In this section, we give our main results. Let
$$ \mu =\limsup_{x\rightarrow +\infty} \frac{\sum_{k=1}^{T}F(k,x)}{x^{2}} $$
(3.1)
and
$$ p_{*}=\min \bigl\{ p_{k},k\in \mathbb{Z}(1,T)\bigr\} ,\qquad p^{*}=\max \bigl\{ p_{k},k\in \mathbb{Z}(1,T)\bigr\} . $$
We have the following result.
Theorem 3.1
Suppose that there are two real sequences \(\{\omega _{n}\}\), \(\{c_{n}\}\) with \(\omega _{n}>0\) and \(\lim_{n\rightarrow +\infty}\omega _{n}=+\infty \) such that
$$ \frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T})c_{n}^{2}< \frac{16p_{*}\omega _{n}^{2}}{(T+1)^{2}(T+3)}\quad \textit{for }n\in \mathbb{Z}(1) $$
(3.2)
and
$$ \rho < \frac{2\mu}{p_{-1}+p_{0}+p_{T-1}+p_{T}}, $$
(3.3)
where
$$ \rho =\liminf_{n\rightarrow \infty} \frac{\sum_{k=1}^{T}\max_{|x|\leq \omega _{n}}F(k,x) -\sum_{k=1}^{T}F(k,c_{n})}{ \frac{16p_{*}\omega _{n}^{2}}{(T+1)^{2}(T+3)}-\frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T})c_{n}^{2}}. $$
Then, for each \(\lambda \in (\frac{p_{-1}+p_{0}+p_{T-1}+p_{T}}{2\mu}, \frac{1}{\rho} )\), problem (1.1) with (1.2) admits an unbounded sequence of solutions.
Proof
It is obvious that
$$ \lim_{ \Vert u \Vert \rightarrow +\infty} \Phi (u)=\lim_{ \Vert u \Vert \rightarrow + \infty} \frac{1}{2} \sum_{k=0}^{T+1}p_{k-1} \bigl(\Delta ^{2}u_{k-1}\bigr)^{2} \geq \lim _{ \Vert u \Vert \rightarrow +\infty}\frac{p_{*}}{2} \Vert u \Vert ^{2} =+ \infty , $$
which means that \(\Phi (u)\) is coercive.
Define
$$ r_{n}=\frac{16p_{*}\omega _{n}^{2}}{(T+1)^{2}(T+3)}. $$
If \(u\in E\) and \(\Phi (u)< r_{n}\), then we have the following inequality:
$$ \frac{1}{2}p_{*}\sum_{k=0}^{T+1} \bigl(\Delta ^{2}u_{k-1}\bigr)^{2}< r_{n}. $$
Considering Lemma 2.2, for any \(k\in \mathbb{Z}(1,T)\), we have
$$ \vert u_{k} \vert ^{2}\leq \frac{(T+1)^{2}(T+3) }{32} \sum_{k=0}^{T+1}\bigl( \Delta ^{2}u_{k-1}\bigr)^{2}< \omega _{n}^{2}. $$
Furthermore, according to the definition of ϕ, we have
$$ \phi (r_{n})\leq \inf_{u\in \Phi ^{-1}(-\infty ,r_{n})} \frac{\sum_{k=1}^{T}\max_{|x|\leq \omega _{n}}F(k,x)-\sum_{k=1}^{T}F(k,u_{k})}{ \frac{16p_{*}\omega _{n}^{2}}{(T+1)^{2}(T+3)}-\Phi (u)}. $$
(3.4)
For any \(n\in \mathbb{Z}(1)\), take \((q_{n})_{k}=c_{n}\) for \(k\in \mathbb{Z}(1,T)\) and \((q_{n})_{-1}=(q_{n})_{0}=(q_{n})_{T}=(q_{n})_{T+1}=0\), then \(q_{n}\in E\) and
$$ \Phi (q_{n})=\frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T})c_{n}^{2} \leq r_{n} $$
by exploiting (3.2). Therefore, from (3.4), we have
$$ \begin{aligned} \phi (r_{n})&\leq \frac{\sum _{k=1}^{T}\max _{|x|\leq \omega _{n}}F(k,x) -\sum _{k=1}^{T}F(k,(q_{n})_{k})}{ \frac{16p_{*}\omega _{n}^{2}}{(T+1)^{2}(T+3)}-\Phi (q_{n})} \\ &= \frac{\sum _{k=1}^{T}\max _{|x|\leq \omega _{n}}F(k,x) -\sum _{k=1}^{T}F(k,c_{n})}{ \frac{16p_{*}\omega _{n}^{2}}{(T+1)^{2}(T+3)}-\frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T})c_{n}^{2}}. \end{aligned} $$
Moreover, combining (3.3), it is clear that \(\alpha \leq \liminf_{n\rightarrow +\infty}\phi (r_{n})\leq \rho <+ \infty \).
We assert that \(J_{\lambda}\) is unbounded from below. In fact, when \(\mu <+\infty \), since
$$ 2\lambda \mu >p_{-1}+p_{0}+p_{T-1}+p_{T}, $$
there exists \(\varepsilon _{0}>0\) such that
$$ 2\lambda (\mu -\varepsilon _{0})>p_{-1}+p_{0}+p_{T-1}+p_{T}. $$
From (3.1), we know that there exists a positive sequence \(\{a_{n}\} \) with \(\lim_{n\rightarrow +\infty}a_{n}=+\infty \) such that
$$ \sum_{k=1}^{T}F(k,a_{n}) \geq (\mu -\varepsilon _{0})a_{n}^{2}. $$
For each \(n\in \mathbb{Z}(1)\), define \(\upsilon _{n}\in E\) with \((\upsilon _{n})_{k}=a_{n}\) for \(k\in \mathbb{Z}(1,T)\), then we have the following inequality:
$$ \begin{aligned}[b] J_{\lambda}(\upsilon _{n})&=\frac{1}{2}\sum _{k=0}^{T+1}p_{k-1} \bigl( \Delta ^{2}(\upsilon _{n})_{k-1} \bigr)^{2} -\lambda \sum _{k=1}^{T} F \bigl(k,(\upsilon _{n})_{k}\bigr) \\ &\leq \frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T})a_{n}^{2}- \lambda (\mu - \varepsilon _{0})a_{n}^{2} \\ &=\frac{1}{2} \bigl(p_{-1}+p_{0}+p_{T-1}+p_{T}-2 \lambda (\mu - \varepsilon _{0}) \bigr)a_{n}^{2}. \end{aligned} $$
(3.5)
The above inequality implies \(\lim_{n\rightarrow +\infty}J_{\lambda}(\upsilon _{n})=-\infty \). If \(\mu =+\infty \), it can be seen that there is a sequence of positive number \(\{\bar{a}_{n}\}\) with \(\lim_{n\rightarrow +\infty}\bar{a}_{n}=+\infty \) such that
$$ \sum_{k=1}^{T}F(k,\bar{a}_{n}) \geq \frac{p_{-1}+p_{0}+p_{T-1}+p_{T}}{\lambda}\bar{a}_{n}^{2} $$
from the definition of μ. Define \(\bar{\upsilon}_{n}\in E\) as \((\bar{\upsilon}_{n})_{k}=\bar{a}_{n}\) for \(k\in \mathbb{Z}(1,T)\), then
$$ \begin{aligned}[b] J_{\lambda}(\bar{ \upsilon}_{n})&=\frac{1}{2}\sum _{k=0}^{T+1}p_{k-1}\bigl(\Delta ^{2}( \bar{\upsilon}_{n})_{k-1}\bigr)^{2} -\lambda \sum _{k=1}^{T} F\bigl(k,(\bar{ \upsilon}_{n})_{k}\bigr) \\ &\leq -\frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T}) \bar{a}_{n}^{2} \rightarrow -\infty \quad \text{as }n \rightarrow +\infty .\end{aligned} $$
(3.6)
By combining (3.5) with (3.6), we can conclude that condition (\(I_{1}\)) of Lemma 2.1 does not hold. Therefore, the functional \(J_{\lambda}\) has a sequence of critical points with \(\lim_{n\rightarrow +\infty}\Phi (u_{n})=+\infty \), which means that the problem (1.1) with (1.2) admits an unbounded sequence of solutions. □
Corollary 3.2
If there is a sequence of positive numbers \(\{\tilde{\omega}_{n}\}\) with \(\tilde{\omega}_{n}\rightarrow +\infty \) as \(n\rightarrow +\infty \) such that
$$ \tilde{\rho}< \frac{2\mu}{p_{-1}+p_{0}+p_{T-1}+p_{T}}, $$
(3.7)
where
$$ \tilde{\rho}=\liminf_{n\rightarrow \infty} \frac{(T+1)^{2}(T+3)\sum_{k=1}^{T}\max_{|x|\leq \tilde{\omega}_{n}}F(k,x) }{16p_{*}\tilde{\omega}_{n}^{2}},$$
then, for each \(\lambda \in (\frac{p_{-1}+p_{0}+p_{T-1}+p_{T}}{2\mu}, \frac{1}{\tilde{\rho}} )\), problem (1.1) with (1.2) admits an unbounded sequence of nontrivial solutions.
Proof
Taking \(c_{n}=0\) for all \(n\in \mathbb{Z}(1)\), it can be easily proved by Theorem 3.1. □
In particular, if the nonlinear function f in (1.1) with the form \(f(k,u)=a_{k}g(u)\), where \(a_{k}>0\) for \(k\in \mathbb{Z}(1,T)\), and \(p_{k}\equiv 1\) for \(k\in \mathbb{Z}(-1,T)\). Then (1.1) reads
$$ \triangle ^{2} \bigl( p_{k-2} \triangle ^{2} u_{k-2} \bigr)= \lambda a_{k}g(u_{k}), \quad k \in \mathbb{Z}(1,T). $$
(3.8)
Define
$$ \bar{\mu}=\limsup_{x\rightarrow +\infty}\frac{\bar{G}(x)}{x^{2}}, $$
where
$$ \bar{G}(x)= \int _{0}^{x}g(s)\,ds. $$
Then we have the following.
Corollary 3.3
Suppose that there are two real sequences \(\{\bar{\omega}_{n}\}\), \(\{\bar{c}_{n}\}\) with \(\bar{\omega}_{n}>0\) and \(\lim_{n\rightarrow +\infty}\bar{\omega}_{n}=+\infty \) such that
$$ \frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T}) \bar{c}_{n}^{2}< \frac{16p_{*}\bar{\omega}_{n}^{2}}{(T+1)^{2}(T+3)}\quad \textit{for }n \in \mathbb{Z}(1) $$
(3.9)
and
$$ \bar{\rho}< \frac{2\bar{\mu}}{p_{-1}+p_{0}+p_{T-1}+p_{T}}, $$
(3.10)
where
$$ \bar{\rho}=\liminf_{n\rightarrow \infty} \frac{\max_{|x|\leq \bar{\omega}_{n}}\bar{G}(x) -\bar{G}(\bar{c}_{n})}{ \frac{16p_{*}\bar{\omega}_{n}^{2}}{(T+1)^{2}(T+3)}-\frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T})\bar{c}_{n}^{2}}. $$
Then, for each \(\lambda \in \frac{1}{\sum_{k=1}^{T}a_{k}} (\frac{p_{-1}+p_{0}+p_{T-1}+p_{T}}{2\bar{\mu}} ,\frac{1}{\bar{\rho}})\), problem (3.8) with (1.2) admits an unbounded sequence of nontrivial solutions.
Now, we discuss the existence of infinitely many solutions to the boundary value problem (1.1) with (1.2) by using the oscillatory behavior of the nonlinear function at the origin.
Theorem 3.4
Suppose that there are two real sequences \(\{z_{n}\}\) and \(\{\bar{z}_{n}\}\), where \(\bar{z}_{n}>0\) and \(\lim_{n\rightarrow +\infty}\bar{z}_{n}=0\), such that
$$ \frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T})z_{n}^{2}< \frac{16p_{*}\bar{z}_{n}^{2}}{(T+1)^{2}(T+3)}~{\mathrm{for}}~n\in \mathbb{Z}(1) $$
(3.11)
and
$$ \varrho < \frac{2\mu}{p_{-1}+p_{0}+p_{T-1}+p_{T}}, $$
(3.12)
where
$$ \varrho =\liminf_{n\rightarrow \infty} \frac{\sum_{k=1}^{T}\max_{|x|\leq \bar{z}_{n}}F(k,x) -\sum_{k=1}^{T}F(k,z_{n})}{ \frac{16p_{*}\bar{z}_{n}^{2}}{(T+1)^{2}(T+3)}-\frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T})z_{n}^{2}}. $$
Then, for each \(\lambda \in ( \frac{p_{-1}+p_{0}+p_{T-1}+p_{T}}{2\mu},\frac{1}{\varrho}) \), problem (1.1) with (1.2) has a sequence of nontrivial solutions that converges to 0.
The proof of Theorem 3.4 is similar to that of Theorem 3.1, so we omit it.
Corollary 3.5
Suppose that there is a sequence \(\{\tilde{z}_{n}\}\) where \(\tilde{z}_{n}>0\) and \(\lim_{n\rightarrow +\infty}\tilde{z}_{n}=0\) such that
$$ \bar{\varrho}< \frac{2\mu}{p_{-1}+p_{0}+p_{T-1}+p_{T}}, $$
(3.13)
where
$$ \bar{\varrho}=\liminf_{n\rightarrow \infty} \frac{(T+1)^{2}(T+3) \sum_{k=1}^{T}\max_{|x|\leq \tilde{z}_{n}}F(k,x)}{ 16p_{*}\tilde{z}_{n}^{2}}.$$
Then, for each \(\lambda \in (\frac{p_{-1}+p_{0}+p_{T-1}+p_{T}}{2\mu}, \frac{1}{\bar{\varrho}} )\), problem (1.1) with (1.2) has a sequence of nontrivial solutions that converges to 0.
Considering the boundary value problem (3.8) with (1.2), we have the following result when the nonlinear function g oscillates at the origin.
Corollary 3.6
Suppose there are two real sequences \(\{b_{n}\}\), \(\{\bar{b}_{n}\}\) with \(\bar{b}_{n}>0\) and \(\lim_{n\rightarrow +\infty}\bar{b}_{n}=0\) such that
$$ \frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T})b_{n}^{2}< \frac{16p_{*}\bar{b}_{n}^{2}}{(T+1)^{2}(T+3)}\quad \textit{for } n\in \mathbb{Z}(1) $$
(3.14)
and
$$ {\sigma}< \frac{2\bar{\mu}}{p_{-1}+p_{0}+p_{T-1}+p_{T}}, $$
(3.15)
where
$$ {\sigma}=\liminf_{n\rightarrow \infty} \frac{\max_{|x|\leq \bar{b}_{n}}\bar{G}(x) -\bar{G}(b_{n})}{ \frac{16p_{*}\bar{b}_{n}^{2}}{(T+1)^{2}(T+3)}-\frac{1}{2}(p_{-1}+p_{0}+p_{T-1}+p_{T})b_{n}^{2}}. $$
Then, for each \(\lambda \in \frac{1}{\sum_{k=1}^{T}a_{k}} ( \frac{p_{-1}+p_{0}+p_{T-1}+p_{T}}{2\bar{\mu}} ,\frac{1}{{\sigma}} )\), problem (3.8) with (1.2) admits a sequence of nontrivial solutions that converges to 0.