\gamma }{T+2}\right)}^{1+\alpha }}{a(T+1)}}{\left(\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}\right)}^{-1}.$$

Subsequently,

$${\displaystyle \frac{\gamma }{T+2}}>{\displaystyle \frac{a(T+1)\lambda }{p_0{\left(\frac{\gamma }{T+2}\right)}^{\alpha }}}\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}},$$

which implies that

$$\begin{array}{c}\hfill \gamma >{\displaystyle \frac{a\lambda (T+1)(T+2)}{p_0{\left(\frac{\gamma }{T+2}\right)}^{\alpha }}}\left(\sum _{s=1}^T{\displaystyle \frac{g}{q^{\alpha }}}\right)={\displaystyle \frac{(T+2)(T+1)}{p_0}}\sum _{s=1}^T\left|h_{\lambda }\left(s\right)\right|.\end{array}$$

(18)

Assume that ${\parallel u\parallel }_E\in \left\{\gamma ,R\right\}$ with $R>\gamma .$ From Lemma 4 (ii) and

$${\parallel u\parallel }_E>\gamma \ge {\displaystyle \frac{2(T+2)}{p_0}}\sum _{s=1}^T\left|h_{\lambda }\left(s\right)\right|,$$

we have

$$\begin{array}{c}\hfill u\left(t\right)\ge {\displaystyle \frac{1}{T+2}}{\parallel u\parallel }_{E}q\left(t\right),t\in {[1,T]}_{\mathbb{Z}}.\end{array}$$

(19)

Specially, $u\left(t\right)\ge {\displaystyle \frac{\gamma }{T+2}}q\left(t\right).$ So, $\tilde{u}\equiv u,$ and $u\left(t\right)\ge {\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E$ for $t\in {[\rho ,T+1-\rho ]}_{\mathbb{Z}}$. Thus, u is a solution of the following problem

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-\Delta [p(t-1)\Delta u(t-1)]\ge \lambda g\left(t\right)\left(f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)-{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(t\right)}}\right),t\in {[\rho +1,T-\rho ]}_{\mathbb{Z}},\hfill \\ u\left(\rho \right)\ge 0,u(T+1-\rho )\ge 0.\hfill \end{array}\right.\end{array}$$

(20)

In view of the comparison principle, it is not difficult to see that $u\left(t\right)\ge v\left(t\right)$ on ${[\rho ,T+1-\rho ]}_{\mathbb{Z}},$ where $v\left(t\right)$ is the positive solution of

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-\Delta [p(t-1)\Delta v(t-1)]=\lambda g\left(t\right)\left(f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)-{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(t\right)}}\right),t\in {[\rho +1,T-\rho ]}_{\mathbb{Z}},\hfill \\ v\left(\rho \right)=0,v(T+1-\rho )=0.\hfill \end{array}\right.\end{array}$$

(21)

Let $t_0\in {[\rho +1,T-\rho ]}_{\mathbb{Z}}$ be such that $\Delta v\left(t_0\right)\ge 0$ and $\Delta v(t_0+1)\le 0,$ according to definition of $T,$ ${[\rho +1,T-\rho ]}_{\mathbb{Z}}\ne \varnothing .$ Then, summing from $s=t+1$ to $t_0$, we obtain

$$\Delta v\left(t\right)={\displaystyle \frac{p\left(t_0\right)}{p\left(t\right)}}\Delta v\left(t_0\right)+{\displaystyle \frac{\lambda }{p\left(t\right)}}\sum _{s=t+1}^{t_0}g\left(s\right)\left(f\left({\displaystyle \frac{{\rho \parallel u\parallel }_E}{T+2}}\right)-{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(s\right)}}\right),t\in {[\rho ,t_0-1]}_{\mathbb{Z}},$$

then,

$$\begin{array}{c}\hfill \Delta v\left(t\right)\ge {\displaystyle \frac{\lambda }{p\left(t\right)}}\sum _{s=t+1}^{t_0}g\left(s\right)\left(f\left({\displaystyle \frac{{\rho \parallel u\parallel }_E}{T+2}}\right)-{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(s\right)}}\right),t\in {[\rho ,t_0-1]}_{\mathbb{Z}}.\end{array}$$

(22)

Meanwhile, if $t\in {[t_0+2,T+1-\rho ]}_{\mathbb{Z}},$ then

$$\Delta v\left(t\right)={\displaystyle \frac{p(t_0+1)}{p\left(t\right)}}\Delta v(t_0+1)-{\displaystyle \frac{\lambda }{p\left(t\right)}}\sum _{s=t_0+2}^{t}g\left(s\right)\left(f\left({\displaystyle \frac{{\rho \parallel u\parallel }_E}{T+2}}\right)-{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(s\right)}}\right)$$

and by the fact that $p\left(t\right)>0$ on ${[0,T]}_{\mathbb{Z}}$ and $\Delta v(t_0+1)\le 0$, we obtain

$$\begin{array}{c}\hfill \Delta v\left(t\right)\le -{\displaystyle \frac{\lambda }{p\left(t\right)}}\sum _{s=t_0+2}^{t}g\left(s\right)\left(f\left({\displaystyle \frac{{\rho \parallel u\parallel }_E}{T+2}}\right)-{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(s\right)}}\right).\end{array}$$

(23)

The rest proof will be divided into two cases.

• Case 1. $t_0>\left[{\displaystyle \frac{T+1}{2}}\right].$

By summing (22) from $s=\rho$ to $s=\left[{\displaystyle \frac{3}{8}}(T+1)\right],$ we obtain

$$\begin{array}{cc}v\left(\left[{\displaystyle \frac{3}{8}}(T+1)\right]\right)& \ge {\displaystyle \sum _{s=\rho }^{\left[{\displaystyle \frac{3}{8}}(T+1)\right]-1}}\left({\displaystyle \frac{\lambda }{p\left(s\right)}}{\displaystyle \sum _{\ell =s+1}^{t_0}}g(\ell )\left(f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)-{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(s\right)}}\right)\right)\hfill \\ & \ge {\displaystyle \sum _{s=\rho }^{\left[{\displaystyle \frac{3}{8}}(T+1)\right]-1}}{\displaystyle \frac{\lambda }{p\left(s\right)}}\left({\displaystyle \sum _{s=\left[{\displaystyle \frac{3}{8}}(T+1)\right]}^{\left[{\displaystyle \frac{T+1}{2}}\right]}}g\left(s\right)f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)-{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }}}{\displaystyle \sum _{s=1}^T}{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}\right).\hfill \end{array}$$

Furthermore, since $T>8,$ we know that $\left[{\displaystyle \frac{T+1}{2}}\right]\ge \left[{\displaystyle \frac{3}{8}}(T+1)\right]+1,$ and then,

$$\begin{array}{cc}\hfill v\left(\left[{\displaystyle \frac{3}{8}}(T+1)\right]\right)& \ge \sum _{s=\rho }^{\left[{\displaystyle \frac{3}{8}}(T+1)\right]-1}\left({\displaystyle \frac{\lambda }{p_1}}\sum _{s=\left[{\displaystyle \frac{3}{8}}(T+1)\right]}^{\left[{\displaystyle \frac{T+1}{2}}\right]}{g}_{0}f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)-{\displaystyle \frac{a\lambda }{p_0{\left(\frac{\gamma }{T+2}\right)}^{\alpha }}}\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{T+1}{16}}\left(\lambda \left({\displaystyle \frac{g_0(T+1)}{16p_1}}f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)-{\displaystyle \frac{a}{p_0{\left(\frac{\gamma }{T+2}\right)}^{\alpha }}}\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}\right)\right).\hfill \end{array}$$

(24)

By the assumption $\left(F5\right)$ and

$$f\left(m\right)>{\displaystyle \frac{16a{p}_1{m}^{\beta }}{g_0{p}_0{\rho }^{\beta }{\left(\frac{\gamma }{T+2}\right)}^{\alpha +\beta }}}.\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}},$$

we obtain

$$\begin{array}{l}{\displaystyle \frac{g_0(T+1)}{16p_1}}f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)\ge {\displaystyle \frac{2g_0}{16p_1}}f\left({\displaystyle \frac{\rho \gamma }{T+2}}\right)\\ \ge {\displaystyle \frac{2g_0}{16p_1}}{\left({\displaystyle \frac{\rho \gamma }{m(T+2)}}\right)}^{\beta }f\left(m\right)\ge {\displaystyle \frac{2a}{p_0{\left(\frac{\gamma }{T+2}\right)}^{\alpha }}}{\displaystyle \sum _{s=1}^T}{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}.\end{array}$$

So, it follows from (24) that

$$v\left(\left[{\displaystyle \frac{3}{8}}(T+1)\right]\right)\ge {\displaystyle \frac{(T+1)\lambda g_0}{32p_1}}f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right).$$

Furthermore,

$$\begin{array}{c}\hfill {\parallel u\parallel }_E\ge {\displaystyle \frac{(T+1)\lambda g_0}{32p_1}}f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right).\end{array}$$

(25)

Case 2. $t_0<\left[{\displaystyle \frac{T+1}{2}}\right].$

Summing (23) from $s=\left[{\displaystyle \frac{5}{8}}(T+1)\right]+1$ to $s=T-\rho ,$ then

$$v\left(\left[{\displaystyle \frac{5}{8}}(T+1)\right]\right)\ge \sum _{s=\left[{\displaystyle \frac{5}{8}}(T+1)\right]+1}^{T-\rho }\left({\displaystyle \frac{\lambda }{p\left(s\right)}}\sum _{\ell =t_0+2}^{s}g(\ell )\left(f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)-{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }\left(s\right)}}\right)\right).$$

Furthermore, since $T>8,$ we know that $\left[{\displaystyle \frac{5}{8}}(T+1)\right]\ge \left[{\displaystyle \frac{T+1}{2}}\right]+1,$ and then,

$$\begin{array}{cc}\hfill v\left(\left[{\displaystyle \frac{5}{8}}(T+1)\right]\right)& \ge \sum _{s=\left[{\displaystyle \frac{5}{8}}(T+1)\right]+1}^{T-\rho }\left({\displaystyle \frac{\lambda }{p\left(s\right)}}\sum _{\ell =\left[{\displaystyle \frac{T+1}{2}}\right]+1}^{\left[{\displaystyle \frac{5}{8}}(T+1)\right]}g(\ell )\left(f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)-{\displaystyle \frac{a}{{\left(\frac{\gamma }{T+2}\right)}^{\alpha }{q}^{\alpha }(\ell )}}\right)\right)\hfill \\ \hfill & \ge \sum _{s=\left[{\displaystyle \frac{5}{8}}(T+1)\right]+1}^{T-\rho }\left(\sum _{\ell =\left[{\displaystyle \frac{T+1}{2}}\right]+1}^{\left[{\displaystyle \frac{5}{8}}(T+1)\right]}{\displaystyle \frac{\lambda g_0}{p_1}}f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)-{\displaystyle \frac{a\lambda }{p_0{\left(\frac{\gamma }{T+2}\right)}^{\alpha }}}\sum _{\ell =1}^T{\displaystyle \frac{g(\ell )}{q^{\alpha }(\ell )}}\right)\hfill \\ \hfill & \ge {\displaystyle \frac{T+1}{16}}\left({\displaystyle \frac{\lambda g_0(T+1)}{p_1}}f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right)-{\displaystyle \frac{\lambda a}{p_0{\left(\frac{\gamma }{T+2}\right)}^{\alpha }}}\sum _{s=1}^T{\displaystyle \frac{g\left(s\right)}{q^{\alpha }\left(s\right)}}\right).\hfill \end{array}$$

(26)

The assumption $f\left(m\right)>{\displaystyle \frac{16a{p}_1{m}^{\beta }}{g_0{p}_0{\rho }^{\beta }{\left(\frac{\gamma }{T+2}\right)}^{\alpha +\beta }}}$ implies that (25) holds. Therefore, (25) holds for these above two cases.

If ${\parallel u\parallel }_E=\gamma ,$ by (25), one has $\gamma \ge {\displaystyle \frac{(T+1)\lambda g_0}{32p_1}}f\left({\displaystyle \frac{\rho \gamma }{T+2}}\right)$ and subsequently, $\lambda \le {\displaystyle \frac{32\gamma p_1}{(T+1)g_{0}f\left(\frac{\rho \gamma }{T+2}\right)}}.$ Then, we arrive at a contradiction with $\lambda \in I.$ Thus, ${\parallel u\parallel }_E\ne \gamma$. From (25), we know that

$${\displaystyle \frac{f\left(\frac{\rho }{T+2}{\parallel u\parallel }_E\right)}{{\parallel u\parallel }_E}}\le {\displaystyle \frac{32p_1}{\lambda g_0(T+1)}}.$$

Meanwhile, by $\left(F4\right)$ and $\left(F5\right),$ we know that $\underset{z\to \infty }{lim}{\displaystyle \frac{f\left(\frac{\rho }{T+2}z\right)}{z}}=\infty .$ Hence, ${\parallel u\parallel }_E\ne R$ for $R\gg 1.$ By Lemma 1, $K_{\lambda }$ has two fixed points $u_{i,\lambda },i=1,2,$ such that $\gamma <\parallel u_{1,\lambda }{\parallel }_E<m,$ $m<\parallel u_{2,\lambda }{\parallel }_E<R.$ Since $\parallel u_{i,\lambda }{\parallel }_E\ge \gamma ,$ it follows from (19) with $\xi =0$ that $\parallel u_{i,\lambda }{\parallel }_E\ge {\displaystyle \frac{\gamma }{T+2}}q\left(t\right)$ for $t\in {[1,T]}_{\mathbb{Z}}$ i.e., ${\tilde{u}}_{i,\lambda }=u_{i,\lambda }$ on ${[1,T]}_{\mathbb{Z}}$ for $i=1,2.$ Therefore, $u_{i,\lambda },i=1,2,$ are positive solutions of Theorem 1.

(ii) We will adjust the proof of part (i). Let $\lambda >0$ and satisfy the following inequality

$${\displaystyle \frac{2(T+1)\lambda }{p_0}}\sum _{s=1}^{T}g\left(s\right)\left(f(T+2)+{\displaystyle \frac{a}{q^{\alpha }\left(t\right)}}\right)<T+2.$$

Let $S_{\lambda }v=u$ for $v\in E.$ Here, u is a solution of the following problem

$$\left\{\begin{array}{c}-\Delta [p(t-1)\Delta u(t-1)]=\lambda g\left(t\right)(f\left(\tilde{v}\right)-{\displaystyle \frac{a}{{\tilde{v}}^{\alpha }}}),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ u\left(0\right)=0,\Delta u\left(T\right)+H\left(u\right(T+1\left)\right)=0,\hfill \end{array}\right.$$

where $\tilde{v}\left(t\right)=max(v\left(t\right),q\left(t\right)).$ By the standard argument as in [21], we know that $S_{\lambda }:E\to E$ is a completely continuous operator.

Next, we will prove the following two cases hold.

$\left(c\right)$ Suppose that $u\in E$ with $u=\mu K_{\lambda }u$ for some $\mu \in (0,1].$ Then, ${\parallel u\parallel }_E\ne T+2.$

Suppose on the contrary that ${\parallel u\parallel }_E=T+2.$ Then, similar to the proof of part (a), we obtain

$$T+2={\parallel u\parallel }_E\le {\displaystyle \frac{2(T+1)\lambda }{p_0}}\sum _{s=1}^{T}g\left(s\right)\left(f(T+2)+{\displaystyle \frac{a}{q^{\alpha }\left(s\right)}}\right)<T+2,$$

This contradicts the assumption that $\lambda \in I$. So, ${\parallel u\parallel }_E\ne T+2.$

$\left(d\right)$ Let $u\in E$ satisfy $u=K_{\lambda }u+\xi$ for some $\xi \ge 0.$ Then, ${\parallel u\parallel }_E\ne R$ for $R\gg 1.$

Let ${\parallel u\parallel }_E=R.$ By the similar proof in part $\left(b\right)$ with $\gamma =T+2,$ we arrive at

$$R={\parallel u\parallel }_E>{\displaystyle \frac{2(T+2)}{p_0}}\sum _{s=1}^T\left|h_{\lambda }\left(s\right)\right|,$$

where $h_{\lambda }\left(t\right)={\displaystyle \frac{\lambda ag\left(t\right)}{q^{\alpha }\left(t\right)}}.$ Therefore, Lemma 4 (ii) implies that

$$\begin{array}{c}\hfill u\left(t\right)\ge {\displaystyle \frac{{\parallel u\parallel }_E}{T+2}}q\left(t\right)\ge q\left(t\right),t\in {[0,T+1]}_{\mathbb{Z}},\end{array}$$

(27)

That is to say, $\tilde{u}=u$ on ${[0,T+1]}_{\mathbb{Z}}.$ Similar to (24) and (26), one has

$${\parallel u\parallel }_E\ge {\displaystyle \frac{(T+1)g_0\lambda }{32p_1}}f\left({\displaystyle \frac{\rho }{T+2}}{\parallel u\parallel }_E\right),$$

in another word,

$$\begin{array}{c}\hfill {\displaystyle \frac{(T+1)g_{0}f\left(\frac{\rho }{T+2}{\parallel u\parallel }_E\right)}{32p_1{\parallel u\parallel }_E}}\le {\displaystyle \frac{1}{\lambda }}.\end{array}$$

(28)

It is not difficult to see that the left side of (28) approaches ∞ as ${\parallel u\parallel }_E$ approaches $\infty ,$ so, ${\parallel u\parallel }_E<R$ for a large enough R. This contradicts ${\parallel u\parallel }_E=R$. Thus, $\left(d\right)$ holds.

Now, from Lemma 4, the operator $S_{\lambda }$ has a fixed-point $u_{\lambda }$ such that

$${\parallel u_{\lambda }\parallel }_E>{\displaystyle \frac{2(T+2)}{p_0}}\sum _{s=1}^T\left|h_{\lambda }\left(s\right)\right|.$$

Combining this with (27) (with $\xi =0$), we find that ${\tilde{u}}_{\lambda }=u_{\lambda }$ on ${[0,T+1]}_{\mathbb{Z}}.$ So, by the definition of ${\tilde{u}}_{\lambda }$, we know that $u_{\lambda }$ is a positive solution of the problem (4).

Next, we prove that $\parallel u_{\lambda }{\parallel }_E\to \infty$ as $\lambda \to 0^{+}.$ By using the fact that $H\left(u_{\lambda }(T+1)\right)\ge H\left(0\right)=0$ and the second boundary condition of $u_{\lambda }$, we have $\Delta u_{\lambda }\left(T\right)\le 0.$

Therefore, $u_{\lambda }$ is the solution of

$$\left\{\begin{array}{c}-\Delta [p(t-1)\Delta u_{\lambda }(t-1)]\le \lambda g\left(t\right)f(\parallel u_{\lambda }{\parallel }_E),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ u_{\lambda }\left(0\right)=0,\Delta u_{\lambda }\left(T\right)\le 0.\hfill \end{array}\right.$$

Let $v_{\lambda }$ be the solution of

$$\left\{\begin{array}{c}-\Delta [p(t-1)\Delta v_{\lambda }(t-1)]=\lambda g\left(t\right)f(\parallel u_{\lambda }{\parallel }_E),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ v_{\lambda }\left(0\right)=0,\Delta v_{\lambda }\left(T\right)=0.\hfill \end{array}\right.$$

Then, it follows from the comparison principle that $u_{\lambda }\le v_{\lambda }$ on ${[0,T+1]}_{\mathbb{Z}}.$

Observe that

$$v_{\lambda }\left(t\right)=\sum _{s=0}^{t-1}\left({\displaystyle \frac{\lambda f(\parallel u_{\lambda }{\parallel }_E)}{p\left(s\right)}}\sum _{\ell =s+1}^{T}g(\ell )\right)\le {\displaystyle \frac{\lambda (T+1){\sum }_{s=1}^{T}g\left(s\right)f(\parallel u_{\lambda }{\parallel }_E)}{p_0}},$$

for $t\in {[1,T]}_{\mathbb{Z}}.$ Therefore,

$${\parallel u_{\lambda }\parallel }_E\le {\displaystyle \frac{\lambda (T+1){\sum }_{s=1}^{T}g\left(s\right)f(\parallel u_{\lambda }{\parallel }_E)}{p_0}},$$

in other words,

$${\displaystyle \frac{f(\parallel u_{\lambda }{\parallel }_E)}{\parallel u_{\lambda }{\parallel }_E}}\ge {\displaystyle \frac{p_0}{\lambda (T+1){\sum }_{s=1}^{T}g\left(s\right)}}.$$

Thereby,

$$\underset{\lambda \to 0^{+}}{lim}{\displaystyle \frac{f(\parallel u_{\lambda }{\parallel }_E)}{\parallel u_{\lambda }{\parallel }_E}}=\infty .$$

On the other hand,

$${\parallel u_{\lambda }\parallel }_E>{\displaystyle \frac{2(T+1)}{p_0}}\sum _{s=1}^T\left|h_{\lambda }\left(s\right)\right|.$$

By $\left(F4\right)$, $\underset{\lambda \to 0^{+}}{lim}{\parallel u_{\lambda }\parallel }_E=\infty .$

Then, by Lemma 4 (ii), we obtain

$$u_{\lambda }\left(t\right)\ge \left(\parallel u_{\lambda }{\parallel }_E-{\displaystyle \frac{T+1}{p_0}}\sum _{s=1}^T\left|h_{\lambda }\left(s\right)\right|\right),$$

which implies $\underset{\lambda \to 0^{+}}{lim}{u}_{\lambda }\left(t\right)=\infty$ for $t\in {[1,T]}_{\mathbb{Z}}$. □

Example 1.

Let $g\left(t\right)=\sqrt{t+1}$ and $H\left(u\right)=arctanu$. Consider the boundary value problem

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-\Delta [p(t-1)\Delta u(t-1)]=\lambda \sqrt{t+1}(f\left(u\right)-{\displaystyle \frac{a}{u^{\alpha }}}),t\in {[1,T]}_{\mathbb{Z}},\hfill \\ u\left(0\right)=0,\Delta u\left(T\right)+arctanu(T+1)=0\hfill \end{array}\right.\end{array}$$

(29)

with

$$\begin{array}{c}\hfill f\left(s\right)=\left\{\begin{array}{c}L{s}^{\beta },0\le s\le 1,\hfill \\ L{s}^{\gamma },s>1,\hfill \end{array}\right.\end{array}$$

where the constant $L>0$ is large enough, $0<\beta <1$ and $\gamma >1$ are two constants. Then, f satisfies the condition $\left(F4\right)$ and $\left(F5\right)$ with $m=1$. In fact, according to the definition of $f\left(s\right)$, it is easy to see that $f\left(s\right)$ is continuous and nondecreasing on $[0,\infty )$; meanwhile, it follows from $\gamma >1$ that

$$\underset{s\to \infty }{lim}{\displaystyle \frac{f\left(s\right)}{s}}=\underset{s\to \infty }{lim}{\displaystyle \frac{L{s}^{\gamma }}{s}}=\underset{s\to \infty }{lim}L{s}^{\gamma -1}=\infty .$$

Therefore, f satisfies the condition $\left(F4\right)$. Furthermore, for $z\in (0,m)$ with $m=1$, $f\left(z\right)=L{z}^{\beta }={\left({\displaystyle \frac{z}{m}}\right)}^{\beta }f\left(m\right).$ Therefore, $\left(F5\right)$ holds.

Therefore, the conclusion of Theorem 1 (i) holds, i.e., the boundary value problem (29) has two positive solution for $\lambda$ in a suitable range.

Additionally, let $f\left(u\right)=u^{\gamma }{\pi }^{-{\displaystyle \frac{1}{u}}}$, $\gamma >1$. Then, f satisfies $\left(F4\right)$. Then, the conclusion of Theorem 1 (ii) holds, that is to say, there exists a constant ${\lambda }_0>0$ such that the boundary value problem (29) has a positive solution for $\lambda <{\lambda }_0$.

4. Conclusions

We investigated a discrete nonlinear singular semipositone problem with a nonlinear boundary condition. By constructing auxiliary problems and using the Krasnosel’skii fixed-point theorem, we determined the parameter ranges for the existence of at least one and at least two positive solutions. An example was provided to illustrate the application of the results. Overall, our study contributes to the understanding of the existence and multiplicity of positive solutions for such discrete problems.