We divide the proof into eight steps.

Step 1: We introduce some notations.

• For \(a, b \in Z\), define \(Z [a] = \{a, a+1, \ldots \}\), \(Z [a, b] = \{a, a+1, \ldots , b\}\) for \(a\leq b\).

• Let the set of sequences \(S= \{u=\{u_{n}\}= (\ldots , u_{-n}, \ldots , u_{0}, \ldots , u_{n}, \ldots), u_{n} \in R, n \in Z \}\). For any given positive integer *M*, \(E_{M}\) is defined as a subspace of *S* by

$$ E_{M} = \bigl\{ u=\{u_{n}\} \in S \mid u_{n+M} = u_{n}, n \in Z \bigr\} . $$

• For \(x,y \in S\), \(a, b \in R\), \(ax+by\) is defined by

$$ ax+by = \{ax_{n}+by_{n}\}_{n=-\infty }^{+\infty }, $$

then *S* is a vector space. Clearly, \(E_{M}\) is isomorphic to \(R^{M}\), \(E_{M}\) can be equipped with the inner product

$$ \langle x, y\rangle _{E_{M}} = \sum_{s=1}^{M}x_{s}y_{s}, \quad \forall x, y \in E_{M}, $$

then \(E_{M}\) with the inner product given above is a finite dimensional Hilbert space and linearly homeomorphic to \(R^{M}\). And we denote the norm \(\|x\| = (\sum_{j=1}^{M}x_{j}^{2} )^{\frac{1}{2}}\).

• For a given matrix

B={\left(\begin{array}{cccccc}2& -1& 0& \cdots & 0& -1\\ -1& 2& -1& \cdots & 0& 0\\ 0& -1& 2& \cdots & 0& 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ -1& 0& 0& \cdots & -1& 2\end{array}\right)}_{M\times M},

in view of the results established in [1], all the eigenvalues of *B* are 0, \(\lambda _{1}, \lambda _{2}, \ldots , \lambda _{M-1}\) and \(\lambda _{j}>0\) for all \(j\in Z[1, M-1]\). Moreover,

$$ \lambda _{\min } =2\biggl(1-\cos \frac{2\pi }{M}\biggr),\qquad \lambda _{\max } = \textstyle\begin{cases} 4, & \mbox{when } M \mbox{ is even}, \\ 2(1+\cos {\frac{\pi }{M}}), & \mbox{when } M \mbox{ is odd}. \end{cases} $$

Step 2: Let the functional

$$ \varphi (u)= \frac{1}{2}\sum_{s=1}^{M}( \Delta u_{s})^{2}- F(n, u_{n})-G, $$

(4.1)

where

$$ G=G(u_{1}, u_{2}, \ldots , u_{n-1}, u_{n+1}, u_{n+2}, \ldots , u _{M})=w_{3} \Biggl[ \sum_{s=1}^{n-1} \vert u_{s} \vert ^{3}+\sum_{s=n+1}^{M} \vert u_{s} \vert ^{3} \Biggr]. $$

According to condition (A_{3}), if we let

$$ w=\max \bigl\{ \bigl\vert F(n, x)-w_{3}x^{2}+w_{2} \bigr\vert : n\in Z, \vert x \vert \leq w_{1}\bigr\} $$

and \(\tilde{w}=w+w_{2}\), then

$$ F(n, x)\geq w_{3} \vert x \vert ^{2}-\tilde{w}. $$

Combining with the fact

$$ \sum_{s=1}^{M}(\Delta u_{s})^{2}= \sum_{s=1}^{M}(u_{s+1}-u_{s})^{2}= \sum_{s=1}^{M}\bigl(2u_{s}^{2}-2u_{s}u_{s+1} \bigr), $$

for all \(u\in E_{M}\), there exists a constant \(w^{\prime }> \tilde{w}\) such that

$$\begin{aligned} \varphi (u) =& \frac{1}{2} \Biggl[\sum_{s=1}^{M}( \Delta u_{s})^{2} \Biggr]- F(n, u_{n})-G \\ \leq &\frac{1}{2}\sum_{s=1}^{M} \bigl(2u_{s}^{2}-2u_{s}u_{s+1}\bigr) -w_{3} u ^{2}_{n} +\tilde{w}-w_{3} \sum_{s=1}^{n-1} \vert u_{s} \vert ^{3}-w_{3}\sum_{s=n+1} ^{M} \vert u_{s} \vert ^{3} \\ \leq &\frac{1}{2}\sum_{s=1}^{M} \bigl(2u_{s}^{2}-2u_{s}u_{s+1}\bigr) -w_{3} u ^{2}_{n} +w^{\prime }-w_{3} \sum_{s=1}^{n-1} u^{2}_{s}-w_{3} \sum_{s=n+1} ^{M}u^{2}_{s} \\ =& \frac{1}{2}u^{\top }Bu-w_{3} \Vert u \Vert ^{2} +w^{\prime } \\ \leq & \frac{1}{2}\lambda _{\max } \Vert u \Vert ^{2}-w_{3} \Vert u \Vert ^{2} +w^{ \prime } \\ =& \biggl(\frac{1}{2}\lambda _{\max }-w_{3} \biggr) \Vert u \Vert ^{2} +w^{\prime}. \end{aligned}$$

Notice that if *M* is even then \(w_{3}\in (2, +\infty )\), and if *M* is odd, then \(w_{3}\in (1+\cos \frac{\pi }{M}, +\infty )\), and

$$ \lambda _{\max } = \textstyle\begin{cases} 4, & \mbox{when } M \mbox{ is even}, \\ 2(1+\cos {\frac{\pi }{M}}), & \mbox{when } M \mbox{ is odd}, \end{cases} $$

we have \(\lambda _{\max }/2-w_{3}<0\), which implies that \(\varphi (u) \leq w^{\prime }\). Therefore, \(\varphi (u)\) is bounded from above on \(E_{M}\).

Step 3: Set \(\tilde{c}=\sup_{u\in E_{M}}\varphi (u)\). From \(\lambda _{\max }/2-w_{3}<0\) and

$$ \varphi (u)\leq \biggl(\frac{\lambda _{\max }}{2}-w_{3} \biggr) \Vert u \Vert ^{2} +w^{\prime }, $$

we have \(\varphi (u)\rightarrow -\infty \) as \(\| u \|\rightarrow +\infty \), which implies \(-\varphi (u)\rightarrow +\infty \) as \(\| u \|\rightarrow +\infty \). Hence, for every \(l>|\tilde{c}|\), there is a constant \(P>0\) such that, for every \(\|u\|>P\), \(-\varphi (u)>l>\tilde{c}\). With the help of the continuity of \(\varphi (u)\), there must be a point \(\bar{u}\in E_{M}\) such that \(\varphi (\bar{u})=\tilde{c}= \sup_{u\in E_{M}}\varphi (u)\) and \(\|u\| \leq P\). Therefore, *ū* is a critical point of the functional \(\varphi (u)\) on \(E_{M}\) with the critical value *c̃*.

Next, let us search for the second critical point of the functional \(\varphi (u)\) on \(E_{M}\).

Step 4: Let \(u^{(k)} \in E_{M}\), for all \(k\in N\), be such that \(\{\varphi (u^{(k)})\}\) is bounded. From Step 2, there exists \(M_{1} > 0\), such that

$$ -M_{1} \leq \varphi \bigl(u^{(k)}\bigr) \leq \biggl( \frac{1}{2}\lambda _{\max }-w _{3} \biggr) \bigl\Vert u^{(k)} \bigr\Vert ^{2} +w^{\prime }, $$

which implies that

$$ \bigl\Vert u^{(k)} \bigr\Vert ^{2}\leq \biggl(w_{3}-\frac{1}{2}\lambda _{\max } \biggr)^{-1}\bigl(w^{ \prime }+M_{1}\bigr). $$

That is, \(\{u^{(k)}\}\) is bounded in \(E_{M}\). Since \(E_{M}\) is finite dimensional, there exists a subsequence of \(\{ u^{(k)} \}\) (not labeled), which is convergent in \(E_{M}\), so the \((P.S.)\) condition is verified.

Step 5: Let \(f(t)=t^{2}-w_{3}t^{3}\) for \(t\in [0, +\infty ]\). Then \(f^{\prime }(t)=2t-3w_{3}t^{2}=t(2-3w_{3}t)\). So *f* is increase on \([0, 2/(3w_{3})]\) and decrease on \((2/(3w_{3}), 1/w_{3})\). Combining \(f(0)=0\) and \(f(1/w_{3})=0\), there exist \(\xi \in (0, 2/(3w_{3}))\) and \(\zeta \in ( 2/(3w_{3}), 1/w_{3})\), such that \(f(\xi )=f(\zeta )>0\).

Step 6: By (4.1) and condition (W_{2}), we have \(\varphi (0)=0\). Take

$$ e = \textstyle\begin{cases} u_{n-1}=\xi , \\ u_{i}=0, \quad i=1,2,\ldots ,n-2,n,n+1,\ldots ,M, \end{cases} $$

and

$$ e_{1} = \textstyle\begin{cases} u_{n-1}=\zeta , \\ u_{i}=0, \quad i=1,2,\ldots ,n-2,n,n+1,\ldots ,M. \end{cases} $$

Then it is easy to verify that

$$\begin{aligned} \varphi (e) =& \frac{1}{2} \Biggl[\sum_{s=1}^{M}( \Delta u_{s})^{2} \Biggr]- F(n, u_{n})-G \\ =& u^{2}_{n-1}-w_{3} \vert u_{n-1} \vert ^{3}=\xi ^{2}-w_{3}\xi ^{3}=f( \xi ), \end{aligned}$$

and

$$\begin{aligned} \varphi (e_{1}) =& \frac{1}{2} \Biggl[\sum _{s=1}^{M}(\Delta u_{s})^{2} \Biggr]- F(n, u_{n})-G \\ =& u^{2}_{n-1}-w_{3} \vert u_{n-1} \vert ^{3}=\zeta ^{2}-w_{3}\zeta ^{3}=f(\zeta ). \end{aligned}$$

In view of the fact that \(f(\xi )=f(\zeta )>0\), \(\|e\|=\xi \) and \(\|e_{1}\|=\zeta \), we have \(\varphi (e)=\varphi (e_{1})>0=\varphi (0)\) and \(\|e\|\neq \|e_{1}\|\). Moreover, all the assumptions in Theorem 1.4 are satisfied. Noticing that \(\varphi (u)\) satisfies the \((P.S.)\) condition, then, by Remark 1.1, there exists a critical point *û* such that \(\varphi (\hat{u})=\hat{c}\) (*ĉ* is given in Theorem 1.4).

Step 7: In order to obtain two critical points, we also need to prove that \(\hat{u}\neq \bar{u}\). Since \(\varphi (\hat{u})=\hat{c}\) and \(\varphi (\bar{u})=\tilde{c}\), if we can prove \(\hat{c}\neq \tilde{c}\), that also implies \(\hat{u}\neq \bar{u}\). So in the following, we are ready to prove that \(\hat{c}\neq \tilde{c}\). Since \(u\in E_{M}\) and \(E_{M}\) is linearly homeomorphic to \(R^{M}\), in Theorem 1.4, we can take \(X=E_{M}\) and construct

$$\begin{aligned} \gamma _{1}(t) =& (u_{1},\ldots ,u_{n-2},u_{n-1},u_{n},u_{n+1}, \ldots ,u_{M+1}) \\ =&(0,\ldots , 0,u_{n-1},0,0,\ldots ,0) \\ =&\bigl(0,\ldots , 0,(2\zeta -4\xi )t^{2}+(-\zeta +4\xi )t,0,0,\ldots ,0\bigr), \end{aligned}$$

where \(u_{n-1}=(2\zeta -4\xi )t^{2}+(-\zeta +4\xi )t\), \(t\in [0, 1]\). Obviously, \(\gamma _{1}\in C ([0, 1] )\).

One computes that \(\gamma _{1}(0)=(0,\ldots ,0,0,0,0,\ldots ,0)\),

$$ \gamma _{1}\biggl(\frac{1}{2}\biggr)=(0,\ldots ,0,u_{n-1},0,0,\ldots ,0)= (0,\ldots ,0,\xi ,0,0,\ldots ,0)=e, $$

and

$$ \gamma _{1}(1)=(0,\ldots ,0,u_{n-1},0,0,\ldots ,0)=(0,\ldots ,0,\zeta ,0,0,\ldots ,0)=e_{1}. $$

Hence, \(\gamma _{1}(t)\in \hat{\varGamma }\) where

$$ \hat{\varGamma }:=\biggl\{ \gamma \in C \bigl([0, 1], X \bigr):\gamma (0)=0, \gamma \biggl(\frac{1}{2}\biggr)=e_{1},\gamma (1)=e\biggr\} . $$

Let \(u=\gamma _{1}(t)\), \(t\in [0, 1]\) be in (4.1). Since \(F(n,0)=0\), we have

$$\begin{aligned} \varphi (u) =&\varphi \bigl(\gamma _{1}(t)\bigr)=\frac{1}{2} \Biggl[\sum_{s=1}^{M}(\Delta u_{s})^{2} \Biggr]- F(n, u_{n})-G \\ =&\frac{1}{2}\bigl(u^{2}_{n-1}+u^{2}_{n-1} \bigr)- F(n, 0)-w_{3} \vert u_{n-1} \vert ^{3} \\ =&u^{2}_{n-1}-w_{3} \vert u_{n-1} \vert ^{3}. \end{aligned}$$

(4.2)

Let \(|u_{n-1}|=y\in [0, +\infty ]\), then, by (4.2), we see that when \(y=2/3w_{3}\), \(\varphi (\gamma _{1}(t))=y^{2}-w_{3}y^{3}\) takes the maximum value \(4/(27w^{2}_{3})\). We must point that when \(t\in [0, 1]\), \(|u_{n-1}|\) can take the value of \(2/3w_{3}\). In fact, when \(t=0\),

$$ u_{n-1}=(2\zeta -4\xi )t^{2}+(-\zeta +4\xi )t=0, $$

and when \(t=1\),

$$ u_{n-1}=(2\zeta -4\xi )t^{2}+(-\zeta +4\xi )t=\zeta . $$

Observing that \(\zeta \in (2/(3w_{3}), 1/w_{3})\), then, by the continuity of \(u_{n-1}=(2\zeta -4\xi )t^{2}+(-\zeta +4\xi )t\), there exists \(\tilde{t}\in [0, 1]\) such that \(u_{n-1}=(2\zeta -4\xi )\tilde{t}^{2}+(-\zeta +4\xi )\tilde{t}=2/(3w_{3})\).

Since \(\hat{c}:=\inf_{\gamma \in \hat{\varGamma }}\max_{t\in [0, 1]}\varphi (\gamma (t) )\) and \(\gamma _{1}(t)\in \hat{\varGamma }\), we have \(\hat{c}\leq 4/(27w^{2}_{3})\). In order to obtain two critical point of \(\varphi (u)\) on \(E_{M}\), we will show that \(\tilde{c}>4/(27w^{2}_{3})\).

Since \(E_{M}\) is linearly homeomorphic to \(R^{M}\) and \(M\geq 3\), if we choose

$$ u=(u_{1},\ldots ,u_{n-2},u_{n-1},u_{n},u_{n+1}, \ldots ,u_{M+1})\in E_{M}, $$

then

$$ u_{1}=\cdots =u_{n-2}=u_{n}=u_{n+2}= \cdots =u_{M}=0, \qquad u_{n-1}=-u_{n+1}=- \frac{1}{2w_{3}}. $$

By (4.1) and \(F(n, 0)=0\), we have

$$\begin{aligned} \varphi (u) =& \frac{1}{2}\sum_{s=1}^{M}( \Delta u_{s})^{2}- F(n, u_{n})-G \\ =&\frac{1}{2}\bigl(u^{2}_{n-1}+u^{2}_{n-1}+u^{2}_{n+1}+u^{2}_{n+1} \bigr)- F(n, 0)-w_{3}\bigl(u^{3}_{n-1}+u^{3}_{n+1} \bigr) \\ =&1/\bigl(2w^{2}_{3}\bigr)-0-1/\bigl(4w^{2}_{3} \bigr)=1/\bigl(4w^{2}_{3}\bigr). \end{aligned}$$

Employing \(\tilde{c}=\sup_{u\in E_{M}}\), we have \(\tilde{c}\geq 1/ (4w^{2}_{3})>4/(27w^{2}_{3})\geq \hat{c}\). Combining \(\varphi (\hat{u})=\hat{c}\), \(\varphi (\bar{u})=\tilde{c}\), both *û* and *ū* are critical point of the functional *φ*, we obtain two different critical points of *φ*.

Step 8: (1.1) has at least two nontrivial *M*-periodic solutions. Since \(\varphi \in \mathcal{C}^{2} (E_{M}, R)\), for any \(u = \{u_{n} \}_{n \in Z} \in E_{M}\), according to \(u_{0} = u_{M}\), \(u_{1} = u _{M+1}\), one computes that

$$ \frac{\partial \varphi }{\partial u_{n}}=\Delta ^{2}u_{n-1}+\nabla _{u _{n}} F(n, u_{n}), \quad \forall n \in Z. $$

Therefore, the existence of critical points of *φ* on \(E_{M}\) implies the existence of periodic solutions of system (1.1). Moreover, we obtained two different critical points of \(\varphi (u)\) on \(E_{M}\) in Step 7, so system (1.1) has two different *M*-periodic solutions.

Note that in (4.1), \(\varphi (0)=0\). But \(\varphi (\hat{u})=\hat{c}\geq \varphi (e) >\varphi (0)=0\) and \(\varphi (\bar{u})=\tilde{c}=\sup_{u\in E_{M}}>4/(27w^{2}_{3}) \geq \hat{c}\geq \varphi (e)>0\), so any of the above periodic solutions *û* and *ū* is nontrivial. From this, Theorem 1.5 is proved.

### Remark 4.1

Let \(F(t,x)\) be stated in Example 1.2, from (4.1), we have

$$\begin{aligned} \varphi (u) =& \frac{1}{2} \Biggl[\sum_{s=1}^{M}( \Delta u_{s})^{2} \Biggr]- a\bigl(u^{2}_{n}/2+u_{n}+ \cos u_{n}-1\bigr) \bigl(\phi (n)+K\bigr) \\ &{}-w_{3} \Biggl[ \sum_{s=1}^{n-1} \vert u_{s} \vert ^{3}+\sum _{s=n+1}^{M} \vert u_{s} \vert ^{3} \Biggr], \end{aligned}$$

where \(u= (\ldots , u_{-n}, \ldots , u_{0}, \ldots )\in E_{M}\). We notice that the value of \(\inf_{\|u\|=r}\varphi (u)\) is very difficult to compute, but fortunately the condition in our new mountain pass theorem (Theorem 1.4) is independent of \(\inf_{\|u\|=r} \varphi (u)\), and we need not compute it.