Existence of positive solutions to discrete second-order boundary value problems with indefinite weight

Abstract

Let T > 1 be an integer, $T=\left\{1,2,...,T\right\}$. This article is concerned with the global structure of the set of positive solutions to the discrete second-order boundary value problems

$$\begin{array}{c}{\Delta }^{2}u\left(t-1\right)+rm\left(t\right)f\left(u\left(t\right)\right)=0,t\in T,\\ u\left(0\right)=u\left(T+1\right)=0,\end{array}$$

where r ≠ 0 is a parameter, $m:T\to ℝ$ changes its sign, m(t) ≠ 0 for $t\in T$ and f : ℝ → ℝ is continuous. Also, we obtain the existence of two principal eigenvalues of the corresponding linear eigenvalue problems.

MSC (2010): 39A12; 34B18.

1 Introduction

Let T > 1 be an integer, $T=\left\{1,2,...,T\right\}$. This article is concerned with the global structure of the set of positive solutions to the discrete second-order boundary value problem (BVP)

$${\Delta }^{2}u\left(t-1\right)+rm\left(t\right)f\left(u\left(t\right)\right)=0,t\in T,$$

(1.1)

$$u\left(0\right)=u\left(T+1\right)=0,$$

(1.2)

where r ≠ 0 is a parameter, f : ℝ → ℝ is continuous, m(t) ≠ 0 for $t\in T$ and $m:T\to ℝ$ changes its sign, i.e., there exists a proper subset $T_{+}$ of $T$, such that m(t) > 0 for $t\in T_{+}$ and m(t) < 0 for $t\in T\backslash T_{+}$.

BVPs with indefinite weight arise from a selection-migration model in population genetics, see Fleming [1]. That an allele A₁ holds an advantage over a rival allele A₂ at some points and holds an disadvantage over A₂ at some other points can be presented by changing signs of m. The parameter r corresponds to the reciprocal of the diffusion. The existence and multiplicity of positive solutions of BVPs for second-order differential equations with indefinite weight has been studied by many authors, see, for example [2–5] and the references therein. In [2], using Crandall-Rabinowitz's Theorem and Rabinowitz's global bifurcation theorem, Delgado and Suárez obtained the existence and multiplicity of positive solutions under Dirichlet boundary value condition. In 2006, Afrouzi and Brown [3] also obtained the similar results by using the mountain pass theorem. When f is concave-convex type, similar results were also obtained by Ma and Han [4] and Ma et al. [5], and the main tool they used was the Rabinowitz's global bifurcation theorem.

For the discrete case, there is much literature dealing with different equations similar to (1.1) subject to various boundary value conditions. We refer to [6–14] and the reference therein. In particular, when m(t) > 0 on $T$, fixed point theorems, the discrete Gelfand theorem and the bifurcation techniques have been used to discuss the existence of positive solutions to the discrete problems, see, for example [6–8, 12–14]. However, there are few results on the existence of positive solutions of (1.1) and (1.2) when m(t) changes its sign on $T$. Maybe the main reason is that the spectrum of the following linear eigenvalue problems

$${\Delta }^{2}u\left(t-1\right)+\lambda m\left(t\right)u\left(t\right)=0,t\in T,$$

(1.3)

$$u\left(0\right)=u\left(T+1\right)=0$$

(1.4)

is not clear when m changes its sign on $T$.

It is another aim of our article to give some information of the spectrum of (1.3) and (1.4). In this article, we will show that (1.3) and (1.4) has two principal eigenvalues λ_m,-< 0 < λ_m,+, and the corresponding eigenfunctions which we denote by ψ_m,-and ψ_m,+don't change their signs on $T$. Based on this result, using Rabinowitz's global bifurcation theorem [15], we will discuss the global structure of the set of positive solutions of (1.1), (1.2), and obtain the existence of positive solutions of (1.1) and (1.2). Moreover, we can also obtain the existence of negative solutions of (1.1) and (1.2).

Now, we give the definition of a positive solution and a negative solution of (1.1) and (1.2).

Definition 1.1. A positive solution of problem (1.1) and (1.2) refers to a pair (r, u), where r ≠ 0, u is a solution of (1.1) with u > 0 on $T$ and u satisfies (1.2). Meanwhile m (r, u) is called a negative solution of (1.1) and (1.2), if (r, -u) is a positive solution of (1.1) and (1.2).

The article is arranged as follows. In Section 2, we state the Rabinowitz's global bifurcation theorem. In Section 3, the existence of two principal eigenvalues of (1.3) and (1.4) will be discussed. In Section 4, we state the main result and provide the proof.

2 Preliminaries

For the readers' convenience, we state the Rabinowitz's global bifurcation theorem [15] here.

Let E be a real Banach space. Consider the equation

$$u=\lambda Tu+H\left(\lambda ,u\right),$$

(2.1)

which possesses the line of solutions {(λ,0)|λ ∈ ℝ} henceforth referred to as the trivial solutions, where T : E → E is a bounded linear operator and H(λ, u) is continuous on ℝ × E with H(λ, u) = o⊠ u⊠ near u = 0 uniformly on bounded λ intervals. Moreover, we assume that T and H are compact on E and ℝ × E, respectively, i.e., are continuous and they map bounded sets into relatively compact sets.

we will say μ is a characteristic value of T if there exists v ∈ E, v ≠ 0, such that v = μTv, i.e., μ^-1 is a nonzero eigenvalue of T. Let r(T) denote the set of real characteristic values of T and Γ denote the closure of the set of nontrivial solutions of (2.1).

Theorem 2.1 ([15, Theorem 1.3]). If μ ∈ r(T) is of odd multiplicity, then Γ contains a maximum subcontinuum $\mathcal{C}$ such that $\left(\mu ,0\right)\in \mathcal{C}$ and either

(i) meets ∞ in ℝ × E,

or

(ii) meets $\left(\tilde{\mu },0\right)$ where $\mu \ne \tilde{\mu }\in r\left(T\right)$.

From [15], there exist two connected subsets, ${\mathcal{C}}^{+}$ and ${\mathcal{C}}^{-}$, of $\mathcal{C}$ such that $\mathcal{C}={\mathcal{C}}^{+}\cup {\mathcal{C}}^{-}$ and ${\mathcal{C}}^{+}\cap {\mathcal{C}}^{-}=\left\{\left(\mu ,0\right)\right\}$. Furthermore, Rabinowitz also shows that

Theorem 2.2 ([15, Theorem 1.40]). Each of ${\mathcal{C}}^{+},{\mathcal{C}}^{-}$ meets (μ, 0) and either

(i) meets ∞ in ℝ × E,

or

(ii) meets $\left(\tilde{\mu },0\right)$ where $\mu \ne \tilde{\mu }\in r\left(T\right)$.

3 Existence of two principal eigenvalues to (1.3) and (1.4)

Recall that $T=\left\{1,2,...,T\right\}$. Let $\widehat{T}=\left\{0,1,...,T+1\right\}$. Let $X=\left\{u:\widehat{T}\to ℝ|u\left(0\right)=u\left(T+1\right)=0\right\}$. Then X is a Banach space under the norm ${∥u∥}_X=\underset{t\in \widehat{T}}{\text{max}}\left|u\left(t\right)\right|$. Let $Y=\left\{u|u:T\to ℝ\right\}$. Then Y is a Banach space under the norm ${∥u∥}_Y=\underset{t\in \mathsf{\text{T}}}{\text{max}}\left|u\left(t\right)\right|$.

Define the operator L : X → Y by

$$Lu\left(t\right)=-{\Delta }^{2}u\left(t-1\right),t\in T.$$

In this section, we will discuss the existence of principal eigenvalues for the BVP (1.3) and (1.4). At first, we give the definition of principal eigenvalue of (1.3) and (1.4).

Definition 3.1. An eigenvalue λ for (1.3) and (1.4) is called principal if there exists a nonnegative eigenfunction corresponding to λ, i.e., if there exists a nonnegative u ∈ X \ {0} such that (λ, u) is a solution of (1.3) and (1.4).

The main idea we will use arises from [16, 17]. For the reader's convenience, we state them here. At first, it is necessary to provide the definition of simple eigenvalue.

Definition 3.2. An eigenvalue λ of (1.3) and (1.4) is called simple if dim ${\bigcup }_{n=1}^{\infty }ker{\left(I-\lambda L^{-1}\right)}^n=1$, where kerA denotes the kernel of A.

Theorem 3.1. (1.3) and (1.4) has two simple principal eigenvalues.

Proof. Consider, for fixed λ, the eigenvalue problems

$$Lu-\lambda m\left(t\right)u\left(t\right)=\mu u\left(t\right),t\in T,$$

(3.1)

$$u\left(0\right)=u\left(T+1\right)=0.$$

(3.2)

By Kelley and Peterson [18, Theorem 7.6], for fixed λ, (3.1), and (3.2) has T simple eigenvalues

$${\mu }_{m,1}\left(\lambda \right)<{\mu }_{m,2}\left(\lambda \right)<\cdots <{\mu }_{m,T}\left(\lambda \right),$$

and the corresponding eigenfunction ψ_{m, k}(λ, t) has exactly k - 1 simple generalized zeros.

Thus, λ is a principal eigenvalue of (1.3) and (1.4), if and only if μ_m,1(λ) = 0.

On the other hand, let

$$S_{m,\lambda }=\left\{\sum _{t=0}^T|\Delta \varphi \left(t\right){|}^2-\lambda \sum _{t=1}^{T}m\left(t\right)\varphi {\left(t\right)}^2:\varphi \in X,\sum _{t=1}^T\varphi {\left(t\right)}^2=1\right\}.$$

Clearly, S_m,λis bounded below and μ_m,1(λ) = inf_ϕ∈XS_m,λ, see [18, Theorem 7.7].

For fixed $\varphi \in X,\lambda \to {\sum }_{t=0}^T|\Delta \varphi \left(t\right){|}^2-\lambda {\sum }_{t=1}^{T}m\left(t\right){\varphi }^2\left(t\right)$ is an affine function and so a concave function. As the infimum of any collection of concave functions is concave, it follows that λ → μ_m,1(λ) is a concave function. Also, by considering test functions ϕ₁, ϕ₂ ∈ X such that ${\sum }_{t=1}^{T}m\left(t\right){\varphi }_1^2\left(t\right)<0$ and ${\sum }_{t=1}^{T}m\left(t\right){\varphi }_2^2\left(t\right)>0$, it is easy to see that μ_m,1(λ) → -∞ as λ → ±∞. Thus, λ → μ_m,1(λ) is an increasing function until it attains its maximum, and is a decreasing function thereafter.

Since μ_m,1(0) > 0, λ → μ_m,1(λ) must have exactly two zeros. Thus, (1.3) and (1.4) has exactly two principal eigenvalues, λ_m,+> 0 and λ_m,-< 0, and the corresponding eigenfunctions don't change sign on $\widehat{T}$.

Now, we give a property for the above two principal eigenvalues.

Theorem 3.2. If $m,m_1:T\to ℝ$ change their signs, and m(t) ≤ m₁(t) for $t\in T$, then ${\lambda }_{m_1,-}\le {\lambda }_{m,-},{\lambda }_{m_1,+}\le {\lambda }_{m,+}$.

Proof. It can be seen that for λ < 0, $S_{m,\lambda }\ge S_{m_1,\lambda }$, which implies ${\mu }_{m,1}\left(\lambda \right)\ge {\mu }_{m_1,1}\left(\lambda \right)$ and consequently, ${\lambda }_{m,+}\ge {\lambda }_{m_1,+}$.

On the other hand, for λ < 0, $S_{m,\lambda }\le S_{m_1,\lambda }$, which indicates ${\mu }_{m,1}\left(\lambda \right)\le {\mu }_{m_1,1}\left(\lambda \right)$ and consequently, ${\lambda }_{m,-}\ge {\lambda }_{m_1,-}$.

4 Main result

We make the following assumptions.

(H1) f : ℝ → ℝ is continuous and sf(s) > 0 for s ≠ 0.

(H2) $f_0={\text{lim}}_{\left|s\right|\to 0}\frac{f\left(s\right)}{s}\in \left(0,\infty \right),f_{\infty }={\text{lim}}_{\left|s\right|\to +\infty }\frac{f\left(s\right)}{s}\in \left(0,\infty \right)$.

Theorem 4.1. Suppose that (H1) and (H2) hold Assume that

$$r\in \left(\frac{{\lambda }_{m,+}}{f_{\infty }},\frac{{\lambda }_{m,+}}{f_0}\right)\cup \left(\frac{{\lambda }_{m,-}}{f_0},\frac{{\lambda }_{m,-}}{f_{\infty }}\right),$$

(4.1)

or

$$r\in \left(\frac{{\lambda }_{m,+}}{f_0},\frac{{\lambda }_{m,+}}{f_{\infty }}\right)\cup \left(\frac{{\lambda }_{m,-}}{f_{\infty }},\frac{{\lambda }_{m,-}}{f_0}\right).$$

(4.2)

Then (1.1) and (1.2) has two solutions u⁺ and u^- such that u⁺ is positive on $T$ and u^- is negative on $T$.

Obviously, we can get the following lemma with ease.

Lemma 4.1. Suppose that u ∈ X and $u\not\equiv 0$ on $T$ satisfies (1.1) (or (1.3)) and there exists $t_0\in T$ such that u(t₀) = 0, then u(t₀ - 1)u(t₀ + 1) < 0.

Proof of Theorem 4.1. First, we deal with the case r > 0.

Let ζ, ξ ∈ C(ℝ, ℝ) such that

$$f\left(u\right)=f_{0}u+\zeta \left(u\right),f\left(u\right)=f_{\infty }u+\xi \left(u\right).$$

Clearly

$$\underset{\left|u\right|\to 0}{\text{lim}}\frac{\zeta \left(u\right)}{u}=0,\underset{\left|u\right|\to \infty }{\text{lim}}\frac{\xi \left(u\right)}{u}=0.$$

(4.3)

Let

$$\tilde{\xi }\left(u\right)=\underset{0\le \left|s\right|\le u}{\text{max}}\left|\xi \left(s\right)\right|.$$

Then $\tilde{\xi }$ is nondecreasing and

$$\underset{\left|u\right|\to \infty }{\text{lim}}\frac{\tilde{\xi }\left(u\right)}{u}=0.$$

(4.4)

Let us consider

$$Lu-\lambda m\left(t\right)r{f}_{0}u-\lambda m\left(t\right)r\zeta \left(u\right)=0,$$

(4.5)

as a bifurcation problem from the trivial solution u ≡ 0.

Equation (4.5) can be converted to the equivalent equation

$$\begin{array}{ll}\hfill u\left(t\right)& =\lambda L^{-1}\left[m\left(\cdot \right)r{f}_{0}u\left(\cdot \right)+m\left(\cdot \right)r\zeta \left(u\left(\cdot \right)\right)\right]\left(t\right)\\ :=\lambda Tu+H\left(\lambda ,u\right).\end{array}$$

(4.6)

It is easy to see that T : X → X is compact. Further we note that H(λ, u) = λL^-1[m(·)ζ( u (·))] = o⊠u⊠near λ = 0 uniformly on bounded λ intervals, since

$$\begin{array}{ll}\hfill \left|\right|L^{-1}\left[m\left(\cdot \right)\zeta \left(u\left(\cdot \right)\right)\right]\left|\right|& =\underset{t\in \mathsf{\text{T}}}{\text{max}}\left|\sum _{s=1}^{T}G\left(t,s\right)m\left(s\right)\zeta \left(u\left(s\right)\right)\right|\\ \le C\cdot \underset{s\in \mathsf{\text{T}}}{\text{max}}\left|m\left(s\right)\right|\left|\right|\zeta \left(u\left(\cdot \right)\right)\left|\right|,\end{array}$$

where $C=\underset{t\in \widehat{T}}{\text{max}}{\sum }_{s=1}^{T}G\left(t,s\right)$ and

$$G\left(t,s\right)=\frac{1}{T+1}\left\{\begin{array}{cc}\hfill \left(T+1-t\right)s,\hfill & \hfill 0\le s\le t\le T+1,\hfill \\ \hfill \left(T+1-t\right)t,\hfill & \hfill 0\le t\le s\le T+1.\hfill \end{array}\right.$$

Let $E=ℝ\times X$ under the product topology. Let $S^{+}:=\left\{u\in X|u\left(t\right)>0\mathsf{\text{for}}t\in T\right\}$. Set S^- = -S⁺, S = S⁺ ∪ S^-. Then S⁺ and S^- are disjoint in X. Finally let Ψ^± = ℝ × S^± and Ψ = ℝ × S. Let Σ be the closure of the set of nontrivial solutions of (1.1) and (1.2).

It is easy to see that $\frac{{\lambda }_{m,+}}{r{f}_0}\in r\left(T\right)$ is simple. Now applying Theorems 2.1 and 2.2, we get the result as follows: Σ contains a maximum subcontinuum ${\mathcal{C}}_{+}$ which is composed of two distinct connected set ${\mathcal{C}}_{+}^{+}$ and ${\mathcal{C}}_{+}^{-}$ such that ${\mathcal{C}}_{+}={\mathcal{C}}_{+}^{+}\cup {\mathcal{C}}_{+}^{-}$ and ${\mathcal{C}}_{+}^{+}\cap {\mathcal{C}}_{+}^{-}=\left\{\left(\frac{{\lambda }_{m,+}}{r{f}_0},0\right)\right\}$. Moreover, Lemma 4.1 guarantees the second case in Theorems 2.1 and 2.2 cannot happen. Otherwise, there will exist $\left(\eta ,y\right)\in C_{+}^v$, such that y has a multiple zero point t₀, (i.e., t₀ satisfies y(t₀) = 0 and y(t₀ - 1)y(t₀ + 1) > 0). However, this contradicts Lemma 4.1. Thus, for each $\nu \in \left\{+,-\right\},{\mathcal{C}}_{+}^{\nu }$ joins $\left(\frac{{\lambda }_{m,+}}{r{f}_0},0\right)$ to infinity in Ψ^vand ${\mathcal{C}}_{+}^{\nu }\backslash \left\{\left(\frac{{\lambda }_{m,+}}{r{f}_0},0\right)\right\}\subset {\Psi }^{\nu }$.

It is obvious that any solution to (4.5) of the form (1, u) yields a solution u to (1.1) and (1.2). We will show that ${\mathcal{C}}_{+}^{\nu }$ crosses the hyperplane {1} × X in ℝ × X. To achieve this goal, it will be enough to show that

$$\left(\frac{{\lambda }_{m,+}}{r{f}_{\infty }},\frac{{\lambda }_{m,+}}{r{f}_0}\right)\subseteq {\text{Proj}}_{ℝ}{\mathcal{C}}_{+}^{\nu },$$

(4.7)

or

$$\left(\frac{{\lambda }_{m,+}}{r{f}_0},\frac{{\lambda }_{m,+}}{r{f}_{\infty }}\right)\subseteq {\text{Proj}}_{ℝ}{\mathcal{C}}_{+}^{\nu },$$

(4.8)

where ${\mu }_n+\left|\right|y_n|{|}_X\to \infty .$ denotes the projection of ${\mathcal{C}}_{+}^{\nu }$ on ℝ.

Let $\left({\mu }_n,y_n\right)\in {\mathcal{C}}_{+}^{\nu }$ satisfy

$${\mu }_n+\left|\right|y_n|{|}_X\to \infty .$$

We note that μ _n > 0 for all n ∈ ℕ since (0,0) is the only solution of (4.5) for λ = 0 and ${\mathcal{C}}_{+}^{\nu }\cap \left(\left\{0\right\}\times X\right)=\varnothing$.

Case 1. $\frac{{\lambda }_{m,+}}{r{f}_{\infty }}<1<\frac{{\lambda }_{m,+}}{r{f}_0}$.

We divide the proof into two steps.

Step 1. We show that if there exists a constant number M > 0 such that

$${\mu }_n\in \left(0,M\right],$$

(4.9)

then (4.7) holds.

In this case it follows that

$$\left|\right|y_n|{|}_X\to \infty .$$

(4.10)

We divide the equation

$$L{y}_n-{\mu }_{n}m\left(t\right)r{f}_{\infty }{y}_n-{\mu }_{n}m\left(t\right)r\xi \left(y_n\right)=0$$

(4.11)

by ⊠y _n ⊠ _x and set ${ȳ}_n=\frac{y_n}{\left|\right|y_n|{|}_X}$. Since ${ȳ}_n$ is bounded in X and μ _n is bounded in ℝ, after taking the subsequence if necessary, we have that ${ȳ}_n\to ȳ$ for some $ȳ\in X$ with $\left|\right|ȳ|{|}_X=1$ and ${\mu }_n\to \bar{\mu }$ for some μ ∈ ℝ. Moreover, from (4.4) and the fact that $\tilde{\xi }$ is nondecreasing, we have that

$$\underset{n\to \infty }{\text{lim}}\frac{\left|\xi \left(y_n\left(t\right)\right)\right|}{\parallel y_n{\parallel }_X}=0,$$

(4.12)

since $\frac{\left|\xi \left(y_n\left(t\right)\right)\right|}{\parallel y_n{\parallel }_X}\le \frac{\tilde{\xi }\left(\left|y_n\left(t\right)\right|\right)}{\parallel y_n{\parallel }_X}\le \frac{\xi \left(\parallel y_n{\parallel }_X\right)}{\parallel y_n{\parallel }_X}$. Thus,

$$ȳ\left(t\right)=\sum _{s=1}^{T}G\left(t,s\right)\bar{\mu }m\left(s\right)r{f}_{\infty }ȳ\left(s\right),$$

which implies that

$$Lȳ-\bar{\mu }m\left(t\right)r{f}_{\infty }ȳ=0.$$

(4.13)

We claim that

$$\left(\bar{\mu },ȳ\right)\in {\mathcal{C}}_{+}^{\nu }.$$

We only prove that if $y_n\in {\mathcal{C}}_{+}^{+}$, then ${ȳ}_n\in {\mathcal{C}}_{+}^{+}$. The other case that if $y_n\in {\mathcal{C}}_{+}^{-}$, then ${ȳ}_n\in {\mathcal{C}}_{+}^{-}$ can be treated similarly.

Obviously when $y_n\in {\mathcal{C}}_{+}^{+}$, then $ȳ\left(t\right)\ge 0$ on $\widehat{T}$. Furthermore, $ȳ\left(t\right)>0$ on $T$. In fact, if there exists a $t_0\in T$ such that $ȳ\left(t_0\right)=0$, then, by Lemma 4.1, we obtain $ȳ\left(t_0-1\right)ȳ\left(t_0+1\right)<0$ which contradicts the fact that $ȳ\left(t\right)\ge 0$ on $\widehat{T}$. Thus, $ȳ\left(t\right)>0$ on $T$. This together with the fact ${\mathcal{C}}_{+}$ is a closed set in $E$ implies that $ȳ\in {\mathcal{C}}_{+}^{+}$. Moreover, $\bar{\mu }r{f}_{\infty }={\lambda }_{m,+}$, so that

$$\bar{\mu }=\frac{{\lambda }_{m,+}}{r{f}_{\infty }}.$$

Thus, (4.7) holds.

Step 2. We show that there exists a constant M > 0 such that μ _n ∈ (0, M] for all n.

Since {(μ _n , y _n )} are the solutions to (4.5), they follow that

$$L{y}_n={\mu }_{n}rm\left(t\right){\Gamma }_n\left(t\right)y_n,$$

(4.14)

where ${\Gamma }_n\left(t\right):=\frac{f\left(y_n\left(t\right)\right)}{y_n\left(t\right)}$. From (H1) and (H2), there exist two positive constants ρ₁ and ρ₂, such that

$${\rho }_1<\frac{f\left(y_n\right)}{y_n}<{\rho }_2.$$

(4.15)

Let η_* > 0 be the positive principal eigenvalue of the following linear eigenvalue problem

$$Lv=\eta {\chi }_1\left(t\right)m\left(t\right)v$$

(4.16)

and η* > 0 the positive principal eigenvalue of the following linear eigenvalue problem

$$Lv=\eta {\chi }_2\left(t\right)m\left(t\right)v,$$

(4.17)

where

$$\begin{array}{cc}\hfill {\chi }_1\left(t\right)=\left\{\begin{array}{cc}\hfill {\rho }_1\hfill & \hfill \mathsf{\text{if}}m\left(t\right)>0,t\in T,\hfill \\ \hfill {\rho }_2\hfill & \hfill \mathsf{\text{if}}m\left(t\right)<0,t\in T,\hfill \end{array}\right.\hfill & \hfill {\chi }_2\left(t\right)=\left\{\begin{array}{cc}\hfill {\rho }_2\hfill & \hfill \mathsf{\text{if}}m\left(t\right)>0,t\in T,\hfill \\ \hfill {\rho }_1\hfill & \hfill \mathsf{\text{if}}m\left(t\right)<0,t\in T.\hfill \end{array}\right.\hfill \end{array}$$

By Theorem 3.2, (4.14), (4.15), (4.16), and (4.17), we get

$$\frac{{\eta }^{*}}{r}<{\mu }_n<\frac{{\eta }_{*}}{r}.$$

Case 2. $\frac{{\lambda }_{m,+}}{r{f}_0}<1<\frac{{\lambda }_{m,+}}{r{f}_{\infty }}$.

From Step 2 of Case 1, there exists M > 0 such that for all n ∈ ℕ,

$${\mu }_n\in \left(0,M\right].$$

Applying a similar argument to that used in Step 1 of Case 1 (after taking a subsequence and relabeling, if necessary), we get

$${\mu }_n\to \frac{{\lambda }_{m,+}}{r{f}_{\infty }},y_n\to \infty \mathsf{\text{as}}n\to \infty ,$$

which implies that (4.8) holds.

At last, we deal with the case r < 0.

Let us consider

$$Lu-\lambda rm\left(t\right)f_{0}u-\lambda rm\left(t\right)\zeta \left(u\right)=0,$$

(4.18)

as a bifurcation problem from the trivial solution u ≡ 0. Now, applying Theorems 2.1 and 2.2, we get the following results: Σ contains a maximum subcontinuum ${\mathcal{C}}_{-}$ which is composed of two distinct connected set ${\mathcal{C}}_{-}^{+}$ and ${\mathcal{C}}_{-}^{-}$ such that ${\mathcal{C}}_{-}={\mathcal{C}}_{-}^{+}\cup {\mathcal{C}}_{-}^{-}$ and ${\mathcal{C}}_{+}^{-}\cap {\mathcal{C}}_{-}^{-}=\left\{\left(\frac{{\lambda }_{m,-}}{-r{f}_0},0\right)\right\}$. Moreover, by Lemma 4.1, for each $\nu \in \left\{+,-\right\},{\mathcal{C}}_{-}^{\nu }$ joins $\left(\frac{{\lambda }_{m,-}}{-r{f}_0},0\right)$ to infinity in Ψ^vand ${\mathcal{C}}_{-}^{\nu }\backslash \left\{\left(\frac{{\lambda }_{m,-}}{-r{f}_0},0\right)\right\}\subset {\Psi }^{\nu }$, where Σ and Ψ^vare defined as in the case r > 0.

It is clear that any solution to (4.18) of the form (-1, u) yields a solutions u of (1.1) and (1.2). We will show ${\mathcal{C}}_{-}^{\nu }$ crosses the hyperplane {-1} × X in ℝ × X. To achieve this goal, it will be enough to show that

$$\left(\frac{{\lambda }_{m,-}}{-r{f}_{\infty }},\frac{{\lambda }_{m,-}}{-r{f}_0}\right)\subseteq {\text{Proj}}_{ℝ}{\mathcal{C}}_{-}^{\nu },$$

(4.19)

or

$$\left(\frac{{\lambda }_{m,-}}{-r{f}_0},\frac{{\lambda }_{m,-}}{-r{f}_{\infty }}\right)\subseteq {\text{Proj}}_{ℝ}{\mathcal{C}}_{-}^v.$$

(4.20)

Let $\left({\mu }_n,y_n\right)\in {\mathcal{C}}_{-}^{\nu }$ satisfy

$$\left|{\mu }_n\right|+\parallel y_n{\parallel }_X\to \infty .$$

We note that μ _n < 0 for all n ∈ ℕ since (0, 0) is the only solution to (4.18) for λ = 0 and ${\mathcal{C}}_{-}^{\nu }\cap \left(\left\{0\right\}\times X\right)=\varnothing$.

The rest of the proof is similar to the proof of the case r > 0, so we omit it.

5 Example

Let T = 5, then $T=\left\{1,2,3,4,5\right\}$. Consider the following discrete second-order BVPs

$${\Delta }^{2}u\left(t-1\right)+\lambda m\left(t\right)f\left(u\left(t\right)\right)=0,t\in T,$$

(5.1)

$$u\left(0\right)=u\left(6\right)=0,$$

(5.2)

where $m:T\to ℝ$ which is defined by

$$m\left(1\right)=1,m\left(2\right)=2,m\left(3\right)=1,m\left(4\right)=-1,m\left(5\right)=-3,$$

and

$$f\left(s\right)=\frac{s^3+s}{s^2+2}.$$

By using Matlab 7.0, we get the following eigenvalue problem

$${\Delta }^{2}u\left(t-1\right)+\lambda m\left(t\right)u\left(t\right)=0,t\in T,$$

(5.3)

$$u\left(0\right)=u\left(6\right)=0$$

(5.4)

has two principal eigenvalues λ_m,-= -0.5099 and λ_m,+= 0.2867. The corresponding eigenfunctions

ψ _m,- (t) and ψ_m,+(t) satisfy

$$\begin{array}{c}{\psi }_{m,-}\left(0\right)=0,{\psi }_{m,-}\left(1\right)=0.0471,{\psi }_{m,-}\left(2\right)=0.1182,\\ {\psi }_{m,-}\left(3\right)=0.3099,{\psi }_{m,-}\left(4\right)=0.6595,{\psi }_{m,-}\left(5\right)=0.6729,{\psi }_{m,-}\left(6\right)=0,\end{array}$$

and

$$\begin{array}{c}{\psi }_{m,+}\left(0\right)=0,{\psi }_{m,+}\left(1\right)=0.3867,{\psi }_{m,+}\left(2\right)=0.6626,\\ {\psi }_{m,+}\left(3\right)=0.5584,{\psi }_{m,+}\left(4\right)=0.2942,{\psi }_{m,+}\left(5\right)=0.1143,{\psi }_{m,+}\left(6\right)=0.\end{array}$$

Moreover,

$$\begin{array}{c}sf\left(s\right)=\frac{s^4+s^2}{s^2+2}>0,\mathsf{\text{for }}s\ne \mathsf{\text{0,}}\\ f_{\mathsf{\text{0}}}=\underset{\left|s\right|\to 0}{\text{lim}}\frac{f\left(s\right)}{s}=\underset{s\to 0}{\text{lim}}\frac{s^3+s}{s^3+2s}=\frac{1}{2}\in \left(0,\infty \right),f_{\infty }=\underset{\left|s\right|\to \infty }{\text{lim}}\frac{f\left(s\right)}{s}=\underset{s\to \infty }{\text{lim}}\frac{s^3+s}{s^3+2s}=1\in \left(0,\infty \right).\end{array}$$

Obviously, f(s) satisfies (H1) and (H2). Thus, for

$$r\in \left({\lambda }_{m,+},2{\lambda }_{m,+}\right)$$

or

$$r\in \left(2{\lambda }_{m,-},{\lambda }_{m,-},\right),$$

(5.1) and (5.2) has a positive solution u⁺ and a negative solution u^-.