Variational approach to anti-periodic boundary value problems involving the discrete p-Laplacian

Abstract

Using critical point theory, we obtain the existence and multiplicity of nonzero solutions to anti-periodic boundary value problems with p-Laplacian in the case where the nonlinearities are p-sublinear at zero. Some examples are given to illustrate the results.

1 Introduction

Difference equations occur in many fields [1, 20], such as economics, discrete optimization, computer science. In the past decade, discrete p-Laplacian problems and difference equations have become a hot topic; see [11–19, 21, 22] and [25, 26]. Among the methods used are the method of upper and lower solutions, fixed point theory, Leray–Schauder degree, mountain pass lemma and the linking theorem. Recently, a lot of new results [5–11, 16, 23, 24] have been established by using variational methods.

In these last years, the existence and multiplicity of solutions for nonlinear discrete problems subject to various boundary value conditions have been widely studied by using different methods (see, e.g. [2–4] and [12–19, 21]). Bai et al. [2, 3] studied the second-order difference equations with Neumann boundary value conditions. D’Aguì et al. [16] investigated the existence of positive solutions for a discrete two point nonlinear boundary value problem with p-Laplacian in the case where the nonlinear term is p-sublinear at zero. However, little work has been done that has referred to anti-periodic boundary value problems with the discrete p-Laplacian operators in the case where the nonlinearities are p-sublinear at zero.

The idea of this paper comes from the method in [6, 9, 16]. One obtained two distinct critical points for functionals unbounded from below without p-superlinear nonlinearities at zero. The loss of p-superlinear condition at zero puts some critical points theorems cannot be immediately used. Therefore, In this paper, we mainly deal with the existence and multiplicity of solutions for anti-periodic boundary value problems

$$ \textstyle\begin{cases} -\Delta[a(k-1)\phi_{p}(\Delta u(k-1))]=\lambda f(k,u(k)), \\ u(0)=-u(T),\qquad u(1)=-u(T+1), \end{cases} $$

(1.1)

for $k \in[1,T]$, where $p>1$ is a fixed real number and $\phi _{p}(t)=|t|^{p-2}t$ for all $t \in R$. $a(k)> 0$ and $a(0)=a(T)$, $f:[1,T] \times R \to R$, is continuous and is p-sublinear at zero in the second variable for all $k \in[1,T]$. Moreover, Δ is the forward difference operator defined by $\Delta u(k)=u(k+1)-u(k)$, $\Delta^{2} u(k)=\Delta(\Delta u(k))$.

The rest of this paper is organized as follows. In Sect. 2, we establish the variational structure associated with (1.1), and provide some preliminary results. In Sect. 3, we state our main results and give examples. In Sect. 4, we provide the proofs of the main results.

2 Variational structure and some preliminaries

In this section, we establish a variational structure which reduces the existence of solutions for (1.1) to the existence of critical points of the corresponding functional.

Throughout this paper, we always assume that the following conditions are satisfied:

• (a) $a(k)>0$ for all $k\in[1,T]$ and $a(0)=a(T)$. Let ā and $a_{*}$ be the maximum and minimum of $\{a(k)\}$, respectively.

• (f) $f(k,u)$ is continuous in u and $F(k,u)=\int _{0}^{u}f(k,s)\,ds$ for $u \in R$ and $k \in[1,T]$.

We define the set E as

$$E=\bigl\{ u=\bigl\{ u(k)\bigr\} | u(T+1)=-u(1), u(k) \in R \text{ for } k \in[1,T+1] \bigr\} . $$

Then E is a vector space with $au+bv=\{au(k)+bv(k)\}$ for $u,v \in E$ and $a,b \in R$. Obviously, E is isomorphic to $R^{T}$ and hence E can be equipped with the norm $\|\cdot\|_{p}$ as

$$\Vert u \Vert _{p}= \Biggl(\sum_{k=1}^{T} \bigl\vert u(k) \bigr\vert ^{p} \Biggr)^{\frac{1}{p}} \quad\text{for } u \in E. $$

We also define norms $\|\cdot\|_{\infty}$ and $\|\cdot\|$ in E by

$$\Vert u \Vert _{\infty}=\max\bigl\{ \bigl\vert u(k) \bigr\vert :1 \leq k \leq T\bigr\} $$

and

$$\Vert u \Vert = \Biggl[a(T) \bigl\vert u(1)+u(T) \bigr\vert ^{p}+\sum_{k=1}^{T-1}a(k) \bigl\vert \Delta u(k) \bigr\vert ^{p} \Biggr]^{1/p} $$

respectively. Consider the functionals $\Phi(u)$, $\Psi(u)$ and $I_{\lambda}(u)$on E defined by

$$\begin{aligned}& \Phi(u)=\frac{a(T)}{p} \bigl\vert u(1)+u(T) \bigr\vert ^{p}+\sum_{k=1}^{T-1} \frac {a(k)}{p} \bigl\vert \Delta u(k) \bigr\vert ^{p}, \end{aligned}$$

(2.1)

$$\begin{aligned}& \Psi(u)=\sum_{k=1}^{T}F \bigl(k,u(k)\bigr) \end{aligned}$$

(2.2)

and

$$ \begin{aligned}[b] I_{\lambda}(u)&=\Phi(u)-\lambda\Psi(u)\\&=\frac {a(T)}{p} \bigl\vert u(1)+u(T) \bigr\vert ^{p}+\sum_{k=1}^{T-1} \frac{a(k)}{p} \bigl\vert \Delta u(k) \bigr\vert ^{p}-\lambda\sum _{k=1}^{T}F\bigl(k,u(k)\bigr).\end{aligned} $$

(2.3)

Then the partial derivatives of $\Phi(u)$ are given by

$$ \textstyle\begin{cases}\frac{ \partial\Phi(u)}{\partial u(1)}=-a(1)\phi _{p}(\Delta u(1))+a(T)\phi_{p}(u(1)+u(T)), \\ \frac{ \partial\Phi(u)}{\partial u(2)}=a(1)\phi_{p}(\Delta u(1))-a(2)\phi_{p}(\Delta u(2)), \\ \dots, \\ \frac{ \partial\Phi(u)}{\partial u(T-1)}=a(T-2)\phi_{p}(\Delta u(T-2))-a(T-1)\phi_{p}(\Delta u(T-1)), \\ \frac{ \partial\Phi(u)}{\partial u(T)}=a(T-1)\phi_{p}(\Delta u(T-1))+a(T)\phi_{p}(u(1)+u(T)). \end{cases} $$

(2.4)

This, combined with $a(0)=a(T)$, $u(0)=-u(T)$ and $u(1)=-u(T+1)$, gives us

$$ \textstyle\begin{cases}\frac{ \partial\Phi(u)}{\partial u(1)}=-\Delta [a(0)\phi_{p}(\Delta u(0))], \\ \frac{ \partial\Phi(u)}{\partial u(2)}=-\Delta[a(1)\phi _{p}(\Delta u(1))], \\ \dots, \\ \frac{ \partial\Phi(u)}{\partial u(T-1)}=-\Delta[a(T-2)\phi _{p}(\Delta u(T-2))], \\ \frac{ \partial\Phi(u)}{\partial u(T)}=-\Delta[a(T-1)\phi _{p}(\Delta u(T-1))]. \end{cases} $$

(2.5)

Then Φ has continuous Gâteaux derivatives in finite dimensional space and $\Phi\in C^{1}(E,R)$, the Fréchet derivative is given by

$$ \bigl\langle \Phi'(u),v \bigr\rangle =-\sum _{k=1}^{T}\Delta \bigl[a(k-1)\phi_{p}\bigl( \Delta u(k-1)\bigr) v(k) \bigr] $$

(2.6)

for $u,v \in E$. By direct computation, we have

$$ \bigl\langle \Phi'(u),v \bigr\rangle =\sum _{k=1}^{T} \bigl[a(k)\phi_{p}\bigl(\Delta u(k)\bigr) \bigr]\Delta v(k). $$

(2.7)

Similarly, we have $\Psi\in C^{1}(E,R)$. The Fréchet derivative is given by

$$ \bigl\langle \Psi'(u),v \bigr\rangle =\sum _{k=1}^{T} f\bigl(k,u(k)\bigr)v(k) $$

(2.8)

for $u,v \in E$. Therefore, $I_{\lambda} \in C^{1}(E,R)$, the Fréchet derivative is given by

$$\bigl\langle I_{\lambda}'(u),v \bigr\rangle =\bigl\langle \Phi'(u),v \bigr\rangle -\lambda\bigl\langle \Psi'(u),v \bigr\rangle . $$

The partial derivatives of $I_{\lambda}$ are given by

$$ \textstyle\begin{cases}\frac{ \partial I_{\lambda}}{\partial u(1)}=-\Delta [a(0)\phi_{p}(\Delta u(0))]-\lambda f(1,u(1)), \\ \frac{ \partial I_{\lambda}}{\partial u(2)}=-\Delta[a(1)\phi _{p}(\Delta u(1))]-\lambda f(2,u(2)), \\ \dots, \\ \frac{ \partial I_{\lambda}}{\partial u(T-1)}=-\Delta[a(T-2)\phi _{p}(\Delta u(T-2))]-\lambda f(T-1,u(T-1)), \\ \frac{ \partial I_{\lambda}}{\partial u(T)}=-\Delta[a(T-1)\phi _{p}(\Delta u(T-1))]-\lambda f(T,u(T)). \end{cases} $$

(2.9)

Equations (2.3) and (2.9) imply that a nonzero critical point of the functional $I_{\lambda}$ on E is a nontrivial solution of (1.1).

Definition 2.1

Let $I \in C^{1}(H,R)$. A sequence $\{x_{j}\} \subset H$ is called a Palais–Smale sequence (P.S. sequence) for I if $\{I(x_{j})\}$ is bounded and $I'(x_{j}) \to0$ as $j \to+\infty$. We say I satisfies the Palais–Smale condition (P.S. condition) if any P.S. sequence for I possesses a convergent subsequence.

Our main tool is taken from [9], which we recall here for the reader’s convenience.

Theorem 2.1

([9])

Let X be a real Banach space and let $\Phi,\Psi: X \to R$ be two functionals of class $C^{1}$ such that $\inf_{X}\Phi=\Phi(0)=\Psi (0)=0$. Assume that there are $r \in R$ and $u^{*} \in X$, with $0< \Phi (u^{*})<r$, such that

$$ \frac{1}{r}\sup_{u\in\Phi ^{-1}(-\infty,r]}\Psi(u)< \frac{\Psi(u^{*})}{\Phi(u^{*})}, $$

(2.10)

and for each

$$\lambda\in\Lambda= \biggl(\frac{\Phi(u^{*})}{\Psi(u^{*})},r \Bigl(\sup_{u\in\Phi^{-1}(-\infty,r]} \Psi(u) \Bigr)^{-1} \biggr) , $$

the functional $I_{\lambda}=\Phi-\lambda\Psi$ satisfies the P.S. condition and it is unbounded from below. Then for each $I_{\lambda}$ it admits at least two nonzero critical points $u_{\lambda1}$, $u_{\lambda2}$ such that $I_{\lambda}(u_{\lambda1})<0<I_{\lambda}(u_{\lambda2})$.

3 Main results and examples

Theorem 3.1

Assume that the conditions (a) and (f) hold. There exist two positive constants b and ρ such that

$$ F(k,u)\geq b \vert u \vert ^{p} \quad\textit{for } k \in [1,T] \textit{ and } \vert u \vert \geq\rho. $$

(3.1)

There also exist two positive constants $c_{*}$ and $d_{*}$ with

$$d_{*}< \biggl(\frac{a_{*}}{\bar{a}} \biggr)^{1/p} \biggl(\frac{1}{T} \biggr)^{1/q}c_{*} $$

such that

$$ \frac{pT^{p-1}}{a_{*}(2c_{*})^{p}} \sum_{k=1}^{T} \Bigl(\max_{|\xi|\leq c_{*}}F(k,\xi) \Bigr)< \min \Biggl\{ \frac{p}{\bar{a}(2d_{*})^{p}} \sum_{k=1}^{T} F(k,d_{*}), \frac{pb}{ \bar{a}2^{(p+1)}} \Biggr\} , $$

(3.2)

where $1/p+1/q=1$. Then, for each $\lambda\in\Lambda$ with

$$\Lambda= \Biggl( \max \Biggl\{ \frac{\bar{a}2^{(p+1)}}{pb},\frac{\bar {a}(2d_{*})^{p}}{p} \Biggl[ \sum _{k=1}^{T} F(k,d_{*}) \Biggr]^{-1} \Biggr\} , \frac{a_{*}(2c_{*})^{p}}{pT^{p-1}} \Biggl[ \sum_{k=1}^{T} \Bigl(\max_{|\xi |\leq c_{*}}F(k,\xi) \Bigr) \Biggr]^{-1} \Biggr), $$

(1.1) admits at least two nonzero solutions $u_{\lambda1}$, $u_{\lambda2}$ such that $I_{\lambda}(u_{\lambda1})<0<I_{\lambda }(u_{\lambda2})$.

Remark 3.1

If all the conditions of Theorem 3.1 are satisfied and $f(k,u)$ is odd in u for each $k \in[1,T]$, then (1.1) admits at least four nonzero solutions $\pm u_{\lambda1}$, $\pm u_{\lambda 2}$ such that $I_{\lambda}(-u_{\lambda1})=I_{\lambda}(u_{\lambda 1})<0<I_{\lambda}(u_{\lambda2})=I_{\lambda}(-u_{\lambda2})$.

Corollary 3.1

Assume that the conditions (a) and (f) hold. If $f(k,u)$ is odd in u for each $k \in[1,T]$, and

$$ \lim_{ \vert s \vert \to+\infty} \frac {F(k,s)}{ \vert s \vert ^{p}}=+\infty,\qquad \lim _{s \to0^{+}} \frac {F(k,s)}{s^{p}}=+\infty $$

(3.3)

for all $k \in[1,T]$, then, for each $\lambda\in\Lambda^{*}$ with

$$\Lambda^{*}= \Biggl(0,\frac{a_{*}(2c_{*})^{p}}{pT^{p-1}} \Biggl[ \sum _{k=1}^{T} \Bigl(\max_{|\xi|\leq c_{*}}F(k,\xi) \Bigr) \Biggr]^{-1} \Biggr), $$

(1.1) admits at least four nonzero solutions $\pm u_{\lambda1}$ and $\pm u_{\lambda2}$.

Example 3.1

Let $p=4$, $T=2$, $a(k)=k$ and

$$f(k,u)=u^{5}+u $$

for all $k \in[1,T]$. Then, for each $\lambda\in (0,3/8 )$, it is easy to check that all the conditions of Corollary 3.1 are satisfied, (1.1) admits at least four nonzero solutions.

Theorem 3.2

Assume that the conditions (a) and (f) hold, $T=2$, $a_{*}=\bar{a}=a$, $f(k,x)\geq0$ for all $x<0$, $k \in[1,T]$. Put

$$ L^{+}_{\infty}(k)=\lim_{s \to+\infty} \frac{F(k,s)}{s^{p}}, \qquad L^{+}_{\infty}=\min_{k \in[1,T]}L^{+}_{\infty}(k). $$

(3.4)

If $L^{+}_{\infty}>0$ and there exist two positive constants $c_{*}$ and $d_{*}$ with

$$d_{*}< \biggl(\frac{1}{2} \biggr)^{1/q}c_{*} $$

such that

$$ \frac{p}{2ac_{*}^{p}} \sum_{k=1}^{T} \Bigl(\max_{|\xi|\leq c_{*}}F(k,\xi ) \Bigr)< \min \Biggl\{ \frac{p}{a(2d_{*})^{p}} \sum_{k=1}^{T} F(k,d_{*}), \frac{pL^{+}_{\infty}}{a 2^{(p+1)}} \Biggr\} , $$

(3.5)

where $1/p+1/q=1$. Then, for each $\lambda\in\Lambda$ with

$$\Lambda= \Biggl( \max \Biggl\{ \frac{a2^{(p+1)}}{pL^{+}_{\infty}},\frac {a(2d_{*})^{p}}{p} \Biggl[ \sum _{k=1}^{T} F(k,d_{*}) \Biggr]^{-1} \Biggr\} , \frac{2ac_{*}^{p}}{p} \Biggl[ \sum_{k=1}^{T} \Bigl(\max_{|\xi|\leq c_{*}}F(k,\xi) \Bigr) \Biggr]^{-1} \Biggr), $$

(1.1) admits at least two nonzero solutions $u_{\lambda1}$, $u_{\lambda2}$ such that $I_{\lambda}(u_{\lambda1})<0<I_{\lambda }(u_{\lambda2})$.

Corollary 3.2

Assume that the conditions (a) and (f) hold, $T=2$, $a_{*}=\bar{a}=a$, $f(k,x)\geq0$ for all $x<0$, $k \in[1,T]$. If

$$ \lim_{s \to+\infty} \frac {F(k,s)}{s^{p}}=+\infty\quad\textit{and}\quad \lim_{s \to0^{+}} \frac {F(k,s)}{s^{p}}=+\infty, $$

(3.6)

for all $k \in[1,T]$, then, for each $\lambda\in\Lambda^{*}$ with

$$\Lambda^{*}= \Biggl(0,\frac{2ac_{*}^{p}}{p} \Biggl[ \sum _{k=1}^{T} \Bigl(\max_{|\xi|\leq c_{*}}F(k,\xi) \Bigr) \Biggr]^{-1} \Biggr), $$

(1.1) admits at least two nonzero solutions.

Example 3.2

Let $p=2$, $T=2$, $a(k)=4$ and

$$f(k,u)=e^{u} $$

for $k \in[1,T]$. Then, for each $\lambda\in (0,\frac{2}{e-1} )$, it is easy to check that all the conditions of Corollary 3.2 are satisfied, (1.1) admits at least two nonzero solutions.

4 Proofs of main results

In order to prove Theorem 3.1, we need the following lemmas.

Lemma 4.1

If $u \in E$ and $p>1$, then

$$\frac{a_{*}}{p} \biggl(\frac{2}{T} \biggr)^{p} \sum _{k=1}^{T} \bigl\vert u(k) \bigr\vert ^{p} \leq\Phi(u) \leq\frac{\bar{a}2^{(p+1)}}{p}\sum _{k=1}^{T} \bigl\vert u(k) \bigr\vert ^{p} $$

and

$$\frac{2}{T} a_{*}^{1/p} \Vert u \Vert _{p} \leq \Vert u \Vert \leq2(2\bar{a})^{1/p} \Vert u \Vert _{p}. $$

Proof

On the one hand,

$$ \begin{aligned}[b] \Phi(u)&= \frac{a(T)}{p} \bigl\vert u(1)+u(T) \bigr\vert ^{p}+\sum_{k=1}^{T-1} \frac {a(k)}{p} \bigl\vert \Delta u(k) \bigr\vert ^{p} \\ &\leq\frac{\bar{a}}{p} \Biggl[ 2^{p} \bigl( \bigl\vert u(1) \bigr\vert ^{p}+ \bigl\vert u(T) \bigr\vert ^{p} \bigr)+\sum _{k=1}^{T-1}2^{p} \bigl( \bigl\vert u(k) \bigr\vert ^{p}+ \bigl\vert u(k+1) \bigr\vert ^{p} \bigr) \Biggr] \\ &\leq\frac{\bar{a}}{p}2^{p}\sum_{k=1}^{T}2 \bigl\vert u(k) \bigr\vert ^{p} = \frac{\bar{a}2^{(p+1)}}{p}\sum_{k=1}^{T} \bigl\vert u(k) \bigr\vert ^{p}. \end{aligned} $$

(4.1)

On the other hand, $u(1)=-u(T+1)$, for each $k \in[1,T]$,

$$ \begin{aligned}[b] 2u(k)&=u(2)-u(1) +\cdots+u(k)-u(k-1)+u(k)-u(k+1)+ \cdots+u(T)-u(T+1) \\ &\leq \bigl\vert u(2)-u(1) \bigr\vert +\cdots+ \bigl\vert u(k)-u(k-1) \bigr\vert + \bigl\vert u(k)-u(k+1) \bigr\vert +\cdots + \bigl\vert u(T)+u(1) \bigr\vert \\ &= \bigl\vert u(2)-u(1) \bigr\vert +\cdots+ \bigl\vert u(k)-u(k-1) \bigr\vert + \bigl\vert u(k+1)-u(k) \bigr\vert +\cdots + \bigl\vert u(T)+u(1) \bigr\vert \\ &\leq \Biggl[ \bigl\vert u(T)+u(1) \bigr\vert ^{p}+\sum _{k=1}^{T-1} \bigl\vert \Delta u(k) \bigr\vert ^{p} \Biggr]^{1/p}T^{1/q}, \end{aligned} $$

(4.2)

where $1/p+1/q=1$, that is,

$$ \Vert u \Vert _{\infty} \leq\frac{1}{2} \Biggl[ \bigl\vert u(T)+u(1) \bigr\vert ^{p}+\sum _{k=1}^{T-1} \bigl\vert \Delta u(k) \bigr\vert ^{p} \Biggr]^{1/p}T^{1/q}. $$

(4.3)

Since

$$\sum_{k=1}^{T} \bigl\vert u(k) \bigr\vert ^{p} \leq T \bigl( \Vert u \Vert _{\infty} \bigr)^{p}, $$

this, combined with (4.3), gives us

$$\sum_{k=1}^{T} \bigl\vert u(k) \bigr\vert ^{p} \leq \biggl(\frac{T}{2} \biggr)^{p} \Biggl[ \bigl\vert u(T)+u(1) \bigr\vert ^{p}+\sum _{k=1}^{T-1} \bigl\vert \Delta u(k) \bigr\vert ^{p} \Biggr] $$

and

$$\Phi(u)\geq\frac{a_{*}}{p} \Biggl[ \bigl\vert u(T)+u(1) \bigr\vert ^{p}+\sum_{k=1}^{T-1} \bigl\vert \Delta u(k) \bigr\vert ^{p} \Biggr] \geq\frac{a_{*}}{p} \biggl( \frac {2}{T} \biggr)^{p} \sum_{k=1}^{T} \bigl\vert u(k) \bigr\vert ^{p}. $$

The proof is complete. □

Lemma 4.2

If the condition (3.1) holds, then the functional $I_{\lambda}$ satisfies the P.S. condition and it is unbounded from below for all $\lambda\in (\frac{\bar{a} 2^{(p+1)}}{pb},+\infty )$.

Proof

Let $\{I_{\lambda}(u_{j})\}$ be a bounded sequence and $\{ u_{j}\}$ be a sequence in E, i.e., there exists a positive constant M such that

$$\bigl\vert I_{ \lambda}(u_{j}) \bigr\vert \leq M \quad\text{for } j \in Z^{+}. $$

Let

$$M_{\rho}=\max_{1\leq k \leq T}\bigl\{ \bigl\vert F(k,u)- b \vert u \vert ^{p} \bigr\vert : \vert u \vert \leq\rho\bigr\} . $$

It is easy to check that

$$ F(k,u)\geq b \vert u \vert ^{p} -M_{\rho}\quad \text{for } k \in[1,T] \text{ and } u \in R. $$

(4.4)

By (4.4) and (4.1), we have

$$ I_{\lambda}(u_{j})=\Phi(u_{j})-\lambda \Psi (u_{j})\leq \biggl(\frac{\bar{a}2^{(p+1)}}{p}- \lambda b \biggr) \Vert u_{j} \Vert ^{p}_{p} +T\lambda M_{\rho} $$

(4.5)

for $j \in Z^{+}$. Now, we claim $\{u_{j}\}$ is bounded. In fact, $\|u_{j}\| \to+\infty$, $\|u_{j}\|_{p} \to+\infty$ and $\frac{\bar {a}2^{(p+1)}}{p}- \lambda b<0$, one has $I_{\lambda}(u_{j}) \to-\infty$ and this is absurd. Hence, $I_{\lambda}$ satisfies the P.S. condition. Next, we prove that $I_{\lambda}$ is unbounded from below. By (4.5), we have $I_{\lambda}(u_{n}) \to-\infty$ as $\|u_{n}\| \to +\infty$. □

Proof of Theorem 3.1

Put Φ and Ψ as in (2.1) and (2.2), it is easily checked that Φ and Ψ satisfy all regularity assumptions required in Theorem 2.1. So, our end is to verify condition (2.10) in Theorem 2.1. Let $u \in\Phi^{-1}(-\infty,r]$; this means that

$$\frac{a_{*}}{p} \Biggl[ \bigl\vert u(T)+u(1) \bigr\vert ^{p}+ \sum_{k=1}^{T-1} \bigl\vert \Delta u(k) \bigr\vert ^{p} \Biggr] \leq\frac{a(T)}{p} \bigl\vert u(1)+u(T) \bigr\vert ^{p}+\sum_{k=1}^{T-1} \frac {a(k)}{p} \bigl\vert \Delta u(k) \bigr\vert ^{p} \leq r, $$

this, combined with (4.2), gives us

$$\bigl\vert u(k) \bigr\vert \leq\frac{1}{2} \biggl(\frac{rp}{a_{*}} \biggr)^{1/p}T^{1/q} $$

for $k \in[1,T]$. Let

$$c_{*}=\frac{1}{2} \biggl(\frac{rp}{a_{*}} \biggr)^{1/p}T^{1/q}, $$

then

$$r=\frac{a_{*}(2c_{*})^{p}}{pT^{p-1}} $$

and

$$ \begin{aligned}[b] \frac{1}{r}\sup _{u\in\Phi^{-1}(-\infty,r]}\Psi(u)& \leq\frac {1}{r}\sum _{k=1}^{T}\max_{|\xi|\leq c_{*}}F(k,\xi) \\ &=\frac{pT^{p-1}}{a_{*}(2c_{*})^{p}} \sum_{k=1}^{T}\max _{|\xi|\leq c_{*}}F(k,\xi). \end{aligned} $$

(4.6)

Now, we define $u_{*} \in E$ by $u_{*}=\{u_{*}(k)\}$ and

$$u_{*}(k)=d_{*}< \biggl(\frac{a_{*}}{\bar{a}} \biggr)^{1/p} \biggl( \frac {1}{T} \biggr)^{1/q}c_{*} $$

for $k \in[1,T]$. It is easy to check that $\Phi(u_{*})< r$ and

$$\frac{\Psi(u_{*})}{\Phi(u_{*})} \geq\frac{p}{\bar{a}(2d_{*})^{p}}\sum_{k=1}^{T}F(k,d_{*}). $$

This, combined with (4.6) and (3.2), produces at once (2.10). Therefore, Theorem 2.1 ensures that (1.1) has at least two nonzero critical points $u_{\lambda1}$ and $u_{\lambda2}$. The proof is complete. □

Proof of Theorem 3.2

Let $\{u_{j}\}$ be a sequence in E such that $\{I_{\lambda}(u_{j})\}$ is bounded and $I'_{\lambda}(u_{j}) \to0$ as $j \to+\infty$. Put $u^{+}_{j}(k)=\max\{u_{j}(k),0\}$ and $u^{-}_{j}(k)=\max\{-u_{j}(k),0\}$ for all $k \in[1,T]$, then $u^{+}_{j}=\{u^{+}_{j}(k)\}$ and $u^{-}_{j}=\{u^{-}_{j}(k)\}$ for $k \in[1,T]$. Therefore, $u_{j}=u^{+}_{j}-u^{-}_{j}$ for all $j \in Z^{+}$. Considering that $L^{+}_{\infty}>0$ and $\lambda\in (\frac {a2^{(p+1)}}{pL^{+}_{\infty}},+\infty )$, we fix $\lambda> \frac {a2^{(p+1)}}{pL^{+}_{\infty}}$ and fix l such that $L^{+}_{\infty} > l>\frac{a2^{(p+1)}}{p \lambda}$. Now, we claim $\{u^{-}_{j}\}$ is bounded. By direct computation, we have

$$ \begin{aligned}[b] \bigl\Vert u^{-}_{j} \bigr\Vert ^{p}&=a \bigl\vert u^{-}_{j}(1)+u^{-}_{j}(T) \bigr\vert ^{p}+\sum_{k=1}^{T-1}a \bigl\vert \Delta u_{j}^{-}(k) \bigr\vert ^{p} \\ &=a \bigl\vert -u^{-}_{j}(T+1)+u^{-}_{j}(T) \bigr\vert ^{p}+\sum_{k=1}^{T-1}a \bigl\vert \Delta u_{j}^{-}(k) \bigr\vert ^{p} \\ &\leq-\sum_{k=1}^{T}a \bigl[ \phi_{p}\bigl(\Delta u_{j}(k)\bigr) \Delta u_{j}^{-}(k) \bigr] \\ &=-\bigl\langle \Phi'( u_{j}), u^{-}_{j} \bigr\rangle , \end{aligned} $$

(4.7)

where $u_{j}^{-}(T+1)=-u_{j}^{-}(1)$, for all $j \in Z^{+}$. Moreover, by definition of $u^{-}_{j}$ and since $f(k,x)\geq0$ for all $x<0$, we have

$$ \begin{aligned}[b] \bigl\Vert u^{-}_{j} \bigr\Vert ^{p}&\leq-\bigl\langle \Phi'( u_{j}), u^{-}_{j} \bigr\rangle + \lambda\sum_{k=1}^{T}f \bigl(k,u_{j}(k)\bigr)u^{-}_{j}(k) \\ &= - \bigl\langle \Phi'( u_{j}), u^{-}_{j} \bigr\rangle + \lambda\bigl\langle \Psi'( u_{j}), u^{-}_{j} \bigr\rangle \\ &=-\bigl\langle I'_{\lambda}( u_{j}), u^{-}_{j} \bigr\rangle \end{aligned} $$

(4.8)

for all $j \in Z^{+}$. This, combined with the formulas

$$\lim_{j \to+\infty}I'_{\lambda}( u_{j})=0,\qquad \lim_{j \to+\infty} \frac{-\langle I'_{\lambda}( u_{j}), u^{-}_{j} \rangle}{ \Vert u^{-}_{j} \Vert }=0, $$

gives us

$$\lim_{j \to+\infty} \bigl\Vert u^{-}_{j} \bigr\Vert =0. $$

Hence, our claim is proved. Therefore, there exists $Q>0$ such that $\| u^{-}_{j}\|\leq Q$ for all $j \in Z^{+}$. Using a similar argument to (4.2) produces at once

$$ \bigl\vert u^{-}_{j}(k) \bigr\vert \leq \frac{Q}{a} \biggl(\frac {1}{2} \biggr)^{1/p}=L $$

(4.9)

for all $k \in[1,T]$ and $j \in Z^{+}$.

Now, arguing by contradiction, assume that $\{u_{j}\}$ is unbounded, that is, $\{u^{+}_{j}\}$ is unbounded. From

$$\lim_{s \to+\infty} \frac{F(k,s)}{s^{p}}\geq L^{+}_{\infty} > l, $$

there exists $\delta>0$ such that $F(k,s)>l s^{p}$ for all $s> \delta$ and $k \in[1,T]$. Let

$$M_{\delta}=\max_{1\leq k \leq T}\bigl\{ \bigl\vert F(k,u)-l \vert u \vert ^{p} \bigr\vert :-L \leq u\leq\delta\bigr\} . $$

Then it is easy to check that

$$ F\bigl(k,u_{j}(k)\bigr)\geq l \bigl\vert u_{j}(k) \bigr\vert ^{p} -M_{\delta}\quad \text{for } k \in[1,T] \text{ and } j\in Z^{+}. $$

(4.10)

This, combined with (4.1), gives us

$$I_{\lambda}(u_{j})=\frac{1}{p} \Vert u_{j} \Vert ^{p}-\lambda\Psi(u_{j})\leq\frac {a2^{(p+1)}}{p} \Vert u_{j} \Vert ^{p}_{p}-\lambda l \Vert u_{j} \Vert ^{p}_{p}+ T\lambda M_{\delta}, $$

that is,

$$I_{\lambda}(u_{j}) \leq \biggl(\frac{a2^{(p+1)}}{p}- \lambda l \biggr) \Vert u_{j} \Vert ^{p}_{p} +T\lambda M_{\delta} $$

for $j \in Z^{+}$. Since $\|u_{j}\| \to+\infty$ and $\frac{\bar {a}2^{(p+1)}}{p}- \lambda l<0$, one has $I_{\lambda}(u_{j}) \to-\infty$ and this is absurd. Hence $I_{\lambda}$ satisfies the P.S. condition.

Finally, we prove that $I_{\lambda}$ is unbounded from below. Arguing as before, we have $I_{\lambda}(u_{n}) \to-\infty$ as $\|u_{n}\| \to +\infty$. The rest of the proof is similar to Theorem 3.1 and is omitted. The proof is complete. □