Global bifurcation and constant sign solutions of discrete boundary value problem involving p-Laplacian

Abstract

We study the unilateral global bifurcation result for the one-dimensional discrete p-Laplacian problem

$$ \textstyle\begin{cases} -\Delta [\varphi _{p}(\Delta u(t-1))]=\lambda a(t)\varphi _{p}(u(t))+g(t,u(t), \lambda ),\quad t\in [1,T+1]_{Z}, \\ \Delta u(0)=u(T+2)=0, \end{cases} $$

where $\Delta u(t)=u(t+1)-u(t)$ is a forward difference operator, $\varphi _{p}(s)=|s|^{p-2}s$ ($1< p<+\infty$) is a one-dimensional p-Laplacian operator. λ is a positive real parameter, $a: [1,T+1]_{Z}\to [0,+\infty )$ and $a(t_{0})>0$ for some $t_{0}\in [1,T+1]_{Z}$, $g :[1,T+1]_{Z}\times \mathbb{R}^{2}\to \mathbb{R}$ satisfies the Carathéodory condition in the first two variables. We show that $(\lambda _{1},0)$ is a bifurcation point of the above problem, and there are two distinct unbounded continua $\mathscr{C}^{+}$ and $\mathscr{C}^{-}$, consisting of the bifurcation branch $\mathscr{C}$ from $(\lambda _{1},0)$, where $\lambda _{1}$ is the principal eigenvalue of the eigenvalue problem corresponding to the above problem. Let $T>1$ be an integer, Z denote the integer set for $m, n\in Z$ with $m< n$, $[m, n]_{Z}:=\{m, m+1,\ldots , n\}$.

As the applications of the above result, we prove more details about the existence of constant sign solutions for the following problem:

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

where $f\in C(\mathbb{R},\mathbb{R})$ with $sf(s)>0$ for $s\neq 0$.

1 Introduction

In this paper, we establish a Dancer-type unilateral global bifurcation result for one-dimensional discrete p-Laplacian problem

$$ \textstyle\begin{cases} -\Delta [\varphi _{p}(\Delta u(t-1))]=\lambda a(t)\varphi _{p}(u(t))+g(t,u(t), \lambda ), \quad t\in [1,T+1]_{Z}, \\ \Delta u(0)=u(T+2)=0, \end{cases} $$

(1.1)

where $a: [1,T+1]_{Z}\to [0,+\infty )$ and $a(t_{0})>0$ for some $t_{0}\in [1,T+1]_{Z}$, $g :[1,T+1]_{Z}\times \mathbb{R}^{2}\to \mathbb{R}$ satisfies the Carathéodory condition in the first two variables and

$$ \lim_{ \vert s \vert \rightarrow 0}\frac{g(t,s,\lambda )}{ \vert s \vert ^{p-1}}=0 $$

uniformly for a.e. $t\in [1,T+1]_{Z}$ and λ on bounded sets. Under these assumptions, we shall show that $(\lambda _{1},0)$ is a bifurcation point of (1.1) and there are two distinct unbounded continua $\mathscr{C}^{+}$ and $\mathscr{C}^{-}$, consisting of the bifurcation branch $\mathscr{C}$ from $(\lambda _{1},0)$, where $\lambda _{1}$ is the principal eigenvalue of the eigenvalue problem corresponding to (1.1).

When $p=2$, Ma and Ma [17] considered the optimal intervals of r, for which they gave a complete description of the global behavior of solutions set for the problem

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

(1.2)

under some suitable assumptions on f and g. Using the bifurcation theory of Rabinowitz [5, 18], they proved that if $\frac{\lambda _{1}}{f_{\infty }}< r<\frac{\lambda _{1}}{f_{0}}$ or $\frac{\lambda _{1}}{f_{0}}< r<\frac{\lambda _{1}}{f_{\infty }}$, (1.2) has two solutions $u^{+}$ and $u^{-}$ such that $u^{+}$ is positive in $[0,T]_{Z}$ and $u^{-}$ is negative in $[0,T]_{Z}$, where $f_{0}=\lim_{s\rightarrow 0^{+}}\frac{f(s)}{s}$, $f_{\infty }=\lim_{s\rightarrow +\infty }\frac{f(s)}{s}$, $\lambda _{1}$ is the first eigenvalue of the following linear eigenvalue problem:

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

The idea of using bifurcation methods to study the solvability of nonlinear boundary value problems has been applied to study some Dirichlet, Sturm–Liouville, and periodic boundary value problems, for instance, [10, 17, 21, 24].

When $p\neq 2$, many authors discussed the existence and multiplicity of solutions for the one-dimensional discrete p-Laplacian problem by exploiting various methods, including the method of upper and lower solutions, Leray–Schauder degree, fixed point theory, critical theory, and variational methods, which can be seen in D’Aguì [6], Bereanu [3], Wang et al. [23], and [8, 11–13, 22] and the references therein. In particular, He [11] established the existence of one or two positive solutions for the equation

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

In the paper, he assumed that the nonlinear term f is positive and f is superlinear at 0, sublinear at infinity. In 2015, Bai and Chen [2] discussed the discrete p-Laplacian boundary value problem

$$ \textstyle\begin{cases} -\Delta [\varphi _{p}(\Delta u(t-1))]=\lambda p(t)g(u(t)),\quad t\in [1,T]_{Z}, \\ u(0)=u(T+1)=0, \end{cases} $$

where $\lim_{s\rightarrow \infty }\frac{g(s)}{\varphi _{p}(s)}=0$ and $\frac{g(s)}{\varphi _{p}(s)}$ is strictly decreasing on $(0,\infty )$. They obtained an unbounded continuum $\mathcal{C}$ of positive solutions emanating from $(\lambda ,u)=(0,0)$; while the existence of global continuum of solutions comes from [19].

There is no report on the global structure of solution sets by the bifurcation theory for discrete p-Laplacian problem (1.1). Although there are a great amount of papers researching the bifurcation phenomenon of p-Laplacian problem, but those results are not unilateral, and their conclusions are all obtained in the differential case. As we know, the proofs are based on the local properties of solutions of (1.1) bifurcating from $(\lambda _{1},0)$ (see Lemma 3.4). Although the proof of the above result follows the same steps as for the semilinear case from [7], his methods cannot be applied directly to the quasilinear discrete problem. In addition, the main reason is that the spectrum of the discrete p-Laplacian eigenvalue problem is unknown.

In 2002, Anane [1] discussed the spectra of the following differential p-Laplacian problem:

$$ \textstyle\begin{cases} -(\varphi _{p}(u'))'=\lambda m(t)\varphi _{p}(u), \quad t\in (a,b), \\ u(a)=u(b)=0, \end{cases} $$

(1.3)

where $m\in M(I):=\{m\in L^{\infty } \mid \operatorname{meas}\{t\in I,m(t)>0\}\neq 0, I=[a,b]\}, \lambda$ is the spectral parameter. They obtained the following result.

Theorem A

Assume that $m\in M(I)$ such that $m\not \equiv 1$, $p\neq 2$, we have:

• (i) Every eigenfunction corresponding to the kth eigenvalues $\lambda _{k}(m,I)$ has exactly $k-1$ zeros.

• (ii) For every k, $\lambda _{k}(m,I)$ is simple and verifies the strict monotonicity property with respect to the weight m and the domain I.

• (iii) (1.3) has infinite real eigenvalues, the eigenvalues are ordered as

  $$ 0< \lambda _{1}(m,I)< \lambda _{2}(m,I)< \cdots < \lambda _{k}(m,I)< \cdots \to +\infty\quad \textit{as } k\to +\infty . $$

The first eigenvalue $\lambda _{1}$ is of special importance. However, the method in [1] has no effect on the spectral study of the discrete p-Laplacian problem, whether the first eigenvalue of the discrete p-Laplacian problem is simple or not, and the properties of the corresponding eigenfunction are both unknown.

Of course, the next natural question is: does the unilateral bifurcation version exist for quasilinear difference problem (1.1)? Furthermore, what is the existence of a positive solution or a negative one for a nonlinear p-Laplacian difference problem (1.1)? In this paper, we give a positive answer for those questions.

This paper is organized as follows. In Sect. 2, we show the existence of the principal eigenvalue and the sign of the corresponding eigenfunctions for the one-dimensional discrete p-Laplacian eigenvalue problem, which will be of interest for us. In Sect. 3, we establish the unilateral global bifurcation theory for (1.1). In Sect. 4, as an application, we prove that there exist constant sign solutions for problem (4.1) (see Sect. 4) according to the different behavior of nonlinear term f at 0 and ∞.

2 The existence of the principal eigenvalue

In this section, we consider the existence of the principal eigenvalue and the sign of the corresponding eigenfunction for the discrete p-Laplacian eigenvalue problem

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

(2.1)

Let E be defined by

$$ E= \bigl\{ u\mid u:[1,T+1]_{Z}\rightarrow \mathbb{R} \text{ and } \Delta u(0)=u(T+2)=0 \bigr\} $$

equipped with the norm $\|u\|=\max_{t\in [0,T+2]_{Z}} |u(t) |$, then $(E, \|\cdot \|)$ is a Banach space.

Lemma 2.1

Let $u\in E$. Then

$$ -\sum_{t=1}^{T+1}\Delta \bigl[\varphi _{p} \bigl(\Delta u(t-1) \bigr) \bigr]\cdot u(t)= \sum _{t=1}^{T+1} \bigl\vert \Delta u(t) \bigr\vert ^{p}. $$

Proof

We have

$$\begin{aligned}& \sum_{t=1}^{T+1} \Delta \bigl[\varphi _{p} \bigl(\Delta u(t-1) \bigr) \bigr] \cdot u(t) \\& \quad =\sum_{t=1}^{T+1}\Delta \bigl( \bigl\vert \Delta u(t-1) \bigr\vert ^{p-2}\Delta u(t-1) \bigr)\cdot u(t) \\& \quad =\sum_{j=0}^{T}u(j+1)\Delta \bigl( \bigl\vert \Delta u(j) \bigr\vert ^{p-2}\Delta u(j) \bigr) \quad (j=t-1) \\& \quad =\sum_{j=0}^{T}u(j+1) \bigl( \bigl\vert \Delta u(j+1) \bigr\vert ^{p-2}\Delta u(j+1)- \bigl\vert \Delta u(j) \bigr\vert ^{p-2}\Delta u(j) \bigr) \\& \quad =\sum_{j=0}^{T}u(j+1) \bigl\vert \Delta u(j+1) \bigr\vert ^{p-2}\Delta u(j+1)-\sum _{j=0}^{T}u(j+1) \bigl\vert \Delta u(j) \bigr\vert ^{p-2}\Delta u(j) \\& \quad =\sum_{k=1}^{T+1}u(k) \bigl\vert \Delta u(k) \bigr\vert ^{p-2}\Delta u(k)-\sum _{j=0}^{T}u(j+1) \bigl\vert \Delta u(j) \bigr\vert ^{p-2}\Delta u(j)\quad (k=j+1) \\& \quad =\sum_{k=1}^{T+1}u(k) \bigl\vert \Delta u(k) \bigr\vert ^{p-2}\Delta u(k)-\sum _{j=1}^{T}u(j+1) \bigl\vert \Delta u(j) \bigr\vert ^{p-2}\Delta u(j)-u(1) \bigl\vert \Delta u(0) \bigr\vert ^{p-2}\Delta u(0) \\& \quad =\sum_{k=1}^{T}u(k) \bigl\vert \Delta u(k) \bigr\vert ^{p-2}\Delta u(k)-\sum _{k=1}^{T}u(k+1) \bigl\vert \Delta u(k) \bigr\vert ^{p-2}\Delta u(k) \\& \qquad {}+ u(T+1) \bigl\vert \Delta u(T+1) \bigr\vert ^{p-2}\Delta u(T+1)-u(1) \bigl\vert \Delta u(0) \bigr\vert ^{p-2} \Delta u(0) \\& \quad =-\sum_{k=1}^{T} \bigl\vert \Delta u(k) \bigr\vert ^{p}+u(T+1) \bigl\vert \Delta u(T+1) \bigr\vert ^{p-2} \Delta u(T+1)-u(1) \bigl\vert \Delta u(0) \bigr\vert ^{p-2}\Delta u(0) \\& \quad =-\sum_{k=1}^{T} \bigl\vert \Delta u(k) \bigr\vert ^{p}- \bigl\vert \Delta u(T+1) \bigr\vert ^{p}- \bigl\vert \Delta u(0) \bigr\vert ^{p} \qquad \bigl( \Delta u(0)=u(T+2)=0 \bigr) \\& \quad =-\sum_{k=1}^{T+1} \bigl\vert \Delta u(k) \bigr\vert ^{p}. \end{aligned}$$

 □

Lemma 2.2

$\lambda _{1}(a)$ is the first eigenvalue of (2.1), then the first eigenvalue $\lambda _{1}(a)$ is the minimum of the Rayleigh quotient, that is,

$$ \lambda _{1}(a)=\inf \biggl\{ \frac{\sum^{T+1}_{t=1} \vert \Delta u(t) \vert ^{p}}{\sum^{T+1}_{t=1}a(t) \vert u(t) \vert ^{p}}:u \in E \biggr\} . $$

Furthermore, $\lambda _{1}(a)<\lambda (a)$, where $\lambda (a)$ is some other eigenvalue of (2.1).

Proof

Combining the equation of (2.1) with Lemma 2.1, the conclusion is clearly established. □

Applying a similar method to prove [4, Proposition 1.10] with obvious changes, we can obtain the following theorem.

Lemma 2.3

The first eigenvalue $\lambda _{1}(a)$ is simple. Let $\phi _{1}$ be the eigenfunction corresponding to $\lambda _{1}(a)$, then $\phi _{1}$ does not change sign in $[0,T+2]_{Z}$. Moreover, $\phi _{1}$ does not vanish in $[0, T+2]_{Z}$.

3 Unilateral global bifurcation results for (1.1)

In this section, we establish the unilateral global bifurcation theory for (1.1).

We consider the following auxiliary problem:

$$ \textstyle\begin{cases} \Delta [\varphi _{p}(\Delta u(t-1))]=h(t),\quad t\in [1,T+1]_{Z}, \\ \Delta u(0)=u(T+2)=0, \end{cases} $$

(3.1)

where $h:[1,T+1]_{Z}\rightarrow \mathbb{R}$. It can be easily seen that problem (3.1) is equivalently written as

$$ u(t)=G_{p}(h) (t):=u(1)+\sum_{s=1}^{t-1} \varphi _{p}^{-1} \Biggl[ \sum _{\tau =1}^{s}h(\tau ) \Biggr], \quad t\in [1,T+2]_{Z}, $$

where $G_{p}:\mathbb{R}\rightarrow E$ maps bounded sets of $\mathbb{R}$ into relative compacts of E.

We define the operator $T_{\lambda }^{p}:E\rightarrow E$ by

$$ \begin{aligned} T_{\lambda }^{p}(u) (t)&=u(1)+\sum _{s=1}^{t-1} \varphi _{p}^{-1} \Biggl[-\sum_{\tau =1}^{s}\lambda a\varphi _{p}(u) ( \tau ) \Biggr] \\ &=G_{p} \bigl(-\lambda a\varphi _{p}(u) \bigr) (t), \end{aligned} $$

then $T_{\lambda }^{p}:E\rightarrow E$ is completely continuous and (2.1) is equivalent to

$$ u=T_{\lambda }^{p}(u). $$

Next, we use Brouwer degree theory to calculate its topological degree. Let $\deg (I-T_{\lambda }^{p},B_{r},0)$ be the Leray–Schauder degree for $I-T_{\lambda }^{p}$ on $B_{r}$, where $B_{r}=\{u\in E:\|u\|< r\}$.

Lemma 3.1

Let λ be a constant, then for arbitrary $r>0$,

$$ \deg \bigl(I-T_{\lambda }^{p},B_{r},0 \bigr)= \textstyle\begin{cases} 1,& 0< \lambda < \lambda _{1}(a), \\ -1, & \lambda \in (\lambda _{1}(a),\lambda _{2}(a)), \end{cases} $$

where $\lambda _{2}(a)$ is the second eigenvalue of problem (2.1).

Proof

Using the similar method of [15, Lemma 2.8], we can get the conclusion of this theorem. □

Define Nemytskii operators $H:\mathbb{R}\times E\rightarrow \mathbb{R}$ by

$$ H(\lambda ,u) (t)=-\lambda a(t)\varphi _{p} \bigl(u(t) \bigr)-g \bigl(t,u(t),\lambda \bigr). $$

Then it is clear that H is a continuous operator which maps bounded sets of $\mathbb{R}\times E$ into the bounded sets of $\mathbb{R}$. Obviously, (1.1) can be equivalently written as

$$ u=G_{p}\circ H(\lambda ,u)=F(\lambda ,u). $$

It is easy to see that $F:\mathbb{R}\times E\rightarrow E$ is completely continuous and $F(\lambda ,0)=0$, $\forall \lambda \in \mathbb{R}$.

For convenience, we abbreviate $\lambda _{1}(a)$ as $\lambda _{1}$. Our first main result for (1.1) is the following theorem.

Theorem 3.1

For $p>1$, $\lambda _{1}$ is a bifurcation point of (1.1) and the associated bifurcation branch $\mathscr{C}$ in $\mathbb{R}\times E$ whose closure contains $(\lambda _{1},0)$, then either

• (i) $\mathscr{C}$ is unbounded in $\mathbb{R}\times E$, or

• (ii) $\mathscr{C}$ contains a pair $(\overline{\lambda },0)$, where λ̅ is an eigenvalue of (2.1) and $\overline{\lambda }\neq \lambda _{1}$.

Proof

Suppose on the contrary that $(\lambda _{1},0)$ is not a bifurcation point of (1.1). Then there exist $\varepsilon >0$, $\rho _{0}>0$ such that, for $|\lambda -\lambda _{1}| \leq \varepsilon$ and $0<\rho <\rho _{0}$, there is no nontrivial solution of the equation

$$ u-F(\lambda ,u)=0 $$

with $\|u\|=\rho$. By the invariance of the degree under a compact homotopy, we obtain

$$ \deg \bigl(I-F(\lambda ,u),B_{\rho },0 \bigr)\equiv \text{constant} $$

(3.2)

for $\lambda \in [\lambda _{1}-\varepsilon ,\lambda _{1}+\varepsilon ]$.

Take ε small enough such that there is no eigenvalue of (2.1) in $(\lambda _{1},\lambda _{1}+\varepsilon ]$. Fix $\lambda \in (\lambda _{1},\lambda _{1}+\varepsilon ]$, we claim that the equation

$$ u-G_{p} \bigl(-\lambda a(t)\varphi _{p} \bigl(u(t) \bigr)- \alpha g \bigl(t,u(t),\lambda \bigr) \bigr)=0 $$

(3.3)

has no solution u such that $\|u\|=\rho$ for every $\alpha \in [0,1]$, where ρ is sufficiently small. Suppose that $\{u_{n}\}$ is the solution of (3.3) with $\|u_{n}\|\rightarrow 0$ as $n\to \infty$.

Let $v_{n}=\frac{u_{n}}{\|u_{n}\|}$, then $v_{n}$ satisfies

$$ \begin{aligned} v_{n}&=G_{p} \biggl(-\lambda a(t)\varphi _{p} \bigl(v_{n}(t) \bigr)- \alpha \frac{g(t,u_{n}(t),\lambda )}{ \Vert u_{n} \Vert ^{p-1}} \biggr) \\ &=u(1)+\sum_{s=1}^{t-1}\varphi _{p}^{-1} \Biggl[ \sum_{ \tau =1}^{s}h_{n}( \tau ) \Biggr], \end{aligned} $$

where $h_{n}(t)=-\lambda a(t)\varphi _{p}(v_{n}(t))-\alpha \frac{g(t,u_{n}(t),\lambda )}{\|u_{n}\|^{p-1}}$. Let

$$ \widetilde{g}(t,u,\lambda )=\max_{0\leq \vert s \vert \leq u} \bigl\vert g(t,s, \lambda ) \bigr\vert , $$

then g̃ is nondecreasing with respect to u and

$$ \lim_{ \vert u \vert \rightarrow 0} \frac{\widetilde{g}(t,u,\lambda )}{ \vert u \vert ^{p-1}}=0. $$

(3.4)

Furthermore, (3.4) implies

$$ \frac{g(t,u,\lambda )}{ \Vert u \Vert ^{p-1}}\leq \frac{\widetilde{g}(t, \vert u \vert ,\lambda )}{ \Vert u \Vert ^{p-1}}\leq \frac{\widetilde{g}(t, \Vert u \Vert ,\lambda )}{ \Vert u \Vert ^{p-1}} \rightarrow 0, \qquad \Vert u \Vert \rightarrow 0, $$

(3.5)

uniformly for a.e. $t\in [1,T+1]_{Z}$ and λ on bounded sets.

It is easy to see that $\{h_{n}\}$ is a bounded sequence in $\mathbb{R}$, thus we can assume that

$$ v_{n}\rightarrow v_{0}\quad \text{and}\quad \Vert v_{0} \Vert =1\quad \text{as } n\to \infty , $$

$v_{0}$ satisfies

$$ -\Delta \bigl[\varphi _{p} \bigl(\Delta v_{0}(t-1) \bigr) \bigr]=\lambda a(t)\varphi _{p} \bigl(v_{0}(t) \bigr). $$

This implies that λ is an eigenvalue of (2.1), this is a contradiction. From the homotopic invariance of the degree (refer to the proof method of Theorem 2.10 in [15]) and Lemma 3.1, we conclude that

$$ \deg \bigl(I-F(\lambda ,\cdot ),B_{r},0 \bigr)=\deg \bigl(I-T_{\lambda }^{p},B_{r},0 \bigr)=-1. $$

(3.6)

Similarly, for $\lambda \in [\lambda _{1}-\varepsilon ,\lambda _{1})$, it follows that

$$ \deg \bigl(I-F(\lambda ,\cdot ),B_{r},0 \bigr)=1. $$

(3.7)

It is easy to see that (3.6) and (3.7) contradict (3.2). Thus $(\lambda _{1},0)$ is a bifurcation point of (1.1). By the global bifurcation theory [18], we can get the existence of a global branch of solutions of (1.1) emanating from $(\lambda _{1},0)$. □

Let $S^{+}=\{u\in E \mid u(t)>0, t\in [1,T+1]_{Z}\}$, and set $S^{-}=-S^{+}$, $S=S^{+}\cup S^{-}$. It is clear that $S^{+}$ and $S^{-}$ are disjoint and open in E. Let $\Phi ^{\pm }=\mathbb{R}\times S^{\pm }$ and $\Phi =\mathbb{R}\times S$ under the product topology.

Lemma 3.2

If $\mathscr{C}\subset (\Phi \cup \{ (\lambda _{1},0)\})$, the second alternative of Theorem 3.1is impossible.

Proof

Suppose on the contrary that there exists $\{(\lambda _{m},u_{m})\}\subset \mathscr{C}$ such that $u_{m}\not \equiv 0$, $(\lambda _{m},u_{m})\rightarrow (\lambda _{k},0)$, $k\neq 1$. With the help of the definition of $\lambda _{1}$, we see that $\lambda _{k}>\lambda _{1}$. According to the Sturm comparison theorem, if $\lambda >\lambda _{1}$, then the eigenfunction u corresponding to λ must change sign in $[0,T+2]_{Z}$. Let $v_{m}=\frac{u_{m}}{\|u_{m}\|}$, then $v_{m}$ satisfies

$$ v_{m}=G_{p} \biggl(-\lambda _{m} a(t) \varphi _{p} \bigl(v_{m}(t) \bigr)- \frac{g(t,u_{m}(t),\lambda _{m})}{ \Vert u_{m} \Vert ^{p-1}} \biggr). $$

(3.8)

From (3.5), (3.8), and the compactness of $G_{p}$, we can get

$$ v_{m}\rightarrow v_{0} \quad \text{and} \quad \Vert v_{0} \Vert =1. $$

In addition, $v_{0}$ satisfies

$$ -\Delta \bigl[\varphi _{p} \bigl(\Delta v_{0}(t-1) \bigr) \bigr]=\lambda _{k} a(t)\varphi _{p} \bigl(v_{0}(t) \bigr). $$

Hence, $v_{0}\in S_{k}$, where $S_{k}$ denotes the set of functions in E which must change sign in $[0,T+2]_{Z}$. Therefore, when m is sufficiently large, $v_{m}\in S_{k}$, combining the definition of $S_{k}$ with Lemma 2.3, it can be seen that this is a contradiction. □

Lemma 3.3

Let $(\lambda ,u)$ be a solution of (1.1). If there exists $t_{0}\in [0,T+1]_{Z}$ such that one of the following cases holds:

• (i) $u(t_{0})=0$, $\Delta u(t_{0})=0$;

• (ii) $u(t_{0})=0$, $u(t_{0}-1)u(t_{0}+1)\geq 0$.

Then $u\equiv 0$ in $[0,T+2]_{Z}$.

Proof

(i) By virtue of the equation of (1.1), we have

$$ \varphi _{p} \bigl(\Delta u(t_{0}-1) \bigr)-\varphi _{p} \bigl(\Delta u(t_{0}) \bigr)= \lambda a(t_{0})\varphi _{p} \bigl(u(t_{0}) \bigr)+g \bigl(t,u(t_{0}),\lambda \bigr). $$

Combining $u(t_{0})=0$ with the assumption of g, we obtain

$$\begin{aligned}& \varphi _{p} \bigl(\Delta u(t_{0}-1) \bigr)-\varphi _{p} \bigl(\Delta u(t_{0}) \bigr)=0, \\& \varphi _{p} \bigl(u(t_{0})- u(t_{0}-1) \bigr)-\varphi _{p} \bigl(u(t_{0}+1)- u(t_{0}) \bigr)=0. \end{aligned}$$

Thus, $u(t_{0}-1)=0$. Similarly, in view of

$$ \varphi _{p} \bigl(\Delta u(t_{0}) \bigr)-\varphi _{p} \bigl(\Delta u(t_{0}+1) \bigr)=0, $$

we can get $u(t_{0}+2)=0$. Further,

$$ \Delta u(t_{0}-1)=\Delta u(t_{0}+1)=0, $$

step by step, it follows that $u\equiv 0$.

(ii) Using the same method, we conclude that

$$ \bigl\vert u(t_{0}-1) \bigr\vert ^{p-2}u(t_{0}-1)= \bigl\vert u(t_{0}+1) \bigr\vert ^{p-2}u(t_{0}+1), $$

since $u(t_{0}-1)u(t_{0}+1)\geq 0$, we can only get

$$ u(t_{0}-1)=u(t_{0}+1)=0. $$

Hence, $u\equiv 0$. □

Theorem 3.2

$(\lambda _{1},0)$ bifurcates an unbounded continuum $\mathscr{C}$ of solutions to problem (1.1), with the solutions on $\mathscr{C}$ not changing sign.

Proof

In view of the conclusion of Theorem 3.1 and Lemma 3.2, we only need to prove that $\mathscr{C}\subset (\Phi \cup \{(\lambda _{1},0)\})$.

Suppose on the contrary that $\mathscr{C}\not \subset (\Phi \cup \{(\lambda _{1},0)\})$. Then there exists $(\lambda ,u)\in \mathscr{C}\cap (\mathbb{R}\times \partial S)$ such that

$$ (\lambda ,u)\neq (\lambda _{1},0), \quad u\notin S, $$

and $(\lambda _{n},u_{n})\in \mathscr{C}\cap (\mathbb{R}\times \partial S)$ with

$$ (\lambda _{n},u_{n})\rightarrow (\lambda ,u). $$

Since $u\in \partial S$, by Lemma 3.3, we obtain $u\equiv 0$. Let $\omega _{n}=\frac{u_{n}}{\|u_{n}\|}$, then $\omega _{n}$ satisfies

$$ \omega _{n}=G_{p} \biggl(-\lambda _{n} a(t)\varphi _{p} \bigl(\omega _{n}(t) \bigr)- \frac{g(t,u_{n}(t),\lambda _{n})}{ \Vert u_{n} \Vert ^{p-1}} \biggr). $$

(3.9)

Combining (3.5), (3.9) with the compactness of $G_{p}$, we obtain $\omega _{n}\rightarrow \omega _{0}$ and $\|\omega _{0}\|=1$. Obviously, there is

$$ -\Delta \bigl[\varphi _{p} \bigl(\Delta \omega _{0}(t-1) \bigr) \bigr]=\lambda a(t)\varphi _{p} \bigl( \omega _{0}(t) \bigr). $$

Hence there exists $\lambda =\lambda _{k}$ ($k\neq 1$). Furthermore,

$$ (\lambda _{n},u_{n})\rightarrow (\lambda _{k},0). $$

This contradicts Lemma 3.2. □

Now, we will show more details about the bifurcation from Theorem 3.2. Let $\mathbb{E}=\mathbb{R}\times E$, $\Phi (\lambda ,u)=u-F(\lambda ,u)$ and $S:= \overline{\{(\lambda ,u)\in \mathbb{E}: \Phi (\lambda ,u)=0, u\neq 0\}}^{\mathbb{E}}$. For convenience, let us introduce a few notations. Given any $\lambda \in \mathbb{R}$ and $0< s<+\infty$, we consider an open neighborhood of $(\lambda _{1},0)$ in $\mathbb{E}$ defined by

$$ \mathbb{B}_{s}(\lambda _{1},0):= \bigl\{ (\lambda ,u) \in \mathbb{E} \mid \Vert u \Vert + \vert \lambda -\lambda _{1} \vert < s \bigr\} . $$

And $B_{s}(0)$ denotes $\{u\in E \mid \|u\|< s\}$. Let $E_{0}$ be a closed subspace of E such that

$$ E=\operatorname{span}\{\phi _{1}\}\oplus E_{0}. $$

By the Hahn–Banach theorem [26], there exists a linear functional $l\in E^{\ast }$, where $E^{\ast }$ denotes the dual space of E such that

$$ l(\phi _{1})=1\quad \text{and}\quad E_{0}= \bigl\{ u\in E\mid l(u)=0 \bigr\} . $$

Finally, for any $0<\eta <1$, we define

$$ K_{\eta }= \bigl\{ (\lambda ,u)\in \mathbb{E} \mid \bigl\vert l(u) \bigr\vert >\eta \Vert u \Vert \bigr\} . $$

Since $u\mapsto |l(u)|-\|u\|$ is continuous, $K_{\eta }$ is an open subset of $\mathbb{E}$ consisting of two disjoint components $K_{\eta }^{+}$ and $K_{\eta }^{-}$, where

$$ K_{\eta }^{+}= \bigl\{ (\lambda ,u)\in \mathbb{E} \mid l(u)> \eta \Vert u \Vert \bigr\} ,\qquad K_{\eta }^{-}= \bigl\{ ( \lambda ,u) \in \mathbb{E} \mid l(u)< -\eta \Vert u \Vert \bigr\} . $$

In particular, both $K_{\eta }^{+}$ and $K_{\eta }^{-}$ are convex cones, $K_{\eta }^{+}=-K_{\eta }^{-}$.

Before proving our main result, we need the following lemma, which localizes the possible solutions of (1.1) bifurcating from $(\lambda _{1},0)$.

Lemma 3.4

For every $\eta \in (0,1)$, there is $\delta _{0}>0$ such that, for each $0<\delta <\delta _{0}$,

$$ \bigl( \bigl(S\setminus \bigl\{ (\lambda _{1},0) \bigr\} \bigr)\cap \overline{\mathbb{B}}_{\delta }( \lambda _{1},0) \bigr)\subset K_{\eta }. $$

Moreover, for each

$$ (\lambda ,u)\in \bigl(S\setminus \bigl\{ (\lambda _{1},0) \bigr\} \bigr)\cap \bigl( \overline{\mathbb{B}}_{\delta }(\lambda _{1},0) \bigr), $$

there are $s\in \mathbb{R}$ and $y\in E_{0}$ (unique) such that

$$ u=s\phi _{1}+y\quad \textit{and}\quad \vert s \vert >\eta \Vert u \Vert . $$

Furthermore, for these solutions $(\lambda ,u)$,

$$ \lambda =\lambda _{1}+o(1)\quad \textit{and}\quad y=o(s) $$

as $s\to 0$.

Proof

This conclusion can be obtained by using the similar method in López-Gómez [16, Lemma 6.4.1]. □

Remark 3.1

From the proof of Lemma 6.4.1 of [14], we can see that if $g(t,u,\lambda )$ is replaced by $g_{n}(t,u,\lambda )$, which satisfies

$$ \lim_{ \Vert u \Vert \to 0}\frac{g_{n}(t,u,\lambda )}{ \Vert u \Vert ^{p-1}}=0 $$

uniformly for all $n\in \mathbb{N}$, then $\delta _{0}$ can be chosen uniformly with respect to n.

Let $\delta >0$ be the constant from Lemma 3.4. For $0<\varepsilon \leq \delta$, we define $\mathscr{D}_{\lambda _{1},\varepsilon }^{v}$ to be the component of $\{(\lambda _{1},0)\}\cup (S\cap \overline{\mathbb{B}}_{\varepsilon }\cap K_{\eta }^{v})$ containing $(\lambda _{1},0)$, $\mathscr{C}_{\lambda _{1},\varepsilon }^{v}$ to be the component of $\overline{\mathscr{C}\setminus \mathscr{D}_{\lambda _{1},\varepsilon }^{-v}}$ containing $(\lambda _{1},0)$, and $\mathscr{C}^{v}$ to be the closure of $\bigcup_{0<\varepsilon \leq \delta }\mathscr{C}_{\lambda _{1}, \varepsilon }^{v}$. Obviously, $\mathscr{C}^{v}$ is connected. By Lemma 3.4, the definition of $\mathscr{C}^{v}$ is independent of the choice of η and $\mathscr{C}=\mathscr{C}^{+}\cup \mathscr{C}^{-}$.

Similar to Dancer’s result [9, Theorem 2], we can obtain the following unilateral global bifurcation result.

Theorem 3.3

Either $\mathscr{C}^{+}$ and $\mathscr{C}^{-}$ are both unbounded, or else $\mathscr{C}^{+}\cap \mathscr{C}^{-}\neq \{(\lambda _{1},0)\}$.

Remark 3.2

Although the proof of the above result is similar to that of the semilinear case in [9], the method cannot be used directly in this paper. In fact, the proof of Lemma 3 in [9] strictly depends on the linear characteristics of L. Therefore, we use Lemma 3.4 and functional analysis to prove the above result.

To show Theorem 3.3, we need the following three lemmas.

Lemma 3.5

Suppose $\delta _{1}, \delta _{2}>0$ such that $0<\delta _{1}+\delta _{2}<\delta$ and $\Phi (\lambda ,u)\neq 0$ if $\|u\|=\delta _{1}$ and $|\lambda -\lambda _{1}|\leq \delta _{2}$. If $0<\sigma <\delta _{2}$ and $\beta (\sigma )$ is sufficiently small and positive, then

$$ \deg \bigl(\Phi (\lambda _{1}+\sigma ,\cdot ), W^{v}, 0 \bigr)-\deg \bigl(\Phi ( \lambda _{1}-\sigma ,\cdot ), W^{v}, 0 \bigr)=1, $$

where $W^{v}=\{u\in E \mid (\lambda ,u)\in K_{\eta }^{v}, \beta (\sigma )< \|u\|<\delta _{1}\}$.

Proof

Recall that $u=l(u)\varphi _{1}+y$. We define

$$ \widehat{g}(t,u,\lambda )= \textstyle\begin{cases} g(t,u,\lambda ), & l(u)\leq -\eta \Vert u \Vert , \\ \frac{-l(u)}{\eta \Vert u \Vert }g(t,-\eta \Vert u \Vert \phi _{1}+y,\lambda ),& {-}\eta \Vert u \Vert < l(u)\leq 0, \\ -g(t,-u,\lambda ),& l(u)>0. \end{cases} $$

It is easy to verify that ĝ is odd with respect to u. Let

$$ \widehat{\Phi }(\lambda ,u)=u-G_{p} \bigl(-\lambda a(t)\varphi _{p} \bigl(u(t) \bigr)- \widehat{g} \bigl(t,u(t),\lambda \bigr) \bigr), $$

then the mapping $\widehat{\Phi }(\lambda ,u)$ is odd with respect to u.

By Lemma 3.4 and our assumptions, the equation $\Phi (\lambda _{1}+\sigma ,u)=0$ has no solution in $B_{\delta _{1}}\setminus (W^{+}\cup W^{-}\cup B_{\beta })$. By Lemma 3.4, $\widehat{\Phi }(\lambda _{1}+\sigma ,u)=\Phi (\lambda _{1}+\sigma ,u)$ on $\partial B_{\delta _{1}}\cup \partial B_{\beta }$. It follows that

$$\begin{aligned}& \deg \bigl(\widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),B_{\delta _{1}},0 \bigr)- \deg \bigl(\widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),B_{\beta },0 \bigr) \\& \quad = \deg \bigl( \widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),W^{+},0 \bigr)+ \deg \bigl( \widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),W^{-},0 \bigr). \end{aligned}$$

The oddness of $\widehat{\Phi }(\lambda _{1}+\sigma ,\cdot )$ and the definition of the degree in Schwartz [20] ensure that

$$ \deg \bigl(\widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),W^{+},0 \bigr)= \deg \bigl( \widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),W^{-},0 \bigr). $$

Then the definition of Φ̂ implies

$$ \deg \bigl(\Phi (\lambda _{1}+\sigma ,\cdot ),W^{v},0 \bigr)= \deg \bigl( \widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),W^{v},0 \bigr). $$

Thus

$$ 2\deg \bigl(\Phi (\lambda _{1}+\sigma ,\cdot ),W^{v},0 \bigr)=\deg \bigl( \widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),B_{\delta _{1}},0 \bigr)- \deg \bigl( \widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),B_{\beta },0 \bigr). $$

(3.10)

A similar result holds with $\lambda _{1}+\sigma$ replaced by $\lambda _{1}-\sigma$,

$$ 2\deg \bigl(\Phi (\lambda _{1}-\sigma ,\cdot ),W^{v},0 \bigr)=\deg \bigl( \widehat{\Phi }(\lambda _{1}-\sigma ,\cdot ),B_{\delta _{1}},0 \bigr)- \deg \bigl( \widehat{\Phi }(\lambda _{1}-\sigma ,\cdot ),B_{\beta },0 \bigr). $$

(3.11)

Connecting Lemma 3.4 with the definition of Φ̂, we obtain

$$ \deg \bigl(\widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),B_{\beta },0 \bigr)= \deg \bigl( \Phi (\lambda _{1}+\sigma ,\cdot ),B_{\beta },0 \bigr) $$

and

$$ \deg \bigl(\widehat{\Phi }(\lambda _{1}-\sigma ,\cdot ),B_{\beta },0 \bigr)= \deg \bigl( \Phi (\lambda _{1}-\sigma ,\cdot ),B_{\beta },0 \bigr). $$

From the proof of Theorem 3.1, it is obvious that

$$ \deg \bigl(\Phi (\lambda _{1}-\sigma ,\cdot ),B_{\beta },0 \bigr)=1 \quad \text{and}\quad \deg \bigl(\Phi (\lambda _{1}+\sigma , \cdot ),B_{\beta },0 \bigr)=-1. $$

(3.12)

By Lemma 3.4 and the definition of Φ̂, it follows that

$$ \deg \bigl(\widehat{\Phi }(\lambda _{1}+\sigma ,\cdot ),B_{\delta _{1}},0 \bigr)= \deg \bigl(\Phi (\lambda _{1}+\sigma , \cdot ),B_{\delta _{1}},0 \bigr) $$

and

$$ \deg \bigl(\widehat{\Phi }(\lambda _{1}-\sigma ,\cdot ),B_{\delta _{1}},0 \bigr)= \deg \bigl(\Phi (\lambda _{1}-\sigma , \cdot ),B_{\delta _{1}},0 \bigr). $$

By our hypothesis, for $\lambda \in [\lambda _{1}-\sigma ,\lambda _{1}+\sigma ]$, the homotopy $\Phi (\lambda ,\cdot )$ is admissible on $B_{\delta _{1}}$. The homotopy invariance of the degree ensures that

$$ \deg \bigl(\Phi (\lambda _{1}+\sigma ,\cdot ),B_{\delta _{1}},0 \bigr)= \deg \bigl( \Phi (\lambda _{1}-\sigma ,\cdot ),B_{\delta _{1}},0 \bigr). $$

Subtracting (3.10) from (3.11) and using (3.12), we have

$$ \deg \bigl(\Phi (\lambda _{1}+\sigma ,\cdot ),W^{v},0 \bigr)- \deg \bigl(\Phi ( \lambda _{1}-\sigma ,\cdot ),W^{v},0 \bigr)=1. $$

The proof of this lemma is complete. □

Define $T_{\lambda _{1},\varepsilon }^{-}$ to be the component of $\mathscr{C}\setminus (\mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+})$ containing $(\lambda _{1},0)$.

Lemma 3.6

If $0<\varepsilon <\delta$, zero is an isolated solution of $\Phi (\lambda _{1},u)=0$ and $T_{\lambda _{1},\varepsilon }^{-}$ is bounded in $\mathbb{E}$, then

$$ \partial \mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+}\cap T_{ \lambda _{1},\varepsilon }^{-}\neq \emptyset . $$

Proof

The proof of Lemma 2 in Dancer [9] is also valid for the quasilinear case, so we omit the proof. □

Lemma 3.7

Lemma 3.6holds without the assumption that zero is an isolated solution of $\Phi (\lambda _{1},u)=0$.

Proof

Because when $\|u\|\to 0$, there is

$$ \frac{g(t,u,\lambda _{1})}{ \Vert u \Vert ^{p-1}}\to 0 $$

uniformly for a.e. $t\in [1,T+1]_{Z}$. So we can choose a continuous function $\rho :[0,+\infty )\to \mathbb{R}$ such that $\rho (0)=0$, and for any $0<\iota <\delta$,

$$ \rho (\iota ) \Vert \phi _{1} \Vert >\sup \biggl\{ \frac{ \Vert g(t,u,\lambda _{1}) \Vert }{ \Vert u \Vert ^{p-1}}: \Vert u \Vert =\iota \biggr\} . $$

For every integer n, we can choose continuous functions $f_{n}:[0,+\infty )\to [0,1]$ such that

$$ f_{n}(s)= \textstyle\begin{cases} \varphi _{p}(s), & 0\leq \vert s \vert \leq \frac{1}{2n}, \\ 0, & \vert s \vert \geq \frac{1}{n}. \end{cases} $$

Define

$$ \Phi _{n}(\lambda ,u):=u-G_{p} \bigl(-\lambda a(t) \varphi _{p} \bigl(u(t) \bigr)-g \bigl(t,u(t), \lambda \bigr)-f_{n} \bigl( \Vert u \Vert \bigr)\rho \bigl( \Vert u \Vert \bigr)\phi _{1} \bigr). $$

Since $\lim_{\|u\|\to 0}\frac{g(t,u,\lambda )}{\|u\|^{p-1}}=0$, we can see that

$$ \lim_{ \Vert u \Vert \to 0} \frac{g(t,u,\lambda )+f_{n}( \Vert u \Vert ) \rho ( \Vert u \Vert )\phi _{1}}{ \Vert u \Vert ^{p-1}}=0,\quad n\in \mathbb{N} $$

(3.13)

uniformly for a.e. $t\in [1,T+1]_{Z}$ and λ on bounded sets. Let

$$ S_{n}:= \overline{ \bigl\{ (\lambda ,u)\in \mathbb{E}: \Phi _{n}(\lambda ,u)=0, u\neq 0 \bigr\} }^{ \mathbb{E}}, $$

using Remark 3.1 and (3.13), $S_{n}$ can be chosen such that

$$ \bigl( \bigl(S_{n}\setminus \bigl\{ (\lambda _{1},0) \bigr\} \bigr)\cap \overline{\mathbb{B}}_{\delta }(\lambda _{1},0) \bigr)\subset K_{\eta }. $$

We claim that zero is an isolated solution of $\Phi _{n}(\lambda _{1},u)=0$ for each $n\in \mathbb{N}^{+}$.

Suppose on the contrary that u is a nontrivial solution of $\Phi _{n}(\lambda _{1},u)=0$ such that

$$ 0< \Vert u \Vert :=\iota < \delta . $$

We divide the proof into two cases.

Case 1. u∈ span $\{\phi _{1}\}$.

$\Phi _{n}(\lambda _{1},u)=0$ implies

$$ u=G_{p} \bigl(-\lambda _{1} a(t)\varphi _{p} \bigl(u(t) \bigr)-g \bigl(t,u(t),\lambda _{1} \bigr)-f_{n} \bigl( \Vert u \Vert \bigr)\rho \bigl( \Vert u \Vert \bigr)\phi _{1} \bigr), $$

i.e.,

$$ -\Delta \bigl[\varphi _{p} \bigl(\Delta u(t-1) \bigr) \bigr]=\lambda _{1} a(t)\varphi _{p} \bigl(u(t) \bigr)+g \bigl(t,u(t), \lambda _{1} \bigr)+f_{n} \bigl( \Vert u \Vert \bigr) \rho \bigl( \Vert u \Vert \bigr)\phi _{1}. $$

Obviously, there is

$$ g(t,u,\lambda _{1})+f_{n} \bigl( \Vert u \Vert \bigr)\rho \bigl( \Vert u \Vert \bigr)\phi _{1}=0. $$

(3.14)

However,

$$ \begin{aligned} \bigl\Vert g(t,u,\lambda _{1})+f_{n} \bigl( \Vert u \Vert \bigr)\rho \bigl( \Vert u \Vert \bigr)\phi _{1} \bigr\Vert &\geq \bigl\Vert f_{n} \bigl( \Vert u \Vert \bigr)\rho \bigl( \Vert u \Vert \bigr)\phi _{1} \bigr\Vert - \bigl\Vert g(t,u,\lambda _{1}) \bigr\Vert \\ &\geq \iota ^{p-1}\rho (\iota ) \Vert \phi _{1} \Vert - \bigl\Vert g(t,u,\lambda _{1}) \bigr\Vert \\ &>0. \end{aligned} $$

This contradicts (3.14).

Case 2. u∉ span $\{\phi _{1}\}$.

By $\Phi _{n}(\lambda _{1},u)=0$ and Lemma 2.1, we have

$$ \sum_{t=1}^{T+1} \bigl\vert \Delta u(t) \bigr\vert ^{p}=\lambda _{1}\sum _{t=1}^{T+1}a(t) \vert u \vert ^{p}+ \sum_{t=1}^{T+1} g(t,u,\lambda _{1})u+f_{n} \bigl( \Vert u \Vert \bigr)\rho \bigl( \Vert u \Vert \bigr)\sum_{t=1}^{T+1} \phi _{1} u. $$

Let

$$ f(u)=\sum_{t=1}^{T+1} \bigl\vert \Delta u(t) \bigr\vert ^{p}-\lambda _{1}\sum _{t=1}^{T+1}a(t) \vert u \vert ^{p}- \sum_{t=1}^{T+1} g(t,u,\lambda _{1})u-f_{n} \bigl( \Vert u \Vert \bigr)\rho \bigl( \Vert u \Vert \bigr)\sum_{t=1}^{T+1} \phi _{1} u, $$

then $f(u)=0$. u∉ span $\{\phi _{1}\}$ implies that there is $\gamma >0$ such that

$$ \Biggl\vert \sum_{t=1}^{T+1} \bigl\vert \Delta u(t) \bigr\vert ^{p}-\lambda _{1}\sum _{t=1}^{T+1}a(t) \vert u \vert ^{p} \Biggr\vert \geq \gamma \Vert u \Vert ^{p-1}. $$

From (3.13), we know that

$$ \lim_{ \Vert u \Vert \to 0} \frac{g(t,u,\lambda _{1})u+f_{n}( \Vert u \Vert ) \rho ( \Vert u \Vert )\phi _{1}u}{ \Vert u \Vert ^{p-1}}=0, \quad n\in \mathbb{N} $$

(3.15)

uniformly for a.e. $t\in [1,T+1]_{Z}$. By (3.15), it can be easily seen that when ι is sufficiently small,

$$ \Biggl\vert \sum_{t=1}^{T+1} g(t,u, \lambda _{1})u+f_{n} \bigl( \Vert u \Vert \bigr)\rho \bigl( \Vert u \Vert \bigr) \sum_{t=1}^{T+1} \phi _{1} u \Biggr\vert < \gamma \Vert u \Vert ^{p-1}. $$

Thus $|f(u)|>0$. This contradicts $f(u)=0$.

For any $0<\varepsilon <\delta$, we assume that $T_{\lambda _{1},\varepsilon }^{-}$ is bounded in $\mathbb{E}$. Define $T_{n}$ to be the component of $S_{n}\setminus (\mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+})$ containing $(\lambda _{1},0)$. It is easy to see that the limit of $T_{n}$ is $T_{\lambda _{1},\varepsilon }^{-}$, so $T_{n}$ is bounded in $\mathbb{E}$. Suppose that Lemma 3.7 is false, then

$$ \partial \mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+}\cap T_{ \lambda _{1},\varepsilon }^{-}= \emptyset . $$

The definition of $T_{\lambda _{1},\varepsilon }^{-}$ implies

$$ \mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+}\cap T_{ \lambda _{1},\varepsilon }^{-}= \emptyset . $$

Since $T_{\lambda _{1},\varepsilon }^{-}$ is bounded, we can find a constant $R>0$ such that $T_{\lambda _{1},\varepsilon }^{-}\subset \mathbb{B}_{R}(\lambda _{1},0)$.

By these facts and the classical topological result from Whyburn [21], we obtain that

$$ K:= \bigl(S\cap \overline{\mathbb{B}}_{R}(\lambda _{1},0) \bigr) \setminus \bigl(\mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+} \bigr)=k_{1} \cup k_{2}, $$

where $k_{1}$, $k_{2}$ are disjoint compact subsets of K and $T_{\lambda _{1},\varepsilon }^{-}\subset k_{1}$,

$$ \bigl(S\cap \partial \mathbb{B}_{R}(\lambda _{1},0) \bigr)\cup \bigl(S \cap \partial \mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+} \bigr)\subset k_{2}. $$

Therefore, there exists a bounded open set U in $\mathbb{E}$ such that $k_{1}\subset U$, $k_{2}\cap \overline{U}=\emptyset$, $(\lambda _{1},0) \in U$, $(\partial U\cap S)\subset (\mathbb{B}_{\varepsilon }(\lambda _{1},0) \cap K_{\eta }^{+})$ and $\partial \mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+} \cap U=\emptyset$.

Applying Lemma 3.6 to $\Phi _{n}$, we can see that

$$ \partial \mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+}\cap T_{n} \neq \emptyset ,\quad n\in \mathbb{N}. $$

By the connectedness of $T_{n}$, there exists $(\lambda _{n},u_{n})\in \partial U\cap T_{n}$. We assume that there are $u_{n}\rightharpoonup u^{\ast }$ in E and $\lambda _{n}\to \lambda ^{\ast }$ in $\mathbb{R}$. Letting $n\to +\infty$ on the both of $\Phi _{n}(\lambda _{n},u_{n})=0$ and using the compact and continuous properties of $G_{p}$, we can show that $\Phi _{n}(\lambda ^{\ast },u^{\ast })=0$. It is easy to see that

$$ \bigl(\lambda ^{\ast },u^{\ast } \bigr)\in (S\cap \partial U ) \setminus \bigl( \mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+} \bigr), $$

this contradicts the definition of U. We have completed the proof. □

Proof of Theorem 3.3

Define $T_{\lambda _{1}}^{-}$ to be the closure of $\bigcup_{0<\varepsilon \leq \delta }T_{\lambda _{1}, \varepsilon }^{-}$, then $T_{\lambda _{1}}^{-}\subseteq \mathscr{C}^{-}$. Suppose that $\mathscr{C}^{-}$ is bounded. Then, by Lemma 3.7, for any $0<\varepsilon \leq \delta$, we obtain

$$ \partial \mathbb{B}_{\varepsilon }(\lambda _{1},0)\cap K_{\eta }^{+}\cap T_{ \lambda _{1}}^{-}\neq \emptyset . $$

It follows that

$$ \bigl(T_{\lambda _{1}}^{-}\setminus \bigl(\mathbb{B}_{\delta }( \lambda _{1},0) \cap K_{\eta }^{-} \bigr) \bigr) \cap \partial \mathbb{B}_{\varepsilon }(\lambda _{1},0) \neq \emptyset . $$

(3.16)

Furthermore, for every open set U in $\mathbb{E}$, which satisfies $(\lambda _{1},0)\in U$ and $U\subseteq \mathbb{B}_{\delta }(\lambda _{1},0)$, (3.16) implies

$$ \bigl(T_{\lambda _{1}}^{-}\setminus \bigl(\mathbb{B}_{\delta }( \lambda _{1},0) \cap K_{\eta }^{-} \bigr) \bigr) \cap \partial U\neq \emptyset . $$

(3.17)

Let $\mathcal{E}=T_{\lambda _{1}}^{-}\setminus (\mathbb{B}_{\delta }( \lambda _{1},0)\cap K_{\eta }^{-})$, T be the component of $T_{\lambda _{1}}^{-}\setminus (\mathbb{B}_{\delta }(\lambda _{1},0) \cap K_{\eta }^{-})$ containing $(\lambda _{1},0)$. It is easy to know that $\mathcal{E}$ is a compact metric space under the induced topology of $\mathbb{E}$ and T is a closed subset of $\mathcal{E}$.

We claim that $T\cap \partial \mathbb{B}_{\delta }(\lambda _{1},0)\neq \emptyset$.

Suppose on the contrary that $T\cap \partial \mathbb{B}_{\delta }(\lambda _{1},0)=\emptyset$. From [25], we know that $K=K_{1}\cup K_{2}$, where $K_{1}$, $K_{2}$ are disjoint compact subsets of K containing T and $\partial \mathbb{B}_{\delta }(\lambda _{1},0)\cap K$, respectively. There exists a bounded open neighborhood $\mathcal{O}$ in $\mathbb{E}$ of $K_{1}$ such that $\mathcal{O}\subseteq \mathbb{B}_{\delta }(\lambda _{1},0)$ and $\partial \mathcal{O}\cap K_{2}=\emptyset$. This contradicts (3.17).

Combining the definition of $\mathscr{C}^{+}$ with $T\cap \partial \mathbb{B}_{\delta }(\lambda _{1},0)\neq \emptyset$, we obtain $T\setminus \{(\lambda _{1},0)\}\neq \emptyset$ and

$$ \mathscr{C}_{\lambda _{1}}^{+}\supseteq \mathscr{C}_{\lambda _{1}, \delta }^{+} \supseteq T. $$

Therefore, $\mathscr{C}^{+}\cap \mathscr{C}^{-}\neq \{(\lambda _{1},0)\}$. Since a similar argument could be used for $\mathscr{C}^{-}$, this completes the proof of Theorem 3.3. □

Combining Theorem 3.2 and Theorem 3.3, we can obtain the following unilateral global bifurcation result.

Theorem 3.4

Let $v\in \{+,-\}$, then $\mathscr{C}^{v}$ is unbounded in $\mathbb{R}\times E$ and

$$ \mathscr{C}^{v}\subset \bigl\{ (\lambda _{1},0) \bigr\} \cup \bigl(\mathbb{R}\times S^{v} \bigr). $$

(3.18)

Proof

We can find a bounded neighborhood $\mathbb{O}$ of $(\lambda _{1},0)$ such that

$$ \bigl(\mathbb{O}\cap \mathscr{C}^{v} \bigr)\subset \bigl\{ (\lambda _{1},0) \bigr\} \cup \bigl( \mathbb{R}\times S^{v} \bigr)\quad \text{or}\quad \bigl(\mathbb{O}\cap \mathscr{C}^{v} \bigr) \subset \bigl\{ (\lambda _{1},0) \bigr\} \cup \bigl(\mathbb{R}\times S^{-v} \bigr). $$

Without loss of generality, we may suppose that

$$ \bigl(\mathbb{O}\cap \mathscr{C}^{v} \bigr)\subset \bigl\{ (\lambda _{1},0) \bigr\} \cup \bigl( \mathbb{R}\times S^{v} \bigr). $$

By Theorem 3.2, $\mathscr{C}^{v}\setminus \mathbb{O}$ in $\mathbb{R}\times S^{v}$, so (3.18) holds. Next, we only need to prove that both $\mathscr{C}^{+}$ and $\mathscr{C}^{-}$ are unbounded. Suppose on the contrary that $\mathscr{C}^{+}$ is bounded, the case for $\mathscr{C}^{-}$ is similar. By Theorem 3.3, we know that

$$ \bigl(\mathscr{C}^{+}\cap \mathscr{C}^{-} \bigr) \setminus \bigl\{ (\lambda _{1},0) \bigr\} \neq \emptyset , $$

in view of (3.18), there exists $(\lambda _{\ast },u_{\ast })\in \mathscr{C}^{+}\cap \mathscr{C}^{-}$ such that

$$ (\lambda _{\ast },u_{\ast })\neq (\lambda _{1},0) \quad \text{and}\quad u_{\ast }\in S^{+}\cap S^{-}. $$

This contradicts the definitions of $S^{+}$ and $S^{-}$. □

4 Constant sign solutions for nonlinear discrete p-Laplacian problem

In this section, we use Theorem 3.4 to prove the existence of constant sign solutions for the discrete p-Laplacian problem

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

(4.1)

where $a: [1,T+1]_{Z}\to [0,+\infty )$ and $a(t_{0})>0$ for some $t_{0}\in [1,T+1]_{Z}$, $f\in C(\mathbb{R},\mathbb{R})$ with $sf(s)>0$ for $s\neq 0$. We will discuss the existence of constant sign solutions according to the different behavior of nonlinear term f at 0 and ∞. Denote

$$ f_{0}=\lim_{s\rightarrow 0}\frac{f(s)}{\varphi _{p}(s)},\qquad f_{\infty }=\lim_{s\rightarrow \infty } \frac{f(s)}{\varphi _{p}(s)}. $$

Definition 4.1

(see [17])

Let X be a Banach space, $\{C_{n}\mid n=1,2,3,\ldots \}$ be a family of subsets of X. Then the superior D of $C_{n}$ is defined by

$$ D:=\lim \sup_{n\rightarrow \infty }C_{n}=\{x\in X\mid \exists {n_{i}} \subset N \text{ and } x_{n_{i}}\in C_{n_{i}} \text{ such that } x_{n_{i}} \rightarrow x\}. $$

Definition 4.2

(see [20])

The component of M is the largest connected subset in M.

Lemma 4.1

(see [20])

Suppose that X is a compact metric space, A and B are non-intersecting closed subsets of X, and no component of X intersects both A and B. Then there exist two disjoint compact subsets $X_{A}$ and $X_{B}$, such that $X=X_{A}\bigcup X_{B}$, $A\subset X_{A}$, $B\subset X_{B}$.

Lemma 4.2

(see [19])

Let X be a Banach space, $C_{n}$ is a component of X, assume that:

• (i) There exist $z_{n}\in C_{n}$ ($n=1,2,\ldots$) and $z^{\ast }\in X$ such that $z_{n}\rightarrow z^{\ast }$;

• (ii) $\lim_{n\rightarrow \infty }r_{n}=\infty$, where $r_{n}=\sup \{\|x\|: x\in C_{n}\}$;

• (iii) For every $R>0$, $(\bigcup_{n=1}^{\infty }C_{n}) \cap \Omega _{R}$ is a relative compact set of X, where $\Omega _{R}=\{x\in X: \|x\|\leq R\}$.

Then $D:=\lim \sup_{n\rightarrow \infty }C_{n}$ contains an unbounded component C such that $z^{\ast }\in C$.

The main results of this section are the following.

Theorem 4.1

If $f_{0}\in (0,\infty )$ and $f_{\infty }=0$, then (4.1) has at least two solutions $u^{+}$ and $u^{-}$ for $\lambda \in (\frac{\lambda _{1}}{f_{0}},+\infty )$, where $u^{+}$ is positive in $[0,T+1]_{Z}$ and $u^{-}$ is negative in $[0,T+1]_{Z}$ (see Fig. 1).

Figure 1

$f_{0}\in(0,\infty)$, $f_{\infty}=0$

Proof

Let $\zeta \in C(\mathbb{R},\mathbb{R})$ such that $f(s)=f_{0}\varphi _{p}(s)+\zeta (s)$, where ζ satisfies

$$ \lim_{s\rightarrow 0}\frac{\zeta (s)}{\varphi _{p}(s)}=0. $$

Applying Theorem 3.4 to problem (4.1), it can be seen that there exist two different unbounded connected components $C^{+}$ and $C^{-}$, which bifurcate from $(\frac{\lambda _{1}}{f_{0}},0)$, and

$$ C^{v}\subset \biggl( \biggl\{ \biggl(\frac{\lambda _{1}}{f_{0}},0 \biggr) \biggr\} \cup \bigl(\mathbb{R} \times S^{v} \bigr) \biggr), $$

where $v=+$ or −. Obviously, $C^{v}\cap (\{0\}\times E)=\emptyset$.

Next, we prove that $C^{v}$ is unbounded in the direction of the λ axis. Assume on the contrary that

$$ \sup \bigl\{ \lambda :(\lambda ,y)\in C^{v} \bigr\} < \infty . $$

Then there exists a sequence $\{(\mu _{k},y_{k})\}\subset C^{v}$ such that

$$ \lim_{k\rightarrow \infty } \Vert y_{k} \Vert =\infty ,\qquad \vert \mu _{k} \vert \leq C_{0} $$

for some positive constant $C_{0}$ independent of k. This implies that

$$ \lim_{k\rightarrow \infty }y_{k}(t)=\infty \quad \text{uniformly on } t \in [0,T+2]_{Z}. $$

(4.2)

Since $\{(\mu _{k},y_{k})\}\subset C^{v}$, we have that

$$ \textstyle\begin{cases} \Delta [\varphi _{p}(\Delta y_{k}(t-1))]+\mu _{k} a(t)f(y_{k}(t))=0, \quad t\in [1,T+1]_{Z}, \\ \Delta y_{k}(0)=y_{k}(T+2)=0. \end{cases} $$

(4.3)

Set $v_{k}(t)=\frac{y_{k}(t)}{\|y_{k}\|}$, then

$$ \Vert v_{k} \Vert =1. $$

Choosing a subsequence and relabeling if necessary, it follows that there exists $(\mu _{\ast },v_{\ast })\in (0,C_{0}]\times E$ with

$$ \Vert v_{\ast } \Vert =1 $$

(4.4)

such that

$$ \lim_{k\rightarrow \infty }(\mu _{k},v_{k})=(\mu _{\ast },v_{\ast }) \quad \text{in } \mathbb{R}\times E. $$

Moreover, from (4.2), (4.3), and $f_{\infty }=0$, it follows that

$$ \textstyle\begin{cases} \Delta [\varphi _{p}(\Delta v_{\ast }(t-1))]+\mu _{\ast }a(t)\cdot 0=0, \quad t\in [1,T+1]_{Z}, \\ \Delta v_{\ast }(0)=v_{\ast }(T+2)=0. \end{cases} $$

Hence, $v_{\ast }(t)\equiv 0$ for $t\in [0,T+2]_{Z}$. This contradicts (4.4). Therefore,

$$ \sup \bigl\{ \lambda :(\lambda ,y)\in C^{v} \bigr\} =\infty . $$

 □

Theorem 4.2

If $f_{0}\in (0,\infty )$ and $f_{\infty }=+\infty$, then (4.1) has at least two solutions $u^{+}$ and $u^{-}$ for $\lambda \in (0,\frac{\lambda _{1}}{f_{0}})$, where $u^{+}$ is positive in $[0,T+1]_{Z}$ and $u^{-}$ is negative in $[0,T+1]_{Z}$ (see Fig. 2).

Figure 2

$f_{0}\in(0,\infty)$, $f_{\infty}=+\infty$

Proof

In this case, we show that $C^{v}$ joins $(\frac{\lambda _{1}}{f_{0}},0)$ with $(0,\infty )$.

Similar to Theorem 4.1, we know that there exist two different unbounded connected components $C^{+}$ and $C^{-}$, which bifurcate from $(\frac{\lambda _{1}}{f_{0}},0)$.

Let $\{(\mu _{k},y_{k})\}\subset C^{v}$ be such that

$$ \vert \mu _{k} \vert + \Vert y_{k} \Vert \rightarrow \infty ,\quad k\rightarrow \infty . $$

Then

$$ \textstyle\begin{cases} \Delta [\varphi _{p}(\Delta y_{k}(t-1))]+\mu _{k} a(t)f(y_{k}(t))=0,\quad t\in [1,T+1]_{Z}, \\ \Delta y_{k}(0)=y_{k}(T+2)=0. \end{cases} $$

If $\{\|y_{k}\|\}$ is bounded, i.e., there exists a constant $M_{1}$ depending not on k such that $\|y_{k}\|\leq M_{1}$, then we may assume that

$$ \lim_{k\rightarrow \infty }\mu _{k}=\infty . $$

Combining this with the fact

$$ \frac{f(y_{k}(t))}{\varphi _{p}(y_{k}(t))}\geq \inf \biggl\{ \frac{f(s)}{\varphi _{p}(s)}\Bigm| 0\leq s\leq M_{1} \biggr\} >0, $$

we obtain

$$ \lim_{k\rightarrow \infty }\mu _{k} \frac{f(y_{k}(t))}{\varphi _{p}(y_{k}(t))}=\infty ,\quad \forall t\in [0,T+2]_{Z}, $$

using the relation

$$ \Delta \bigl[\varphi _{p} \bigl(\Delta y_{k}(t-1) \bigr) \bigr]+\mu _{k} a(t) \frac{f(y_{k}(t))}{\varphi _{p}(y_{k}(t))}\varphi _{p} \bigl(y_{k}(t) \bigr)=0,\quad t\in [1,T+1]_{Z}. $$

(4.5)

We deduce that $y_{k}$ must change its sign on $[0,T+2]_{Z}$ if k is large enough, and this contradicts the fact that $y_{k}$ does not change sign. Hence,

$$ \Vert y_{k} \Vert \rightarrow \infty ,\quad k\rightarrow \infty . $$

Now, taking $\{(\mu _{k},y_{k})\}\subset C^{v}$ such that

$$ \Vert y_{k} \Vert \rightarrow \infty ,\quad k\rightarrow \infty , $$

we show that $\lim_{k\rightarrow \infty }\mu _{k}=0$.

Suppose on the contrary that choosing a subsequence and relabeling if necessary, $\mu _{k}\geq M_{2}$ for some constant $M_{2}>0$. By (4.2), we obtain

$$ \lim_{k\rightarrow \infty }y_{k}(t)=\infty , \quad \text{uniformly on } t \in [0,T+2]_{Z}. $$

Consequently, we have

$$ \lim_{k\rightarrow \infty }\mu _{k} \frac{f(y_{k}(t))}{\varphi _{p}(y_{k}(t))} = \infty ,\quad \forall t\in [0,T+2]_{Z}. $$

Hence, we have from (4.5) that $y_{k}$ must change its sign on $[0,T+2]_{Z}$ for k large enough. However, this is impossible. Thus, $\lim_{k\rightarrow \infty }\mu _{k}=0$.

So, we prove that $C^{v}$ joins $(\frac{\lambda _{1}}{f_{0}},0)$ with $(0,\infty )$. □

Theorem 4.3

If $f_{0}=0$ and $f_{\infty }\in (0,+\infty )$, then (4.1) has at least two solutions $u^{+}$ and $u^{-}$ for $\lambda \in (\frac{\lambda _{1}}{f_{\infty }},+\infty )$, where $u^{+}$ is positive in $[0,T+1]_{Z}$ and $u^{-}$ is negative in $[0,T+1]_{Z}$ (see Fig. 3).

Figure 3

$f_{0}=0$, $f_{\infty}\in(0,+\infty)$

Proof

If $(\lambda ,u)$ is a solution of (4.1) and $\|u\|\neq 0$, dividing (4.1) by $\|u\|^{2(p-1)}$ and setting $\omega =\frac{u}{\|u\|^{2}}$, we obtain

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

(4.6)

Define

$$ \widetilde{f}(\omega )= \textstyle\begin{cases} \Vert \omega \Vert ^{2(p-1)}f(\frac{\omega }{ \Vert \omega \Vert ^{2}}),&\omega \neq 0, \\ 0,&\omega =0. \end{cases} $$

Clearly, (4.6) is equivalent to

$$ \textstyle\begin{cases} -\Delta [\varphi _{p}(\Delta \omega (t-1))]=\lambda a(t)\widetilde{f}( \omega (t)), \quad t\in [1,T+1]_{Z}, \\ \Delta \omega (0)=\omega (T+2)=0. \end{cases} $$

(4.7)

It is obvious that $(\lambda ,0)$ is always a solution of (4.7).

By simple calculation, we know $\widetilde{f_{0}}=f_{\infty }$, $\widetilde{f_{\infty }}=f_{0}$. By applying Theorem 4.1, we can get the conclusions of this theorem under the inversion $\omega \rightarrow \frac{\omega }{\|\omega \|^{2}}=u$. □

Theorem 4.4

If $f_{0}=+\infty$ and $f_{\infty }\in (0,+\infty )$, then (4.1) has at least two solutions $u^{+}$ and $u^{-}$ for $\lambda \in (0,\frac{\lambda _{1}}{f_{\infty }})$, where $u^{+}$ is positive in $[0,T+1]_{Z}$ and $u^{-}$ is negative in $[0,T+1]_{Z}$ (see Fig. 4).

Figure 4

$f_{0}=+\infty$, $f_{\infty}\in(0,+\infty)$

Proof

Using the conclusion of Theorem 4.2 and the method of Theorem 4.3, we can easily prove the conclusion of this theorem. □

Theorem 4.5

If $f_{0}=+\infty$ and $f_{\infty }=+\infty$, then there exists $\lambda ^{\ast }>0$ such that (4.1) has at least two solutions $u_{1}^{+}$ and $u_{2}^{+}$ for $\lambda \in (0,\lambda ^{\ast })$; in addition, $u_{1}^{+}$ and $u_{2}^{+}$ are positive in $[0,T+1]_{Z}$ (see Fig. 5(a)). Similarly, there exists $\lambda _{\ast }>0$ such that (4.1) has at least two solutions $u_{1}^{-}$ and $u_{2}^{-}$ for $\lambda \in (0,\lambda _{\ast })$; in addition, $u_{1}^{-}$ and $u_{2}^{-}$ are negative in $[0,T+1]_{Z}$ (see Fig. 5(b)).

Figure 5

$f_{0}=+\infty$, $f_{\infty}=+\infty$

Proof

Define

$$ f_{n}(s)= \textstyle\begin{cases} n\varphi _{p}(s), &s\in [- \frac{1}{n},\frac{1}{n}], \\ (f(\frac{2}{n})-\frac{1}{n^{p-2}})ns+\frac{2}{n^{p-2}}-f(\frac{2}{n}), &s\in (\frac{1}{n},\frac{2}{n}), \\ -(f(-\frac{2}{n})+\frac{1}{n^{p-2}})ns-\frac{2}{n^{p-2}}-f(- \frac{2}{n}), &s\in (-\frac{2}{n},-\frac{1}{n}), \\ f(s), &s\in (- \infty ,-\frac{2}{n}]\cup [\frac{2}{n},+\infty ). \end{cases} $$

(4.8)

Let us consider

$$ \textstyle\begin{cases} \Delta [\varphi _{p}(\Delta u(t-1))]+\lambda a(t)f_{n}(u(t))=0,\quad t \in [1,T+1]_{Z}, \\ \Delta u(0)=u(T+2)=0. \end{cases} $$

(4.9)

There are clearly $\lim_{n\rightarrow +\infty }f_{n}(s)=f(s)$, $(f_{n})_{0}=n$, and $(f_{n})_{\infty }=f_{\infty }=+\infty$. Theorem 4.2 implies that there exists a sequence of unbounded continuum $(C^{v})_{n}$ of solutions of problem (4.9) emanating from $(\frac{\lambda _{1}}{n},0)$ such that $(C^{v})_{n}$ joins $(\frac{\lambda _{1}}{n},0)$ to $(0,\infty )$.

Taking $z_{n}=(\frac{\lambda _{1}}{n},0)$ and $z^{\ast }=(0,0)$, we have that $z_{n}\rightarrow z^{\ast }$. So condition (i) in Lemma 4.2 is satisfied with $z^{\ast }=(0,0)$.

Obviously, $r_{n}=\sup \{\lambda +\|u\|: (\lambda ,u)\in (C^{v})_{n}\} \rightarrow \infty$, and accordingly, (ii) in Lemma 4.2 holds. (iii) can be deduced directly from the Arzéla–Ascoli theorem and the definition of $f_{n}$. Therefore, by Lemma 4.2, $\lim \sup_{n\rightarrow +\infty }(C^{v})_{n}$ contains unbounded connected components $C^{v}$ with

$$ (0,0)\in C^{v}\subset \lim \sup_{n\rightarrow +\infty } \bigl(C^{v} \bigr)_{n} \quad \text{and} \quad (0,\infty )\in C^{v}\subset \lim \sup_{n \rightarrow +\infty } \bigl(C^{v} \bigr)_{n}. $$

For any $\lambda _{0}>0$, it is easy to know that there is at most one $n_{0}$ such that $(\lambda _{0},0)\in (C^{v})_{n_{0}}$. By the definition of upper bound set, we obtain $(\lambda _{0},0)\in C^{v}\subset \lim \sup_{n\rightarrow + \infty }(C^{v})_{n}$. Naturally, $(\lambda _{0},0)\notin C^{v}$.

Consequently, $C^{v}\cap (\mathbb{R}\times \{0\})=\{(0,0)\}$. □

Theorem 4.6

If $f_{0}=0$ and $f_{\infty }=0$, then there exists $\lambda ^{r}>0$ such that (4.1) has at least two solutions $u_{1}^{+}$ and $u_{2}^{+}$ for $\lambda \in (\lambda ^{r},+\infty )$; in addition, $u_{1}^{+}$ and $u_{2}^{+}$ are positive in $[0,T+1]_{Z}$ (see Fig. 6(a)). Similarly, there exists $\lambda _{r}>0$ such that (4.1) has at least two solutions $u_{1}^{-}$ and $u_{2}^{-}$ for $\lambda \in (\lambda _{r},+\infty )$; in addition, $u_{1}^{-}$ and $u_{2}^{-}$ are negative in $[0,T+1]_{Z}$ (see Fig. 6(b)).

Figure 6

$f_{0}=0$, $f_{\infty}=0$

Proof

Define

$$ g_{n}(s)= \textstyle\begin{cases} \frac{1}{n}\varphi _{p}(s), &s\in [-\frac{1}{n},\frac{1}{n}], \\ (f(\frac{2}{n})-\frac{1}{n^{p}})ns+\frac{2}{n^{p}}-f(\frac{2}{n}), &s\in (\frac{1}{n},\frac{2}{n}), \\ -(f(-\frac{2}{n})+\frac{1}{n^{p}})ns-\frac{2}{n^{p}}-f(-\frac{2}{n}), &s\in (-\frac{2}{n},-\frac{1}{n}), \\ f(s), &s\in (- \infty ,-\frac{2}{n}]\cup [\frac{2}{n},+\infty ). \end{cases} $$

(4.10)

Let us consider

$$ \textstyle\begin{cases} \Delta [\varphi _{p}(\Delta u(t-1))]+\lambda a(t)g_{n}(u(t))=0,\quad t \in [1,T+1]_{Z}, \\ \Delta u(0)=u(T+2)=0. \end{cases} $$

Using the conclusion of Theorem 4.5 and the method of Theorem 4.3, we can easily prove the conclusion of this theorem. □

Theorem 4.7

If $f_{0}=0$ and $f_{\infty }=+\infty$, then (4.1) has at least two solutions $u^{+}$ and $u^{-}$ for $\lambda \in (0,+\infty )$, where $u^{+}$ is positive in $[0,T+1]_{Z}$ and $u^{-}$ is negative in $[0,T+1]_{Z}$ (see Fig. 7).

Figure 7

$f_{0}=0$, $f_{\infty}=+\infty$

Proof

Define $g_{n}$ as (4.10). Let us consider

$$ \textstyle\begin{cases} \Delta [\varphi _{p}(\Delta u(t-1))]+\lambda a(t)g_{n}(u(t))=0,\quad t \in [1,T+1]_{Z}, \\ \Delta u(0)=u(T+2)=0. \end{cases} $$

(4.11)

Clearly, $\lim_{n\rightarrow +\infty }g_{n}(s)=f(s)$, $(g_{n})_{0}= \frac{1}{n}$, and $(g_{n})_{\infty }=f_{\infty }=+\infty$. By Theorem 4.2, there exists a sequence of unbounded continuum $(C^{v})_{n}$ of solutions of (4.11) emanating from $(n\lambda _{1},0)$ such that $(C^{v})_{n}$ joins $(n\lambda _{1},0)$ to $(0,\infty )$.

Taking $z_{n}=(n\lambda _{1},0)$ and $z^{\ast }=(\infty ,0)$, we have that $z_{n}\rightarrow z^{\ast }$. So condition (i) in Lemma 4.2 is satisfied with $z^{\ast }=(\infty ,0)$.

Obviously, $r_{n}=\sup \{\lambda +\|u\|: (\lambda ,u)\in (C^{v})_{n}\} \rightarrow \infty$, and accordingly, (ii) in Lemma 4.2 holds.

(iii) can be deduced directly from the Arzéla–Ascoli theorem and the definition of $g_{n}$. Therefore, by Lemma 4.2, $\lim \sup_{n\rightarrow +\infty }(C^{v})_{n}$ contains unbounded connected components $C^{v}$ with

$$ (\infty ,0)\in C^{v}\subset \lim \sup_{n\rightarrow +\infty } \bigl(C^{v} \bigr)_{n} \quad \text{and} \quad (0,\infty )\in C^{v}\subset \lim \sup_{n \rightarrow +\infty } \bigl(C^{v} \bigr)_{n}. $$

 □

Theorem 4.8

If $f_{0}=+\infty$ and $f_{\infty }=0$, then (2) has at least two solutions $u^{+}$ and $u^{-}$ for $\lambda \in (0,+\infty )$ such that $u^{+}$ is positive in $[0,T+1]_{Z}$ and $u^{-}$ is negative in $[0,T+1]_{Z}$ (see Fig. 8).

Figure 8

$f_{0}=+\infty$, $f_{\infty}=0$

Proof

Define $f_{n}$ as (4.8). Let us consider

$$ \textstyle\begin{cases} \Delta [\varphi _{p}(\Delta u(t-1))]+\lambda a(t)f_{n}(u(t))=0,\quad t \in [1,T+1]_{Z}, \\ \Delta u(0)=u(T+2)=0. \end{cases} $$

Using the conclusion of Theorem 4.7 and the method of Theorem 4.3, we can easily prove the conclusion of this theorem. □