# Existence and multiplicity of solutions for second-order Hamiltonian systems satisfying generalized periodic boundary value conditions at resonance

## Abstract

We investigate the existence and multiplicity of solutions for second-order Hamiltonian systems satisfying generalized periodic boundary value conditions at resonance by means of the index theory, the critical point theory without compactness assumptions, the least action principle, the saddle point reduction theorem, and the minimax method. Applying the results to second-order HS satisfying periodic boundary value conditions, we obtain some new results.

## 1 Introduction and main results

Solutions of Hamiltonian systems are very important in applications. In recent years, the existence and multiplicity of solutions for Hamiltonian systems via critical point theory have been studied by many authors (see [2, 5,6,7,8,9,10, 12,13,14,15,16,17,18,19,20,21,22]). In particular, by means of critical point theory, the least action principle, and the minimax method, the existence and multiplicity of periodic solutions for second-order Hamiltonian systems with periodic boundary conditions were extensively studied in the cases where the gradient of the nonlinearity is bounded sublinearly and linearly, and many interesting results are given in [5, 9, 10, 13,14,15,16,17,18,19, 22]. In this paper, we discuss the existence and multiplicity of solutions for the following second-order Hamiltonian systems satisfying generalized periodic boundary value conditions:

$$\textstyle\begin{cases} -x''- B_{1}(t)x=\nabla _{x} V(t,x), \quad \mbox{a.e. } t\in [0,1], \\ x(1) = Mx(0) , \qquad x'(1) = Nx'(0), \end{cases}$$
(1.1)

where $$B_{1}(t)\in L^{\infty }([0, 1], \mathcal{L}_{s} (\mathbf{R}^{n}))= \{B(t)=(b_{jk})_{n\times n} |b_{jk}(t)=b_{kj}(t), t\in [0,1], b_{jk}(t)\in L^{\infty }([0,1])\}$$ with $$\nu _{M}^{s}( B_{1}) \neq 0$$, $$M,N\in GL(n)=\{A=(a_{jk})_{n\times n} |a_{jk}\in \mathbf{R}\mbox{ and } \operatorname{det}(A)\neq 0\}$$, and $$MN^{T}=I_{n}$$, where $$I_{n}$$ is the unit matrix of order n, and $$\nabla _{x} V(t,x)$$ denotes the gradient of $$V(t,x)$$ for $$x\in \mathbf{R}^{n}$$. We suppose that $$V:[0,1]\times \mathbf{R}^{n}\to \mathbf{R}$$ satisfies the following condition:

1. (A)

$$V(t,x)$$ is measurable in t for every $$x\in \mathbf{R}^{n}$$ and continuously differentiable in x for a.e. $$t\in [0,1]$$. Moreover, there exist $$a(x)\in C(\mathbf{R}^{+},\mathbf{R}^{+})$$ and $$b(t)\in L^{1}([0,1],\mathbf{R}^{+})$$ such that

$$\bigl\vert V(t,x) \bigr\vert \leq a\bigl( \vert x \vert \bigr)b(t) \quad \mbox{and} \quad \bigl\vert \nabla _{x} V(t,x) \bigr\vert \leq a \bigl( \vert x \vert \bigr)b(t)$$

for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$, where $$\mathbf{R}^{+}=[0,+ \infty )$$.

Note that if $$M=N=I_{n}$$ and $$B_{1}(t)\equiv 0$$, then $$\nu _{I_{n}} ^{s}(0)\neq 0$$ via (2.5) in the next section. Therefore the periodic solution problem

$$\textstyle\begin{cases} -x''=\nabla _{x} V(t,x) \quad \mbox{a.e. } t\in [0,1], \\ x(1)-x(0)= x'(1)-x'(0)=0, \end{cases}$$
(1.2)

is a particular case of (1.1).

Now we use the index $$(i_{M}^{s}( B),\nu _{M}^{s}( B))\in \mathbf{Z}\times \mathbf{N}$$ defined in [6, 7] (see the next section) for all $$B\in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$ to reach our main results.

### Theorem 1.1

Assume that $$V(t,x)$$ is convex in x for a.e. $$t\in [0,1]$$ and satisfies (A) and

(A1):

$$i_{M}^{s}( B_{1})=0$$;

(A2):
$$\int _{0}^{1}V(t,x)\,dt\to +\infty \quad \textit{as } \Vert x \Vert \to \infty , x\in Z^{0}(B_{1})= \operatorname{ker}(\varLambda -B_{1});$$
(A3):

there exists $$B_{2}\in L^{\infty }([0, 1], \mathcal{L}_{s} (\mathbf{R}^{n}))$$ such that $$B_{2}> B_{1}$$, $$\nu _{M}^{s}(B _{2})\neq 0$$, and $$i_{M}^{s}(B_{2})=i_{M}^{s}(B_{1})+\nu _{M}^{s}(B _{1})$$, and there exists $$\gamma (t)\in L^{1}([0,1],\mathbf{R}^{+})$$ such that

$$V(t,x)\leq \frac{1}{2}\bigl(\bigl(B_{2}(t)-B_{1}(t) \bigr)x,x\bigr)+\gamma (t)$$

for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$, and

$$\operatorname{meas} \biggl\{ t\in [0,1]\Big| V(t,x)-\frac{1}{2}\bigl(\bigl(B _{2}(t)-B_{1}(t)\bigr)x,x\bigr)\to -\infty \textit{ as } \Vert \bar{x} \Vert \to \infty \biggr\} >0,$$

where $$x=\tilde{x}+\bar{x}$$ and $$\bar{x}\in Z^{0}(B_{2})$$.

Then problem (1.1) has at least one solution in $$Z=\{x\in H^{1}([0,1],\mathbf{R}^{n})| x(1) = Mx(0)\}$$.

### Theorem 1.2

Assume that $$V(t,x)$$ satisfies (A), (A1), and

(A4):

there exist $$f,g\in L^{1}([0,1],\mathbf{R}^{+})$$ with $$\nu _{M}^{s}(B_{1}+f(t)I_{n})=0$$ and $$i_{M}^{s}(B_{1}+f(t)I_{n})=i _{M}^{s}(B_{1})+\nu _{M}^{s}(B_{1})$$ such that

$$\bigl\vert \nabla _{x} V(t,x) \bigr\vert \leq f(t) \vert x \vert +g(t)$$

for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$;

(A5):

there exists a function $$\mu (t)\in L^{1}([0,1],\mathbf{R}^{+})$$ with $$\inf_{t\in [0,1]}\mu (t)>0$$ such that $$V(t,x)- \frac{1}{2}\mu (t)|x|^{2}$$ is convex in x for a.e. $$t\in [0,1]$$.

Then problem (1.1) has at least one solution with saddle character in Z (i.e., the solution is a saddle point).

(A6):

there exist $$r>0$$, $$B_{01},B_{02}\in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$ such that $$B_{02}> B_{01}>B_{1}$$ and $$\nu _{M}^{s}(B_{0i})\neq 0$$ ($$i=1,2$$), $$i_{M}^{s}(B_{02})=i_{M}^{s}(B _{01})+\nu _{M}^{s}(B_{01})$$, and for all $$\|x\|\leq r$$,

$$\frac{1}{2}\bigl(\bigl(B_{01}(t)-B_{1}(t)\bigr)x,x \bigr)\leq V(t,x)\leq \frac{1}{2} \bigl(\bigl(B _{02}(t)-B_{1}(t) \bigr)x,x\bigr)$$

for a.e. $$t\in [0,1]$$.

Then problem (1.1) has at least three distinct solutions in Z.

### Theorem 1.3

Assume that $$V(t,x)$$ satisfies (A), (A1), and

($$\mathrm{A}'_{4}$$):

there exist $$\alpha \in [0,1)$$, and $$f\in L^{2}([0,1],\mathbf{R}^{+})$$, and $$g\in L^{1}([0,1],\mathbf{R}^{+})$$ such that

$$\bigl\vert \nabla _{x} V(t,x) \bigr\vert \leq f(t) \vert x \vert ^{\alpha }+g(t)$$

for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$;

(A7):

there exists $$c_{0}>0$$ large enough such that

$$\liminf_{ \Vert x \Vert \to \infty } \Vert x \Vert ^{-2\alpha } \int _{0}^{1}V(t,x)\,dt>c _{0}$$
(1.3)

or

$$\limsup_{ \Vert x \Vert \to \infty } \Vert x \Vert ^{-2\alpha } \int _{0}^{1}V(t,x)\,dt< -c _{0}$$
(1.4)

for $$x\in Z^{0}(B_{1})=\operatorname{ker}(\varLambda -B_{1})$$.

Then problem (1.1) has at least one solution in Z.

($$\mathrm{A}'_{6}$$):

there exist $$\epsilon >0$$, $$r>0$$, $$B_{01},B_{02} \in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$ such that $$B_{02}> B_{01}>B_{1}$$ and $$\nu _{M}^{s}(B_{0i})\neq 0$$ ($$i=1,2$$), $$i_{M}^{s}(B_{02})=i_{M}^{s}(B_{01})+\nu _{M}^{s}(B_{01})$$, and for all $$\|x\|\leq r$$,

$$\frac{1}{2}\bigl(\bigl(\epsilon I_{n}+B_{01}(t)-B_{1}(t) \bigr)x,x\bigr)\leq V(t,x)$$

for a.e. $$t\in [0,1]$$, whereas for all $$x\in \mathbf{R}^{n}$$,

$$V(t,x)\leq \frac{1}{2} \bigl(\bigl(B_{02}(t)-B_{1}(t) \bigr)x,x\bigr)$$

for a.e. $$t\in [0,1]$$.

Then problem (1.1) has at least two distinct solutions in Z.

### Theorem 1.4

Assume that $$V(t,x)$$ satisfies (A), ($$\mathrm{A}'_{4}$$), (A7), and

($$\mathrm{A}_{1}^{\prime }$$):

$$i_{M}^{s}( B_{1})\neq 0$$.

Then problem (1.1) has at least one solution in Z.

(A8):

$$V(t,x)$$ is an even function in x for a.e. $$t\in [0,1]$$;

(A9):

there exist $$\epsilon >0$$, $$r>0$$, $$B_{2}\in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$ such that $$B_{2}> B_{1}$$ and $$\nu _{M}^{s}(B_{2})\neq 0$$, $$i_{M}^{s}(B_{2})>i_{M}^{s}(B_{1})+\nu _{M}^{s}(B_{1})$$, and for all $$\|x\|\leq r$$,

$$V(t,x)-V(t,0)\geq \frac{1}{2}\bigl(\bigl(\epsilon I_{n}+B_{2}(t)-B_{1}(t) \bigr)x,x\bigr)$$

for a.e. $$t\in [0,1]$$.

Then problem (1.1) has at least $$i_{M}^{s}(B_{2})-i_{M} ^{s}(B_{1})-\nu _{M}^{s}(B_{1})$$ pairs of solutions in Z.

We give the proofs in Sect.Â 3, and now we return to some discussions on problem (1.2).

### Corollary 1.5

Assume that $$V(t,x)$$ is convex in x for a.e. $$t\in [0,1]$$ and satisfies (A) and the following conditions:

(H1):
$$\int _{0}^{1}V(t,x)\,dt\to +\infty \quad \textit{as } \vert x \vert \to \infty , x\in \mathbf{R}^{n};$$
(H2):

there exists $$\gamma (t)\in L^{1}([0,1],\mathbf{R}^{+})$$ such that

$$V(t,x)\leq \frac{(2\pi )^{2}}{2} \vert x \vert ^{2}+\gamma (t)$$

for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$, and

$$\operatorname{meas} \biggl\{ t\in [0,1]\Big| V(t,x)-\frac{(2 \pi )^{2}}{2} \vert x \vert ^{2}\to -\infty \textit{ as } |x|\to \infty \biggr\} >0.$$

Then problem (1.2) has at least one solution in $$H^{1}_{0}=\{x\in H^{1}([0,1],\mathbf{R}^{n})| x(1)-x(0)=0\}$$.

### Remark 1.6

For the interval $$[0,T]$$ considered in second-order HS satisfying periodic boundary value conditions, if $$T=1$$, then CorollaryÂ 1.5 reduces to TheoremÂ 3.2 in [17]. By RemarkÂ 1.4 and RemarkÂ 3.2 in [17] we can see that CorollaryÂ 1.5 generalizes TheoremÂ 3.5 in [10] and the corresponding theorem in [13] as $$T=1$$.

### Corollary 1.7

Assume that $$V(t,x)$$ satisfies (A), (A5), and

(H3):

there exist $$f,g\in L^{1}([0,1],\mathbf{R}^{+})$$ with $$0< f(t)<4\pi ^{2}$$ such that

$$\bigl\vert \nabla _{x} V(t,x) \bigr\vert \leq f(t) \vert x \vert +g(t)$$
(1.5)

for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$.

Then problem (1.2) has at least one solution with saddle character in $$H_{0}^{1}$$.

(H4):

there exist $$\delta >0$$ and $$k\in \mathbf{N}\setminus \{0\}$$ such that, for all $$|x|\leq \delta$$,

$$2(k\pi )^{2} \vert x \vert ^{2}\leq V(t,x)\leq 2 \bigl((k+1)\pi \bigr)^{2} \vert x \vert ^{2}$$

for a.e. $$t\in [0,1]$$.

Then problem (1.2) has at least three distinct solutions in $$H_{0}^{1}$$.

### Remark 1.8

As $$T=1$$, in TheoremÂ 2.2 of [18], assume that $$V(t,x)$$ satisfies (A), (H4), and

(H3,1):

there exist $$f,g\in L^{1}([0,1],\mathbf{R}^{+})$$ with $$\int ^{1}_{0}f(t)\,dt<12$$ such that (1.5) holds;

(A5,1):

there exists a function $$\mu (t)\in L^{1}([0,1],\mathbf{R}^{+})$$ with $$\int ^{1}_{0}\mu (t)\,dt>0$$ such that $$V(t,\cdot )$$ is $$\mu (t)$$-monotone.

Then the conclusion of CorollaryÂ 1.7 is also true.

On one hand, by RemarkÂ 1.7 in [17] we know that the $$\mu (t)$$-monotonicity of $$V(t,\cdot )$$ is equivalent to the convexity of $$V(t,\cdot )-\frac{1}{2}\mu (t)$$. Since $$\inf_{t\in [0,1]} \mu (t)>0\Rightarrow \int ^{1}_{0}\mu (t)\,dt>0$$, this shows that $$(\mathrm{A}_{5})\Rightarrow (\mathrm{A}_{5,1})$$.

On the other hand, for $$f\in L^{1}([0,1],\mathbf{R}^{+})$$, we have $$\int ^{1}_{0}f(t)\,dt<12\nRightarrow 0<f(t)<4\pi ^{2}$$ and $$0< f(t)<4\pi ^{2}\nRightarrow \int ^{1}_{0}f(t)\,dt<12$$. Indeed, if $$f(t)= \bigl \{ \scriptsize{ \begin{array}{l@{\quad}l} 4\pi ^{2}, & x\in [0,\frac{1}{4\pi ^{2}}], \\ 0, & x\in (\frac{1}{4\pi ^{2}},1], \end{array}}$$ then $$\int ^{1}_{0}f(t)\,dt=1$$ and $$f(t)\geq 4\pi ^{2}$$ for $$x\in [0,\frac{1}{4\pi ^{2}}]$$; if $$12< f(t)<4\pi ^{2}$$, then $$\int ^{1} _{0}f(t)\,dt>12$$. So CorollaryÂ 1.7 is a new result and in a sense a development of TheoremÂ 2.2 in [18].

Next, we give some examples of a potential function $$V(t,x)$$ satisfying the assumptions of CorollaryÂ 1.7. Let $$\mu (t)=2\pi ^{2}$$ for all $$t\in [0,1]$$, and let

$$V(t,x)= \textstyle\begin{cases} \pi ^{2} \vert x \vert ^{2}+2\pi ^{2} \vert x \vert -\pi ^{2}, & \vert x \vert \geq 1, \\ 2\pi ^{2} \vert x \vert ^{2}, & \vert x \vert \leq 1, \end{cases}$$

for all $$x\in \mathbf{R}^{n}$$. Clearly, assumptions (A), (H3), (H4) hold, and $$F(x)=V(t,x)- \frac{1}{2}\mu (t)|x|^{2}$$ is convex in x because

$$F(x)= g\bigl(h(x)\bigr)$$

is convex, which follows from the facts that

$$g(s)= \textstyle\begin{cases} 2\pi ^{2}s-\pi ^{2}, & s\geq 1, \\ \pi ^{2}s^{2}, & 0\leq s\leq 1, \end{cases}$$

is convex and increasing and

$$h(x)= \vert x \vert , \quad x\in \mathbf{R}^{n},$$

is convex. Thus V satisfies the conditions of CorollaryÂ 1.7. Similarly, we can see that

$$V(t,x)= \textstyle\begin{cases} \pi ^{2}(1+\sin t) \vert x \vert ^{2}+2\pi ^{2}(1+\sin t) \vert x \vert -\pi ^{2}(1+\sin t), & \vert x \vert \geq 1, \\ 2\pi ^{2}(1+\sin t) \vert x \vert ^{2}, & \vert x \vert \leq 1, \end{cases}$$

also satisfies the conditions of CorollaryÂ 1.7.

### Corollary 1.9

Assume that $$V(t,x)$$ satisfies (A), ($$\mathrm{A}'_{4}$$), and

(H5):

there exists $$c_{0}>0$$ large enough such that

$$\liminf_{ \vert x \vert \to \infty } \vert x \vert ^{-2\alpha } \int _{0}^{1}V(t,x)\,dt>c _{0}$$
(1.6)

or

$$\limsup_{ \vert x \vert \to \infty } \vert x \vert ^{-2\alpha } \int _{0}^{1}V(t,x)\,dt< -c _{0}$$
(1.7)

for $$x\in \mathbf{R}^{n}$$.

Then problem (1.2) has at least one solution in $$H^{1}_{0}$$.

($$\mathrm{H}'_{4}$$):

there exist $$\epsilon >0$$, $$r>0$$, and $$k\in \mathbf{N}\setminus \{0\}$$ such that

$$\frac{1}{2}\bigl(\epsilon +(2k\pi )^{2}\bigr) \vert x \vert ^{2}\leq V(t,x)$$

for all $$|x|\leq r$$ and a.e. $$t\in [0,1]$$, and

$$V(t,x)\leq \frac{1}{2} \bigl(2(k+1)\pi \bigr)^{2} \vert x \vert ^{2}$$

for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$.

Then problem (1.2) has at least two distinct solutions in $$H^{1}_{0}$$.

### Remark 1.10

As $$T=1$$, in Theorems 1â€“3 of [14] assume that $$V(t,x)$$ satisfies (A), ($$\mathrm{H}'_{4}$$), and

($$\mathrm{A}''_{4}$$):

there exist $$\alpha \in [0,1)$$ and $$f, g\in L ^{1}([0,1],\mathbf{R}^{+})$$ such that

$$\bigl\vert \nabla _{x} V(t,x) \bigr\vert \leq f(t) \vert x \vert ^{\alpha }+g(t)$$

for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$;

(H5,1):
$$\vert x \vert ^{-2\alpha } \int _{0}^{1}V(t,x)\,dt\to +\infty \quad \mbox{as } \vert x \vert \to \infty$$
(1.8)

or

$$\vert x \vert ^{-2\alpha } \int _{0}^{1}V(t,x)\,dt\to -\infty \quad \mbox{as } \vert x \vert \to \infty .$$
(1.9)

Then the conclusion of CorollaryÂ 1.9 is also true. Clearly, condition (H5,1) is stronger than condition (H5), and condition ($$\mathrm{A}''_{4}$$) is weaker than condition ($$\mathrm{A}'_{4}$$). Moreover, we can see that Examples 3.1â€“3.2 in [16] satisfy the conditions of CorollaryÂ 1.9 but do not satisfy Theorems 1â€“3 in [14]. So CorollaryÂ 1.9 is a new result and in a sense a development of Theorems 1â€“3 in [14].

In addition, if $$T=1$$ and $$c_{0}=\frac{1}{8\pi ^{2}}\int _{0}^{1}f^{2}(t)\,dt$$ or $$c_{0}=-\frac{3}{8 \pi ^{2}}\int _{0}^{1}f^{2}(t)\,dt$$, then CorollaryÂ 1.9 reduces to Theorems 1.1â€“1.2 in [16]. In particular, we need to point out that condition ($$\mathrm{A}''_{4}$$) of Theorems 1.1â€“1.2 in [16] must be amended to condition ($$\mathrm{A}'_{4}$$), because $$\int _{0}^{1}f^{2}(t)\,dt$$ was used in the proof of Theorems 1.1â€“1.2 in [16].

### Remark 1.11

As $$T=1$$, in TheoremÂ 1 of [22] assume that $$V(t,x)$$ satisfies (A), ($$\mathrm{H}'_{4}$$), (H3,1), and

($$\mathrm{H}'_{5,1}$$):
$$\vert x \vert ^{-2} \int _{0}^{1}V(t,x)\,dt\to +\infty \quad \mbox{as } \vert x \vert \to \infty$$
(1.10)

or

$$\vert x \vert ^{-2} \int _{0}^{1}V(t,x)\,dt\to -\infty \quad \mbox{as } \vert x \vert \to \infty .$$
(1.11)

Then the conclusion of CorollaryÂ 1.9 holds. Unfortunately, Tang and Meng [16] pointed out that (1.5) of condition (H3,1) and (1.10) or (1.11) of condition ($$\mathrm{H}'_{5,1}$$) cannot hold together, so that TheoremÂ 1 in [22] is also incorrect.

### Corollary 1.12

Assume that $$V(t,x)$$ satisfies (A), ($$\mathrm{A}'_{4}$$), and

($$\mathrm{H}'_{5}$$):

there exists $$c_{0}>0$$ large enough such that (1.3) or (1.4) hold for $$x\in \operatorname{ker}(\varLambda -(2k\pi )^{2})$$.

Then problem

$$\textstyle\begin{cases} -x''-(2k\pi )^{2}x=\nabla _{x} V(t,x) \quad \textit{a.e. } t\in [0,1], \\ x(1)-x(0)= x'(1)-x'(0)=0, \end{cases}$$
(1.12)

has at least one periodic solution in $$H^{1}_{0}$$. Further, assume that (A8) is satisfied together with

(H6):

there exist $$\epsilon >0$$ and $$r>0$$ such that, for all $$|x|\leq r$$,

$$V(t,x)-V(t,0)\geq \frac{\epsilon +4m(2k+m)\pi ^{2}}{2} \vert x \vert ^{2}$$

for a.e. $$t\in [0,1]$$ and $$k,m\in \mathbf{N}\setminus \{0\}$$ with $$m>1$$.

Then problem (1.12) has at least $$2nm-2n$$ pairs of solutions in $$H^{1}_{0}$$.

### Remark 1.13

As $$T=1$$, in Theorems 1.1â€“1.2 of [19] assume that $$V(t,x)$$ satisfies (A), ($$\mathrm{A}''_{4}$$), (H6), and

($$\mathrm{H}'_{5,2}$$):
$$\Vert x \Vert ^{-2\alpha } \int _{0}^{1}V(t,x)\,dt\to +\infty \quad \mbox{as } \Vert x \Vert \to \infty$$
(1.13)

or

$$\Vert x \Vert ^{-2\alpha } \int _{0}^{1}V(t,x)\,dt\to -\infty \quad \mbox{as } \Vert x \Vert \to \infty$$
(1.14)

for $$x\in \operatorname{ker}(\varLambda -(2k\pi )^{2})$$.

Then the conclusion of CorollaryÂ 1.12 is also true. Clearly, condition ($$\mathrm{H}'_{5,2}$$) is stronger than condition ($$\mathrm{H}'_{5}$$), and condition ($$\mathrm{A}''_{4}$$) is weaker than condition ($$\mathrm{A}'_{4}$$). So CorollaryÂ 1.12 is a new conclusion and in a sense a development of Theorems 1.1â€“1.2 in [19].

### Corollary 1.14

Assume that $$V(t,x)$$ satisfies (A), ($$\mathrm{A}'_{4}$$), and

($$\mathrm{A}''_{1}$$):

$$\operatorname{ker}(\varLambda -A(t))\setminus \{\theta \}\neq \emptyset$$;

($$\mathrm{H}'_{5}$$):

there exists $$c_{0}>0$$ large enough such that (1.3) or (1.4) hold for $$x\in \operatorname{ker}(\varLambda -A(t))$$.

Then problem

$$\textstyle\begin{cases} -x''-A(t)x=\nabla _{x} V(t,x)\quad \textit{a.e. } t\in [0,1], \\ x(1)-x(0)= x'(1)-x'(0)=0, \end{cases}$$
(1.15)

has at least one periodic solution in $$H^{1}_{0}$$, where $$A(t)$$ is a continuous symmetric matrix of order n.

### Remark 1.15

As $$T=1$$, in Theorems 2â€“3 in [15] assume that $$V(t,x)$$ satisfies (A), ($$\mathrm{A}''_{1}$$), ($$\mathrm{A}''_{4}$$), and

($$\mathrm{H}'_{5,3}$$):

there exists $$\gamma (t)\in L^{1}([0,1],\mathbf{R}^{+})$$ such that $$|x|^{-2\alpha }V(t,x)\geq -\gamma (t)$$ for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$, and there exists a subset E of $$[0,1]$$ with $$\operatorname{meas}(E)>0$$ such that

$$\vert x \vert ^{-2\alpha }V(t,x)\to +\infty \quad \textit{as } \vert x \vert \to \infty$$

for a.e. $$t\in E$$; or there exists $$\gamma (t)\in L^{1}([0,1],\mathbf{R}^{+})$$ such that $$|x|^{-2\alpha }V(t,x)\leq \gamma (t)$$ for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$, and there exists a subset E of $$[0,1]$$ with $$\operatorname{meas}(E)>0$$ such that

$$\vert x \vert ^{-2\alpha }V(t,x)\to -\infty \quad \textit{as } \vert x \vert \to \infty$$

for a.e. $$t\in E$$.

Then the conclusion of CorollaryÂ 1.14 is also true. Clearly, from the proof of Theorems 2â€“3 in [15] we can see that $$(\mathrm{H}'_{5,3}) \Rightarrow (\mathrm{H}'_{5,2})$$. Moreover, we know that condition ($$\mathrm{H}'_{5}$$) is weaker than condition ($$\mathrm{H}'_{5,2}$$). So, although condition ($$\mathrm{A}''_{4}$$) is weaker than condition ($$\mathrm{A}'_{4}$$), CorollaryÂ 1.14 is also a new conclusion and in a sense a development of Theorems 2â€“3 in [15].

The proof of Theorems 1.1â€“1.4 and these corollaries will be given in Sect.Â 3, and in Sect.Â 2, we recall some useful results concerning the index theory for linear second-order Hamiltonian systems satisfying generalized periodic boundary value conditions in [6, 7], which will be used in other sections.

## 2 Brief introduction of the index theory

Let $$L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))=\{B(t)\in GL(n)|b _{jk}(t)=b_{kj}(t)\mbox{ for }t\in [0,1] \mbox{ and } b_{jk}(t)\in L ^{\infty }([0,1])\}$$. Index theory in [6, 7] deals with a classification of $$L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$ associated with the following system:

\begin{aligned}& -x''- B(t)x=0, \end{aligned}
(2.1)
\begin{aligned}& x(1) = Mx(0), \qquad x'(1) = Nx'(0) , \end{aligned}
(2.2)

where $$M,N\in GL(n)$$ and $$MN^{T}=I_{n}$$.

Let $$L= L^{2}([0, 1],\mathbf{R}^{n})$$ and $$Z= \{x\in H^{1}([0,1],\mathbf{R}^{n})| x \mbox{ satisfies (2.2)}\}$$. Define $$\varLambda : D(\varLambda )\to L$$ by $$(\varLambda x)(t)= -x''(t)$$. From the Sect.Â 7.1 in [6] we can check that Î› is self-adjoint in L and $$\sigma ( \varLambda )=\sigma _{d}(\varLambda )\subset [0,+\infty )$$. In particular, if $$M=N=I_{n}$$, then $$\sigma (\varLambda )=\sigma _{d}(\varLambda )=\{(2k)^{2} \pi ^{2}|k\in \mathbf{Z}\}$$, and if $$M=N=-I_{n}$$, then $$\sigma (\varLambda )= \sigma _{d}(\varLambda )=\{(2k-1)^{2}\pi ^{2}|k\in \mathbf{Z}\}$$.

For any $$B(t)\in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$, we define

$$q_{B}(x,y)= \int _{0}^{1} \bigl[\bigl(x',y' \bigr)-\bigl(B(t)x,y\bigr) \bigr] \,dt,\quad x,y \in Z,$$
(2.3)

where $$(\cdot ,\cdot )$$ is the usual inner product in $$\mathbf{R}^{n}$$, and Z is a Hilbert space with norm $$\|x\|^{2}=\int _{0}^{1}|x'|^{2}\,dt+ \int _{0}^{1}|x|^{2}\,dt$$ for each $$x\in Z$$. Clearly, the embeddings $$Z\hookrightarrow L$$ and $$Z\hookrightarrow L^{\infty }$$ are compact.

### Proposition 2.1

([7], PropositionÂ 7.2.1)

For any $$B(t)\in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$, the space Z has a $$q_{B}$$-orthogonal decomposition

$$Z=Z^{+}(B)\oplus Z^{0}(B)\oplus Z^{-}(B)$$

such that $$q_{B}$$ is positive definite, null, and negative definite on $$Z^{+}(B)$$, $$Z^{0}(B)$$, and $$Z^{-}(B)$$, respectively. Moreover, $$Z^{0}(B)$$ and $$Z^{-}(B)$$ are finite-dimensional.

### Definition 2.2

([6], DefinitionÂ 2.4.1; [7], DefinitionÂ 7.1.3)

For any $$B(t)\in L^{\infty }([0, 1], \mathcal{L}_{s} (\mathbf{R}^{n}))$$, we define

$$\nu _{M}^{s}(B)=\operatorname{dim}\operatorname{ker}( \varLambda -B), \qquad i_{M}^{s}(B)=\sum _{\lambda < 0}\nu _{M}^{s}(B+\lambda I_{n}).$$

### Definition 2.3

([7], DefinitionÂ 7.2.1)

For any $$B(t)\in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$, we define

$$\nu _{q}(B)=\operatorname{dim}Z^{0}(B), \qquad i_{q}(B)=\operatorname{dim}Z^{-}(B).$$

We call $$\nu _{q}(B)$$ and $$i_{q}(B)$$ the nullity and index of B with respect to the bilinear form $$q_{B}(\cdot ,\cdot )$$, respectively.

### Proposition 2.4

([7], PropositionÂ 7.2.2)

For any $$B(t)\in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$, we have

$$\nu _{M}^{s}(B)=\nu _{q}(B), \qquad i_{M}^{s}(B)=i_{q}(B).$$

For any $$B_{1},B_{2}\in L^{\infty }([0,1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$, we write $$B_{1}\leq B_{2}$$ if $$B_{1}(t)\leq B_{2}(t)$$ for a.e. $$t\in [0,1]$$ and define $$B_{1}< B_{2}$$ if $$B_{1}\leq B_{2}$$ and $$B_{1}(t)< B_{2}(t)$$ on a subset of $$(0,1)$$ of positive measure.

### Proposition 2.5

1. (1)

For any $$B\in L^{\infty }([0,1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$, we have that $$Z^{0}(B)$$ is the solution subspace of systems (2.1)â€“(2.2), and $$\nu _{M}^{s}(B)\in \{0,1,2,\ldots ,2n\}$$ ([6], PropositionÂ 2.4.2(1); [7],Â CorollaryÂ 7.2.2(i)).

2. (2)

For any $$B_{1},B_{2}\in L^{\infty }([0,1],\mathcal{L} _{s}(\mathbf{R}^{n}))$$, if $$B_{1}\leq B_{2}$$, then $$i_{M}^{s}(B_{1})\leq i _{M}^{s}(B_{2})$$ and $$i_{M}^{s}(B_{1})+\nu _{M}^{s}(B_{1}) \leq i_{M} ^{s}(B_{2})+\nu _{M}^{s}(B_{2})$$; if $$B_{1}< B_{2}$$, then $$i_{M}^{s}(B _{1})+\nu _{M}^{s}(B_{1}) \leq i_{M}^{s}(B_{2})$$ ([6],Â PropositionÂ 2.4.2(2); [7], CorollaryÂ 7.2.2(ii)).

3. (3)

For any $$B_{1},B_{2}\in L^{\infty }([0,1],\mathcal{L} _{s}(\mathbf{R}^{n}))$$, if $$B_{1}(t)< B_{2}(t)$$ for a.e. $$t\in [0,1]$$, then

$$i_{M}^{s}(B_{2})-i_{M}^{s}(B_{1})= \sum_{\lambda \in [0,1)}\nu _{M}^{s} \bigl(B_{1}+\lambda (B_{2}-B_{1})\bigr).$$

The summand denoted by $$I_{M}^{s}(B_{1},B_{2})$$ is called the relative Morse index between $$B_{1}$$ and $$B_{2}$$ with respect to $$q_{B}(\cdot ,\cdot )$$ ([7], PropositionÂ 7.2.2(iii)).

4. (4)

(PoincarÃ© inequality) For any $$B\in L^{\infty }([0,1], \mathcal{L}_{s}(\mathbf{R}^{n}))$$, if $$i_{M}^{s}(B)=0$$, then

$$q_{B}(x,x)\geq 0, \quad x\in Z,$$

and the equality holds if and only if $$x\in Z^{0}(B)$$ ([7], PropositionÂ 7.2.2(v)).

5. (5)

For any $$B_{1},B_{2}\in L^{\infty }([0,1],\mathcal{L} _{s}(\mathbf{R}^{n}))$$, if $$B_{1}< B_{2}$$ and $$i_{M}^{s}(B_{2})=i_{M}^{s}(B _{1})+\nu _{M}^{s}(B_{1})$$, then $$Z=Z^{-}(B_{1})\oplus Z^{0}(B_{1}) \oplus Z^{0}(B_{2})\oplus Z^{+}(B_{2})$$, and $$(-q_{B_{1}}(x_{1},x_{1}))^{ \frac{1}{2}}+(q_{B_{2}}(x_{2},x_{2}))^{\frac{1}{2}}$$ is an equivalent norm on Z for $$x=x_{1}+x_{2}$$ with $$x_{1}\in Z^{-}(B_{1})$$ and $$x_{2}\in Z^{+}(B_{2})$$.

### Proof

We only prove (5). Let $$Z_{1}=Z^{-}(B_{1})\oplus Z^{0}(B_{1})$$, $$Z _{2}=Z^{0}(B_{2})\oplus Z^{+}(B_{2})$$. Noticing that $$q_{B_{1}}(x,x) \geq q_{B_{2}}(x,x)$$ for all $$x\in Z$$, $$q_{B_{1}}(x,x)\leq 0$$ for all $$x\in Z_{1}$$, and $$q_{B_{2}}(x,x)\geq 0$$ for all $$x\in Z_{2}$$, if $$x\in Z_{1}\cap Z_{2}$$, we have $$q_{B_{2}}(x,x)=0=q_{B_{1}}(x,x)$$. It follows that $$x\in Z^{0}(B_{2})\cap Z^{0}(B_{1})$$ and $$x(t)=0$$ on a subset of $$[0,1]$$ of positive measure, and hence $$x=0$$ via $$(1)$$. Thus $$Z_{1}\cap Z_{2}=\{\theta \}$$. It remains to prove that $$Z=Z_{1}+ Z_{2}$$. By PropositionÂ 2.1 we have $$Z= Z_{2}\oplus Z ^{-}(B_{2})$$, and for any $$x\in Z$$, there exists a unique pair $$(x_{1},x_{2})\in Z_{2}\times Z^{-}(B_{2})$$ such that $$x=x_{1}+x_{2}$$. Let $$\{e_{j}\}_{j=1}^{k}$$ be a basis of $$Z_{1}, e_{j}=e_{j}^{2}+e_{j} ^{-}$$ with $$e_{j}^{2}\in Z_{2}$$, $$e_{j}^{-}\in Z^{-}(B_{2})$$ for $$j=1, 2,\ldots$$â€‰, $$k=i_{M}^{s}(B_{1})+\nu _{M}^{s}(B_{1})$$. By $$i_{M}^{s}(B _{2})=i_{M}^{s}(B_{1})+\nu _{M}^{s}(B_{1})=k$$, to prove that $$\{e_{j}^{-}\}_{j=1}^{k}$$ is a basis of $$Z^{-}(B_{2})$$, we only need to show that $$\{e_{j}^{-}\}_{j=1}^{k}$$ is linearly independent. In fact, otherwise there would exist not all zero constants $$c_{1},\ldots ,c _{k}$$ such that $$\sum_{j=1}^{k}c_{j}e_{j}^{-}=0$$. This leads to $$\sum_{j=1}^{k}c_{j}e_{j}\in Z_{1}\cap Z_{2}$$, a contradiction. The linear independence shows that there exist constants $$\{\alpha _{j}\}_{j=1}^{k}$$ such that $$x_{2}=\sum_{j=1}^{k} \alpha _{j}e_{j}^{-}$$, and hence $$x=x_{1}+x_{2}=x=x_{1}+\sum_{j=1}^{k}\alpha _{j}e_{j}^{-}=\sum_{j=1}^{k}\alpha _{j}e_{j}+ (x _{1}-\sum_{j=1}^{k}\alpha _{j}e_{j}^{2})$$.

Similarly to the proof of PropositionÂ 7.2.2(iv) in [7], $$(-q_{B_{1}}(x_{1},x_{1}))^{\frac{1}{2}}+(q_{B_{2}}(x _{2},x_{2}))^{\frac{1}{2}}$$ is an equivalent norm on Z for $$x=x_{1}+x_{2}$$ with $$x_{1}\in Z^{-}(B_{1})$$, $$x_{2}\in Z^{+}(B_{2})$$.â€ƒâ–¡

### Proposition 2.6

For any $$B(t)\in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$ and $$Z=Z^{+}(B)\oplus Z^{0}(B)\oplus Z^{-}(B)$$, we have

$$q_{B}(x,y)=0, \quad x,y\in Z^{0}(B).$$

### Proof

By (2.3) and PropositionÂ 2.1, for any $$x,y\in Z^{0}(B)$$, we have

$$0=q_{B}(x+y,x+y)=2 \int ^{1}_{0} \bigl[\bigl(x',y' \bigr)-\bigl(B(t)x,y\bigr) \bigr]\,dt=2q _{B}(x,y),$$

which shows that $$q_{B}(x,y)=0$$ for all $$x,y\in Z^{0}(B)$$.â€ƒâ–¡

### Proposition 2.7

For any $$B(t)\in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$, if $$\nu _{M}^{s}(B)\neq 0$$, then there exists $$\varepsilon _{0}>0$$ such that $$\nu _{M}^{s}(B+\varepsilon _{0} I_{n})=0$$ and $$i_{M}^{s}(B+\varepsilon _{0} I_{n})=i_{M}^{s}(B)+\nu _{M}^{s}(B)$$.

### Proof

Clearly, $$B(t)< B(t)+\varepsilon I_{n}$$ for all $$t\in [0,1]$$ and $$\varepsilon >0$$. By (3) of PropositionÂ 2.5 we have

$$i_{M}^{s}(B+\varepsilon I_{n})-i_{M}^{s}(B)= \sum_{\lambda \in [0,1)}\nu _{M}^{s}(B+\lambda \varepsilon I_{n}).$$

Because $$i_{M}^{s}(B+\varepsilon I_{n})$$ is finite, there are only finitely many Î» such that $$\nu _{M}^{s}(B+\lambda \varepsilon I_{n})\neq 0$$ via (2) of PropositionÂ 2.5. Thus, since $$\nu _{M}^{s}(B)\neq 0$$, we can choose $$\varepsilon _{0}>0$$ such that $$\nu _{M}^{s}(B+\varepsilon _{0} I_{n})=0$$ and $$i_{M}^{s}(B+\varepsilon _{0} I_{n})-i_{M}^{s}(B)=\nu _{M}^{s}(B)$$.â€ƒâ–¡

### Remark 2.8

([6], ExampleÂ 2.4.3; [7], RemarkÂ 7.1.3)

Let $$\alpha _{1}\leq \alpha _{2}\leq \cdots \leq \alpha _{n}$$ be the eigenvalues of a constant $$n\times n$$ symmetric matrix B. Then

\begin{aligned}& i_{I_{n}}^{s}(B) = {}^{\#}\{k:\alpha _{k}>0\}+2\sum_{k=1}^{n} {{}^{\#} \bigl\{ j\in \mathbf{N}:4(j\pi )^{2}< \alpha _{k} \bigr\} }, \end{aligned}
(2.4)
\begin{aligned}& \nu _{I_{n}}^{s}(B) = {}^{\#}\{k:\alpha _{k}=0\}+2\sum_{k=1}^{n} {{}^{\#} \bigl\{ j\in \mathbf{N}:4(j\pi )^{2}=\alpha _{k} \bigr\} }, \end{aligned}
(2.5)
\begin{aligned}& i_{-I_{n}}^{s}(B) = 2\sum_{k=1}^{n}{{}^{\#} \bigl\{ j\in \mathbf{N}:\bigl((2j-1) \pi \bigr)^{2}< \alpha _{k}\bigr\} }, \end{aligned}
(2.6)
\begin{aligned}& \nu _{-I_{n}}^{s}(B) = 2\sum_{k=1}^{n}{{}^{\#} \bigl\{ j\in \mathbf{N}:\bigl((2j-1) \pi \bigr)^{2}=\alpha _{k}\bigr\} }, \end{aligned}
(2.7)

where #A denotes the number of elements in a set A. For $$\eta \in \mathbf{R}\setminus \{\pm 1,0\}$$ with $$\lambda _{0}=\arccos \frac{2}{ \eta ^{-1}+\eta }$$, we have

\begin{aligned}& \begin{aligned} i_{\eta I_{n}}^{s}(B) &=\sum _{k=1}^{n}{{}^{\#} \bigl\{ j\in \mathbf{N}:(2j \pi + \lambda _{0})^{2}< \alpha _{k}\bigr\} } \\ &\quad {}+\sum_{k=1}^{n}{{}^{\#} \bigl\{ j\in \mathbf{N}:(2\pi -\lambda _{0}+2j\pi )^{2}< \alpha _{k}\bigr\} }, \end{aligned} \\& \begin{aligned} \nu _{\eta I_{n}}^{s}(B) &=\sum _{k=1}^{n}{{}^{\#}\bigl\{ j\in \mathbf{N}:(2j \pi + \lambda _{0})^{2}=\alpha _{k}\bigr\} } \\ &\quad {}+\sum_{k=1}^{n}{{}^{\#} \bigl\{ j\in \mathbf{N}:(2\pi -\lambda _{0}+2j\pi )^{2}= \alpha _{k}\bigr\} }. \end{aligned} \end{aligned}

In particular, formulae (2.4) and (2.5) were given first by Mawhin and Willem in the book [10].

## 3 Proof of the main results

In this section, we give proofs of the main results. To this end, we define

$$I(x)= \int _{0}^{1} \biggl[-\frac{1}{2} \bigl\vert x' \bigr\vert ^{2}+\frac{1}{2} \bigl(B_{1}(t)x,x\bigr)+V(t,x) \biggr] \,dt,\quad x\in Z.$$
(3.1)

From assumption (A) it is easy to check that I is continuously differentiable and weakly upper semicontinuous on Z (see [6, 7, 10]), where

$$Z=\bigl\{ x\in H^{1}\bigl([0,1],\mathbf{R}^{n}\bigr)| x(1) = Mx(0) \bigr\}$$

is a Hilbert space with the norm

$$\Vert x \Vert ^{2}= \int _{0}^{1} \bigl\vert x' \bigr\vert ^{2}\,dt+ \int _{0}^{1} \vert x \vert ^{2}\,dt$$

for $$x\in Z$$. Clearly, for $$x\in Z$$, we have

$$\bigl\vert x(t) \bigr\vert \leq \bigl\Vert x(t) \bigr\Vert _{\infty }\leq \bigl\Vert x(t) \bigr\Vert .$$

Moreover, we have

$$I'(x)y= \int _{0}^{1} \bigl[-\bigl(x', y'\bigr)+\bigl(B_{1}(t)x,y\bigr)+\bigl(\nabla _{x}V(t,x), y\bigr) \bigr] \,dt,\quad x,y\in Z,$$

and $$I'$$ is weakly continuous. As in the proof of PropositionÂ 2.4.2(1) in [6], we can find that the critical points of I correspond to the solutions of (1.1) and omit the details.

### 3.1 Proof of Theorem 1.1

To prove TheoremÂ 1.1, we need the following critical point theorem without the compactness assumptions.

### Lemma 3.1

([17], TheoremÂ 1.1)

Let $$X_{1}$$ and $$X_{2}$$ be reflexive Banach spaces, and let $$\varphi \in C^{1}(X_{1}\times X_{2},\mathbf{R})$$ be such that $$\varphi (x_{1},\cdot )$$ is weakly upper semicontinuous for all $$x_{1}\in X_{1}$$, $$\varphi (\cdot ,x_{2}):X_{1}\to \mathbf{R}$$ is convex for all $$x_{2}\in X_{2}$$, and $$\varphi '$$ is weakly continuous. Assume that

$$\varphi (\theta ,x_{2})\to -\infty$$
(3.2)

as $$\|x_{2}\|\to +\infty$$ and, for every $$M>0$$,

$$\varphi (x_{1},x_{2})\to +\infty$$
(3.3)

as $$\|x_{1}\|\to +\infty$$ uniformly for $$\|x_{2}\|\leq M$$. Then Ï† has at least one critical point.

### Proof of TheoremÂ 1.1

By assumption (A1), Propositions 2.1â€“2.4, and DefinitionÂ 2.3 we have $$Z=Z^{0}(B_{1})\oplus Z^{+}(B_{1})$$. Set $$X_{1}=Z^{0}(B_{1})$$, $$X_{2}=Z ^{+}(B_{1})$$, $$x\in Z$$, $$x=x_{1}+x_{2}$$ with $$x_{1}\in X_{1}$$ and $$x_{2}\in X_{2}$$. Next, we divide the proof into three steps.

Step 1. It is obvious that $$V (t, x_{1}(t) + x_{2}(t))$$ is convex in $$x_{1}(t)\in X_{1}$$, so is $$\int _{0}^{1}V(t,x_{1}(t)+x_{2}(t))\,dt$$. From (2.3) and PropositionÂ 2.1 we can see that for every $$x_{2}(t)\in X_{2}$$,

$$I(x_{1}+x_{2})= \int _{0}^{1} \biggl[-\frac{1}{2} \bigl\vert x_{2}'(t) \bigr\vert ^{2}+ \frac{1}{2} \bigl(B_{1}(t)x_{2}(t),x_{2}(t)\bigr)+V \bigl(t,x_{1}(t)+x_{2}(t)\bigr) \biggr] \,dt$$

is convex in $$x_{1}\in X_{1}$$.

Step 2. We prove that (3.3) of LemmaÂ 3.1 holds. By assumption (A) and the convexity of $$V(t,\cdot )$$ we can see that there exists $$c_{1}>0$$ such that

\begin{aligned}& \int _{0}^{1}V\bigl(t,x_{1}(t)+x_{2}(t) \bigr)\,dt \\& \quad \geq 2 \int _{0}^{1}V\biggl(t,\frac{1}{2}x_{1}(t) \biggr)\,dt- \int _{0}^{1}V\bigl(t,-x_{2}(t)\bigr)\,dt \\& \quad \geq 2 \int _{0}^{1}V\biggl(t,\frac{1}{2}x_{1}(t) \biggr)\,dt- \int _{0}^{1}a\bigl( \bigl\vert x_{2}(t) \bigr\vert \bigr)b(t)\,dt \\& \quad \geq 2 \int _{0}^{1}V\biggl(t,\frac{1}{2}x_{1}(t) \biggr)\,dt-\max_{0\leq u\leq \Vert x_{2} \Vert _{\infty }}a(u) \int _{0}^{1}b(t)\,dt \\& \quad \geq 2 \int _{0}^{1}V\biggl(t,\frac{1}{2}x_{1}(t) \biggr)\,dt-\max_{0\leq u\leq c_{1} \Vert x_{2} \Vert }a(u) \int _{0}^{1}b(t)\,dt \\& \quad \geq 2 \int _{0}^{1}V\biggl(t,\frac{1}{2}x_{1}(t) \biggr)\,dt-\max_{0\leq u\leq c_{1}M}a(u) \int _{0}^{1}b(t)\,dt \end{aligned}

for all $$x_{1}\in X_{1}$$ and $$x_{2}\in X_{2}$$ with $$\|x_{2}\|\leq M$$. Note that $$\|x\|^{2}=\|x'\|^{2}_{L^{2}}+\|x\|^{2}_{L^{2}}$$ and $$B_{1}(t)\in L^{\infty }([0, 1],\mathcal{L}_{s} (\mathbf{R}^{n}))$$. By (2.3) and PropositionÂ 2.1 we know that there exists $$c_{2}>0$$ such that

\begin{aligned}& I(x_{1}+x_{2}) \\& \quad \geq -\frac{1}{2} \bigl\Vert x_{2}' \bigr\Vert ^{2}_{L^{2}}-\frac{1}{2}c_{2} \Vert x_{2} \Vert ^{2}_{L^{2}}+ 2 \int _{0}^{1}V\biggl(t,\frac{1}{2}x_{1}(t) \biggr)\,dt-\max_{0\leq u\leq c_{1}M}a(u) \int _{0}^{1}b(t)\,dt \\& \quad \geq -\frac{1+c_{2}}{2}M^{2}+ 2 \int _{0}^{1}V\biggl(t,\frac{1}{2}x_{1}(t) \biggr)\,dt- \max_{0\leq u\leq c_{1}M}a(u) \int _{0}^{1}b(t)\,dt \end{aligned}

for all $$x_{1}\in X_{1}$$ and $$x_{2}\in X_{2}$$ with $$\|x_{2}\|\leq M$$. By assumption (A2) it is easy to see that (3.3) of LemmaÂ 3.1 holds.

Step 3. We check (3.2) of LemmaÂ 3.1. If not, there exist a constant $$c_{3}$$ and a sequence $$x_{2,n}$$ in $$X_{2}$$ such that $$\|x_{2,n}\|\to +\infty$$ as $$n\to \infty$$ and

$$I(x_{2,n})\geq c_{3}$$
(3.4)

for all n. Notice that $$\nu _{M}^{s}(B_{2})\neq 0$$ and $$i_{M}^{s}(B _{2})=i_{M}^{s}(B_{1})+\nu _{M}^{s}(B_{1})$$ in (A3). By (A1) and (5) of PropositionÂ 2.5 we have $$Z= Z^{0}(B_{1})\oplus Z^{0}(B _{2})\oplus Z^{+}(B_{2})$$ and $$X_{2}=Z^{0}(B_{2})\oplus Z^{+}(B_{2})$$. Let $$x_{2,n}=u_{n}+v_{n}$$, $$u_{n}\in Z^{0}(B_{2})$$, $$v_{n}\in Z^{+}(B_{2})$$. Then by (A3), (3.4), (2.3), and PropositionÂ 2.1, we have

\begin{aligned} c_{3} \leq & I(x_{2,n})\leq \int _{0}^{1} \biggl[-\frac{1}{2} \bigl\vert x'_{2,n} \bigr\vert ^{2}+ \frac{1}{2} \bigl(B_{2}(t)x_{2,n},x_{2,n}\bigr) \biggr]\,dt+ \int _{0}^{1}\gamma (t)\,dt \\ =&-q_{B_{2}}(v_{n},v_{n})+ \int _{0}^{1}\gamma (t)\,dt, \end{aligned}

which shows that $$\{v_{n}\}$$ is bounded since $$(-q_{B_{1}}(x_{1},x _{1}))^{\frac{1}{2}}+(q_{B_{2}}(x_{2},x_{2}))^{\frac{1}{2}}$$ is an equivalent norm on Z for $$x=x_{1}+x_{2}$$ with $$x_{1}\in Z^{-}(B_{1})$$and $$x_{2}\in Z^{+}(B_{2})$$, where $$Z^{-}(B _{1})=\{\theta \}$$. Since $$\|x_{2,n}\|\leq \|u_{n}\|+\|v_{n}\|$$, we have $$\|u_{n}\|\to \infty$$ as $$n\to +\infty$$. Set

$$E= \biggl\{ t\in [0,1]\Big| V(t,x)-\frac{1}{2}\bigl(\bigl(B_{2}(t)-B _{1}(t)\bigr)x,x\bigr)\to -\infty \mbox{ as } \Vert \bar{x} \Vert \to \infty \biggr\} ,$$

where $$x=\tilde{x}+\bar{x}$$ and $$\bar{x}\in Z^{0}(B_{2})$$. Noting that $$x_{2,n}\in X_{2}=Z^{0}(B_{2})\oplus Z^{+}(B_{2})$$, we have $$q_{B_{2}}(x_{2,n},x_{2,n})\geq 0$$ for all n via PropositionÂ 2.1. From the Lebesgueâ€“Fatou lemma we have

\begin{aligned}& \limsup_{n\to \infty } I(x_{2,n}) \\& \quad \leq \limsup_{n\to \infty } \int ^{1}_{0} \biggl[V(t,x_{2,n})- \frac{1}{2}\bigl(\bigl(B_{2}(t)-B_{1}(t) \bigr)x_{2,n},x_{2,n}\bigr) \biggr]\,dt \\& \quad \leq \limsup_{n\to \infty } \int _{E} \biggl[V(t,x_{2,n})- \frac{1}{2} \bigl(\bigl(B_{2}(t)-B_{1}(t)\bigr)x_{2,n},x_{2,n} \bigr) \biggr]\,dt+ \int ^{1} _{0}\gamma (t)\,dt \\& \quad \to -\infty \end{aligned}

via (A3), which contradicts (3.4). Hence (3.2) of LemmaÂ 3.1 holds.

By LemmaÂ 3.1 I has at least one critical point. Hence problem (1.1) has at least one solution in Z. The proof is complete.â€ƒâ–¡

### 3.2 Proof of Theorem 1.2

To prove TheoremÂ 1.2, we need the following saddle point reduction theorem under rather general assumptions.

### Lemma 3.2

([1], TheoremÂ 2.3)

Let Y, $$X_{1}$$, $$X_{2}$$ be Hilbert spaces, and let $$\psi \in C^{1}(Y\times X_{1}\times X_{2}, \mathbf{R})$$. Suppose that Ïˆ satisfies the following conditions:

1. (1)

$$D_{1}\psi (\cdot ,x_{1},x_{2}):Y\to Y$$ is Î¼-monotone for all $$(x_{1},x_{2})\in X_{1}\times X_{2}$$, that is, there exists $$\mu >0$$ such that

$$\bigl\langle D_{1}\psi (y_{1},x_{1},x_{2})-D_{1} \psi (y_{2},x_{1},x_{2}),y _{1}-y_{2} \bigr\rangle \geq \mu \Vert y_{1}-y_{2} \Vert ^{2}, \quad y_{1},y_{2}\in Y;$$
2. (2)

$$-D_{2}\psi (y,\cdot ,x_{2}):X_{1}\to X_{1}$$ is Î¼-monotone for all $$(y,x_{2})\in Y\times X_{2}$$.

Then there exists a map $$\phi \in C(X_{2},Y\times X_{1})$$ such that $$\phi (x_{2})=(y(x_{2}),x_{1}(x_{2}))$$ is the unique saddle point of $$\psi (\cdot ,\cdot ,x_{2})$$ for every $$x_{2}\in X_{2}$$. Moreover, the map $$\varphi :X_{2}\to \mathbf{R}$$ defined by

$$\varphi (x_{2})=\psi \bigl(y(x_{2}),x_{1}(x_{2}),x_{2} \bigr)=\min_{y \in Y}\sup_{x_{1}\in X_{1}}\psi (y,x_{1},x_{2})$$
(3.5)

is continuously differentiable, and its derivative is given by

$$\varphi '(x_{2})=D_{3}\psi \bigl(y(x_{2}),x_{1}(x_{2}),x_{2}\bigr) \quad \textit{for every } x_{2}\in X_{2}.$$
(3.6)

### Proof of TheoremÂ 1.2

By assumption (A1), Propositions 2.1â€“2.4, and DefinitionÂ 2.3 we have $$Z=Z^{0}(B_{1})\oplus Z^{+}(B_{1})$$. Set $$Y=\{\theta \}$$, $$X_{1}=Z^{0}(B _{1})$$, $$X_{2}=Z^{+}(B_{1})$$. We define the functional Ï† as follows:

$$\varphi (x_{2})=\sup_{x_{1}\in X_{1}}\psi (x_{1}+x_{2})= \sup_{x_{1}\in X_{1}}-I(x_{1}+x_{2}), \quad x_{2}\in X_{2}.$$

By assumption (A) and the convexity of $$V(t,x)-\frac{1}{2}\mu (t)|x|^{2}$$ in x for a.e. $$t\in [0,1]$$ we have

$$\bigl(\nabla _{x}V(t,x)-\nabla _{x}V(t,y),x-y\bigr)\geq \mu (t) \vert x-y \vert ^{2}, \quad x,y\in Z.$$

Thus for each fixed $$x_{2}\in X_{2}$$ and any $$x_{1,1},x_{1,2}\in X _{1}$$, we have

\begin{aligned}& \int ^{1}_{0}\bigl(\nabla _{x}V(t,x_{1,1}+x_{2})- \nabla _{x}V(t,x_{1,2}+x _{2}),x_{1,1}-x_{1,2} \bigr)\,dt \\& \quad \geq \mu \int ^{1}_{0} \vert x_{1,1}-x_{1,2} \vert ^{2}\,dt \end{aligned}
(3.7)

for all $$x,y\in Z$$, where $$\mu =\inf_{t\in [0,1]}\mu (t)>0$$. Since $$Z=X_{1}\oplus X_{2}=Z^{0}(B_{1})\oplus Z^{+}(B_{1})$$, from (2.3) and Propositions 2.1, and 2.6 we know that

\begin{aligned}& \bigl\langle -\psi '(x_{1,1}+x_{2})-\bigl(-\psi '(x_{1,2}+x_{2})\bigr),x_{1,1}-x _{1,2}\bigr\rangle \\& \quad = \bigl\langle I'(x_{1,1}+x_{2})-I'(x_{1,2}+x_{2}),x_{1,1}-x_{1,2} \bigr\rangle \\& \quad = \int ^{1}_{0}\bigl(\nabla _{x}V(t,x_{1,1}+x_{2})- \nabla _{x}V(t,x_{1,2}+x _{2}),x_{1,1}-x_{1,2} \bigr)\,dt \\& \quad \geq \mu \int ^{1}_{0} \vert x_{1,1}-x_{1,2} \vert ^{2}\,dt. \end{aligned}

Noticing that $$X_{1}=Z^{0}(B_{1})$$ is finite-dimensional, we can see that there exists $$c_{4}>0$$ such that

$$\bigl\langle -\psi '(x_{1,1}+x_{2})-\bigl(-\psi '(x_{1,2}+x_{2})\bigr),x_{1,1}-x_{1,2} \bigr\rangle \geq c_{4}\mu \Vert x_{1,1}-x_{1,2} \Vert ^{2}.$$

By LemmaÂ 3.2 there exists a continuous mapping $$\phi : X_{2} \to X_{1}$$ such that $$\varphi (x_{2})=\psi (\phi (x_{2})+x_{2})$$ for all $$x_{2}\in X_{2}$$, $$\varphi :X_{2}\to \mathbf{R}$$ is continuously differentiable, and $$\varphi '(x_{2})=\psi '(\phi (x_{2})+x_{2})|_{X _{2}}$$ for $$x_{2}\in X_{2}$$. Hence $$x_{2}\in X_{2}$$ is a critical point of Ï†, which shows that $$\phi (x_{2})+x_{2}$$ is a critical point of Ïˆ and I.

Further, for every $$x_{2}\in X_{2}$$, by assumption (A4) we have

\begin{aligned} \biggl\vert \int ^{1}_{0}\bigl(V(t,x_{2})-V(t,\theta ) \bigr)\,dt \biggr\vert =& \biggl\vert \int ^{1}_{0} \int ^{1}_{0}\bigl(\nabla _{x}V(t,sx_{2}),x_{2} \bigr)\,ds\,dt \biggr\vert \\ \leq &\frac{1}{2} \int ^{1}_{0}f(t) \vert x_{2} \vert ^{2}\,dt+ \int ^{1}_{0}g(t) \vert x _{2} \vert \,dt. \end{aligned}

Thus,

\begin{aligned} \varphi (x_{2}) \geq &\psi (x_{2})=-I(x_{2}) \\ \geq &\frac{1}{2} \int _{0}^{1} \bigl[ \bigl\vert x'_{2} \bigr\vert ^{2}-\bigl(B_{1}(t)x_{2},x _{2}\bigr)-f(t) \vert x_{2} \vert ^{2} \bigr] \,dt \\ &{}- \int ^{1}_{0}g(t) \vert x_{2} \vert \,dt- \int ^{1}_{0}V(t,\theta )\,dt. \end{aligned}

Since $$\nu _{M}^{s}(B_{1}+f(t)I_{n})=0$$ and $$i_{M}^{s}(B_{1}+f(t)I_{n})=i _{M}^{s}(B_{1})+\nu _{M}^{s}(B_{1})$$, by (5) of PropositionÂ 2.5 we know that $$(q_{B_{1}+fI_{n}}(x_{2},x_{2}))^{\frac{1}{2}}$$ is an equivalent norm on Z for $$x=x_{2}$$ with $$x_{2}\in X_{2}=Z^{+}(B_{1})=Z ^{+}(B_{1}+fI_{n})$$. Hence there exist $$c_{5}, c_{6}, c_{7}>0$$ such that

\begin{aligned} \varphi (x_{2}) \geq &\frac{c_{5}}{2} \Vert x_{2} \Vert ^{2}- \Vert x_{2} \Vert _{\infty } \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,\theta )\,dt \\ \geq &\frac{c_{5}}{2} \Vert x_{2} \Vert ^{2}-c_{6} \Vert x_{2} \Vert -c_{7} \\ \to &+\infty \end{aligned}

as $$\|x_{2}\|\to +\infty$$. Consequently, there exists $$x_{2,0}\in X _{2}$$ such that $$\varphi (x_{2,0})=\min_{x_{2}\in X_{2}}\varphi (x_{2})$$, and hence $$x_{2,0}+\phi (x_{2,0})$$ is a solution with saddle point character of problem (1.1) in Z.

If condition (A6) holds, then $$Z=Z^{-}(B_{01})\oplus Z^{0}(B_{01}) \oplus Z^{0}(B_{02})\oplus Z^{+}(B_{02})$$ via (5) of PropositionÂ 2.5. Since $$B_{01}>B_{1}$$, there exists $$E\subset [0,1]$$ with $$\operatorname{meas}E>0$$ such that $$B_{01}(t)>B_{1}(t)$$ for all $$t\in E$$. Hence from (2.3) we have

\begin{aligned} q_{B_{1}}(x,x) =& \int _{0}^{1} \bigl\vert x' \bigr\vert ^{2}\,dt- \int _{E}\bigl(B_{1}(t)x,x\bigr) \,dt- \int _{[0,1]\setminus E}\bigl(B_{1}(t)x,x\bigr)\,dt \\ >& \int _{0}^{1} \bigl\vert x' \bigr\vert ^{2}\,dt- \int _{E}\bigl(B_{01}(t)x,x\bigr) \,dt- \int _{[0,1]\setminus E}\bigl(B_{01}(t)x,x\bigr)\,dt \\ =&q_{B_{01}}(x,x) \end{aligned}

for all $$x\in Z\setminus \{\theta \}$$, which implies that $$q_{B_{1}}(x,x)> q_{B_{01}}(x,x)$$ for all $$x\in Z\setminus \{\theta \}$$ and $$q_{B_{1}}(x,x)>0$$ for all $$x\in Z^{0}(B_{01})\setminus \{ \theta \}$$ with $$Z^{0}(B_{01})\subset Z^{+}(B_{1})$$. Let $$X_{2,1}=(Z ^{-}(B_{01})\oplus Z^{0}(B_{01}))\cap Z^{+}(B_{1})$$. Then we can suppose that $$X_{2,2}$$ is the orthogonal complement of $$X_{2,1}$$ in $$X_{2}$$. We claim that $$\phi (\theta )=\theta$$. Indeed, (A6) implies $$V(t,\theta )=0$$ and $$\nabla _{x}V(t,\theta )=\theta$$ for a.e. $$t\in [0,1]$$. From condition (A5) and (3.7) we have

\begin{aligned} 0 =&\bigl\langle \psi '\bigl(\phi (\theta )\bigr),-\phi (\theta ) \bigr\rangle =\bigl\langle -I'\bigl( \phi (\theta )\bigr),-\phi ( \theta )\bigr\rangle \\ =& \int _{0}^{1}\bigl(-\nabla _{x}V\bigl(t, \phi (\theta )\bigr),-\phi (\theta )\bigr)\,dt \\ =& \int _{0}^{1}\bigl(\nabla _{x}V(t,\theta )-\nabla _{x}V\bigl(t,\phi (\theta )\bigr),- \phi (\theta )\bigr)\,dt \\ \geq &\mu \int _{0}^{1} \bigl\vert \phi (\theta ) \bigr\vert ^{2}\,dt \\ \geq &0, \end{aligned}

which shows that $$\phi (\theta )=\theta$$. From the continuity of Ï•, we know that there exists $$0<\delta <r$$ such that $$\|\phi (x _{2})\|< r$$ as $$\|x_{2}\|\leq \delta$$. Consequently, from (A6) and (2.3) we obtain

\begin{aligned} \varphi (x_{2,1}) =&\psi \bigl(x_{2,1}+\phi (x_{2,1}) \bigr)=-I\bigl(x_{2,1}+\phi (x _{2,1})\bigr) \\ \leq &\frac{1}{2} \int ^{1}_{0} \bigl[ \bigl\vert x'_{2,1}+\bigl(\phi (x_{2,1})\bigr)' \bigr\vert ^{2}-\bigl(B _{01}(t) \bigl(x_{2,1}+\phi (x_{2,1})\bigr),x_{2,1}+\phi (x_{2,1})\bigr) \bigr] \,dt \\ =& \frac{1}{2}q_{B_{01}}\bigl(x_{2,1}+\phi (x_{2,1}),x_{2,1}+\phi (x_{2,1})\bigr) \\ \leq &0 \end{aligned}

for all $$x_{2,1}\in X_{2,1}$$ with $$\|x_{2,1}\|\leq \delta$$ via $$B_{01}>B_{1}$$ and $$x_{2,1}+\phi (x_{2,1})\in (Z^{-}(B_{01})\oplus Z ^{0}(B_{01}))\cup Z^{0}(B_{1})$$, and

\begin{aligned} \varphi (x_{2,2}) \geq &\psi (x_{2,2})=-I(x_{2,2}) \\ \geq &\frac{1}{2} \int ^{1}_{0} \bigl[ \bigl\vert x'_{2,2} \bigr\vert ^{2}-\bigl(B_{02}(t)x_{2,2},x _{2,2}\bigr) \bigr]\,dt \\ =&\frac{1}{2} q_{B_{02}}(x_{2,2},x_{2,2}) \\ \geq &0 \end{aligned}

for all $$x_{2,2}\in X_{2,2}$$ with $$\|x_{2,2}\|\leq \delta$$ via $$X_{2,2}=Z^{+}(B_{01})\cap Z^{+}(B_{1})$$ and $$Z^{+}(B_{01})=Z^{+}(B _{02})\oplus Z^{0}(B_{02})$$.

Since I is weakly upper semicontinuous on Z, Ï† id weakly lower semicontinuous on $$X_{2}$$. By the coerciveness and weak lower semicontinuity of Ï† we see that satisfies (PS)-condition and is bounded below.

If $$\inf \{\varphi (x_{2}):x_{2}\in X_{2}\} =0$$, then all $$x_{2,1} \in X_{2,1}$$ with $$\|x_{2,1}\|\leq \delta$$ are minima of Ï†, which shows that Ï† has infinitely many critical points. If $$\inf \{\varphi (x_{2}):x_{2}\in X_{2}\}<0$$, then Ï† has at least two nonzero critical points via TheoremÂ 4 in [2]. Thus problem (1.1) has at least two nontrivial solutions in Z. In addition, since $$V(t,\theta )=0$$ for a.e. $$t\in [0,1]$$, we know that problem (1.1) has trivial solution Î¸. Hence problem (1.1) has three distinct solutions in Z. The proof is complete.â€ƒâ–¡

### 3.3 Proof of Theorem 1.3

In the section, we use the saddle point theorem (see TheoremÂ 4.6, [12] or [10]) and a generalization of the mountain pass theorem (see TheoremÂ 5.29 and ExampleÂ 5.26 in [12]) to prove TheoremÂ 1.3.

### Proof of TheoremÂ 1.3

First, we verify that I satisfies the (PS)-condition. Suppose that $$I'(x_{n})\to 0$$ as $$n\to +\infty$$ and $$I(x_{n})$$ is bounded. From condition (A1) we have $$Z=Z^{0}(B_{1})\oplus Z^{+}(B_{1})$$. Set $$x_{n}=\bar{x}_{n}+\tilde{x} _{n}$$ and $$\bar{x}_{n}\in Z^{0}(B_{1})$$, $$\tilde{x}_{n}\in Z^{+}(B _{1})$$. By assumption ($$\mathrm{A}'_{4}$$) we have

\begin{aligned}& \biggl\vert \int ^{1}_{0}\bigl(\nabla _{x}V(t,x_{n}), \tilde{x}_{n}\bigr)\,dt \biggr\vert \\& \quad \leq \int ^{1}_{0}f(t) \vert \bar{x}_{n}+ \tilde{x}_{n} \vert ^{\alpha } \vert \tilde{x}_{n} \vert \,dt+ \int ^{1}_{0}g(t) \vert \tilde{x}_{n} \vert \,dt \\& \quad \leq \int ^{1}_{0}f(t)2\bigl( \vert \bar{x}_{n} \vert ^{\alpha }+ \vert \tilde{x}_{n} \vert ^{ \alpha }\bigr) \vert \tilde{x}_{n} \vert \,dt+ \int ^{1}_{0}g(t) \vert \tilde{x}_{n} \vert \,dt \\& \quad \leq 2 \biggl( \int ^{1}_{0}f^{2}(t)\,dt \biggr)^{\frac{1}{2}} \biggl( \int ^{1}_{0} \vert \tilde{x}_{n} \vert ^{2} \vert \bar{x}_{n} \vert ^{2\alpha }\,dt \biggr)^{ \frac{1}{2}}+ 2 \int ^{1}_{0}f(t) \vert \tilde{x}_{n} \vert ^{1+\alpha }\,dt \\& \qquad {} + \int ^{1}_{0}g(t) \vert \tilde{x}_{n} \vert \,dt \\& \quad \leq 2\beta _{0} \Vert \bar{x}_{n} \Vert _{\infty }^{\alpha } \biggl( \int ^{1} _{0} \vert \tilde{x}_{n} \vert ^{2}\,dt \biggr)^{\frac{1}{2}}+ 2 \int ^{1}_{0}f(t) \vert \tilde{x}_{n} \vert ^{1+\alpha }\,dt+ \int ^{1}_{0}g(t) \vert \tilde{x}_{n} \vert \,dt \\& \quad \leq \varepsilon \beta _{0} \int ^{1}_{0} \vert \tilde{x}_{n} \vert ^{2}\,dt+\frac{ \beta _{0}}{\varepsilon } \Vert \bar{x}_{n} \Vert _{\infty }^{2\alpha }+ 2 \Vert \tilde{x}_{n} \Vert _{\infty }^{1+\alpha } \int ^{1}_{0}f(t)\,dt \\& \qquad {} + \Vert \tilde{x}_{n} \Vert _{\infty } \int ^{1}_{0}g(t)\,dt \end{aligned}
(3.8)

for all n, where $$\beta _{0}=(\int ^{1}_{0}f^{2}(t)\,dt)^{\frac{1}{2}}$$ and $$\varepsilon >0$$. Thus, from $$\bar{x}_{n}\in Z^{0}(B_{1})$$, $$\tilde{x}_{n}\in Z^{+}(B_{1})$$, (2.3), and PropositionÂ 2.1 we have

\begin{aligned} \Vert \tilde{x}_{n} \Vert \geq& \bigl\langle -I'(x_{n}), \tilde{x}_{n}\bigr\rangle \\ \geq & \int _{0}^{1} \bigl[\bigl(x'_{n}, \tilde{x}'_{n}\bigr)-\bigl(B_{1}(t)x_{n}, \tilde{x}_{n}\bigr)-\varepsilon \beta _{0} \vert \tilde{x}_{n} \vert ^{2} \bigr] \,dt-\frac{ \beta _{0}}{\varepsilon } \Vert \bar{x}_{n} \Vert _{\infty }^{2\alpha } \\ &{}-2 \Vert \tilde{x}_{n} \Vert _{\infty }^{1+\alpha } \int ^{1}_{0}f(t)\,dt- \Vert \tilde{x}_{n} \Vert _{\infty } \int ^{1}_{0}g(t)\,dt \\ =& \int _{0}^{1} \bigl[ \bigl\vert \tilde{x}'_{n} \bigr\vert ^{2}- \bigl(B_{1}(t)\tilde{x}_{n}, \tilde{x}_{n}\bigr)- \varepsilon \beta _{0} \vert \tilde{x}_{n} \vert ^{2} \bigr] \,dt-\frac{ \beta _{0}}{\varepsilon } \Vert \bar{x}_{n} \Vert _{\infty }^{2\alpha } \\ &{}-2 \Vert \tilde{x}_{n} \Vert _{\infty }^{1+\alpha } \int ^{1}_{0}f(t)\,dt- \Vert \tilde{x}_{n} \Vert _{\infty } \int ^{1}_{0}g(t)\,dt \end{aligned}

for n large enough. By PropositionÂ 2.7 we can choose $$\varepsilon _{0}>0$$ such that $$\nu _{M}^{s}(B_{1}+\varepsilon _{0}\beta _{0}I_{n})=0$$ and $$i_{M}^{s}(B_{1}+\varepsilon _{0}\beta _{0}I_{n})=i _{M}^{s}(B_{1})+\nu _{M}^{s}(B_{1})$$. From (5) of PropositionÂ 2.5 we know that $$(q_{B_{1}+\varepsilon _{0}\beta _{0}I_{n}}(x_{2},x_{2}))^{ \frac{1}{2}}$$ is an equivalent norm on Z for $$x=x_{2}$$ with $$x_{2}\in Z^{+}(B_{1})=Z^{+}(B_{1}+\varepsilon _{0}\beta _{0}I_{n})$$. Hence there exist $$c_{8}, c_{9}, c_{10}, c_{11}>0$$ such that

$$\Vert \tilde{x}_{n} \Vert + c_{9} \Vert \bar{x}_{n} \Vert ^{2\alpha }+c_{10} \Vert \tilde{x} _{n} \Vert ^{1+\alpha }+c_{11} \Vert \tilde{x}_{n} \Vert \geq c_{8} \Vert \tilde{x}_{n} \Vert ^{2},$$

which implies that there are $$k_{1}>0$$ and $$k_{2}>0$$ such that

$$k_{1} \Vert \bar{x}_{n} \Vert ^{2\alpha }+k_{2}\geq \Vert \tilde{x}_{n} \Vert ^{2}.$$
(3.9)

In a way similar to (3.8), for all n, we obtain

\begin{aligned}& \biggl\vert \int ^{1}_{0}\bigl(V(t,x_{n})-V(t, \bar{x}_{n})\bigr)\,dt \biggr\vert \\& \quad = \biggl\vert \int ^{1}_{0} \int ^{1}_{0}\bigl(\nabla _{x}V(t, \bar{x}_{n}+s \tilde{x}_{n}),\tilde{x}_{n}\bigr)\,ds \,dt \biggr\vert \\& \quad \leq \int ^{1}_{0} \int ^{1}_{0}f(t) \vert \bar{x}_{n}+s \tilde{x}_{n} \vert ^{ \alpha } \vert \tilde{x}_{n} \vert \,ds\,dt+ \int ^{1}_{0} \int ^{1}_{0}g(t) \vert \tilde{x} _{n} \vert \,ds\,dt \\& \quad \leq \int ^{1}_{0}2f(t) \biggl( \vert \bar{x}_{n} \vert ^{\alpha }+\frac{1}{1+\alpha } \vert \tilde{x}_{n} \vert ^{\alpha }\biggr) \vert \tilde{x}_{n} \vert \,dt+ \int ^{1}_{0}g(t) \vert \tilde{x}_{n} \vert \,dt \\& \quad \leq 2 \biggl( \int ^{1}_{0}f^{2}(t)\,dt \biggr)^{\frac{1}{2}} \biggl( \int ^{1}_{0} \vert \tilde{x}_{n} \vert ^{2} \vert \bar{x}_{n} \vert ^{2\alpha }\,dt \biggr)^{ \frac{1}{2}} +2 \int ^{1}_{0}f(t) \vert \tilde{x}_{n} \vert ^{1+\alpha }\,dt \\& \qquad {} + \int ^{1}_{0}g(t) \vert \tilde{x}_{n} \vert \,dt \\& \quad \leq 2\beta _{0} \Vert \bar{x}_{n} \Vert _{\infty }^{\alpha } \biggl( \int ^{1} _{0} \vert \tilde{x}_{n} \vert ^{2}\,dt \biggr)^{\frac{1}{2}} +2 \int ^{1}_{0}f(t) \vert \tilde{x}_{n} \vert ^{1+\alpha }\,dt + \int ^{1}_{0}g(t) \vert \tilde{x}_{n} \vert \,dt \\& \quad \leq \frac{\varepsilon _{0}\beta _{0}}{2} \int ^{1}_{0} \vert \tilde{x}_{n} \vert ^{2}\,dt+\frac{2 \beta _{0}}{\varepsilon _{0}} \Vert \bar{x}_{n} \Vert _{\infty }^{2\alpha } +2 \Vert \tilde{x}_{n} \Vert _{\infty }^{1+\alpha } \int ^{1}_{0}f(t)\,dt \\& \qquad {} + \Vert \tilde{x}_{n} \Vert _{\infty } \int ^{1}_{0}g(t)\,dt. \end{aligned}
(3.10)

Notice that by the boundedness of $$\{I(x_{n})\}$$ and $$\alpha \in [0,1)$$, the equivalence of the norm $$(q_{B_{1}+\varepsilon _{0}\beta _{0}I_{n}}(x _{2},x_{2}))^{\frac{1}{2}}$$ on Z for $$x=x_{2}$$ with $$x_{2}\in Z ^{+}(B_{1})=Z^{+}(B_{1}+\varepsilon _{0}\beta _{0}I_{n})$$, and (3.9) we can see that there exist $$c_{12}\in \mathbf{R}$$ and $$c_{13},c_{14},c_{15},c _{16}>0$$ such that

\begin{aligned} c_{12} \leq &-I(x_{n}) \\ =& \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x'_{n} \bigr\vert ^{2}-\bigl(B _{1}(t)x_{n},x_{n}\bigr) \bigr] \,dt \\ &{}- \int ^{1}_{0}\bigl(V(t,x_{n})-V(t, \bar{x}_{n})\bigr)\,dt- \int ^{1}_{0}V(t, \bar{x}_{n})\,dt \\ \leq & \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert \tilde{x}'_{n} \bigr\vert ^{2}- \bigl(B_{1}(t) \tilde{x}_{n},\tilde{x}_{n}\bigr)- \varepsilon _{0}\beta _{0} \vert \tilde{x}_{n} \vert ^{2} \bigr] \,dt+\varepsilon _{0}\beta _{0} \int ^{1}_{0} \vert \tilde{x}_{n} \vert ^{2}\,dt \\ &{}+\frac{2\beta _{0}}{\varepsilon _{0}} \Vert \bar{x}_{n} \Vert _{\infty }^{2 \alpha } +2 \Vert \tilde{x}_{n} \Vert _{\infty }^{1+\alpha } \int ^{1}_{0}f(t)\,dt + \Vert \tilde{x}_{n} \Vert _{\infty } \int ^{1}_{0}g(t)\,dt \\ &{}- \int ^{1}_{0}V(t,\bar{x}_{n})\,dt \\ \leq &c_{13} \Vert \tilde{x}_{n} \Vert ^{2}+c_{14} \Vert \tilde{x}_{n} \Vert ^{2}+2c_{9} \Vert \bar{x}_{n} \Vert ^{2\alpha } +c_{10} \Vert \tilde{x}_{n} \Vert ^{1+\alpha }+c_{11} \Vert \tilde{x}_{n} \Vert \\ &{}- \int ^{1}_{0}V(t,\bar{x}_{n})\,dt \\ \leq &c_{15} \Vert \tilde{x}_{n} \Vert ^{2}+c_{16}+2c_{9} \Vert \bar{x}_{n} \Vert ^{2 \alpha } - \int ^{1}_{0}V(t,\bar{x}_{n})\,dt \\ \leq &(c_{15}k_{1}+2c_{9}) \Vert \bar{x}_{n} \Vert ^{2\alpha }+c_{15}k_{2}+c _{16} - \int ^{1}_{0}V(t,\bar{x}_{n})\,dt \\ \leq & \Vert \bar{x}_{n} \Vert ^{2\alpha } \biggl((c_{15}k_{1}+2c_{9}) - \Vert \bar{x}_{n} \Vert ^{-2\alpha } \int ^{1}_{0}V(t,\bar{x}_{n})\,dt \biggr)+c _{15}k_{2}+c_{16} \end{aligned}

for n large enough. Taking $$c_{0}>c_{15}k_{1}+2c_{9}$$, by this inequality and (1.3) of condition (A7) we obtain that $$\{\|\bar{x}_{n}\|\}$$ is bounded. If (1.4) of condition (A7) holds, similarly to this inequality, by (3.9) and (3.10) we have

\begin{aligned} -I(x_{n}) \geq &-2c_{9} \Vert \bar{x}_{n} \Vert ^{2\alpha } -c_{10} \Vert \tilde{x} _{n} \Vert ^{1+\alpha }-c_{11} \Vert \tilde{x}_{n} \Vert - \int ^{1}_{0}V(t,\bar{x} _{n})\,dt \\ \geq &-2c_{9} \Vert \bar{x}_{n} \Vert ^{2\alpha } -(c_{10}+c_{11}) \Vert \tilde{x} _{n} \Vert ^{2}- \int ^{1}_{0}V(t,\bar{x}_{n}) \,dt-(c_{10}+c_{11}) \\ \geq & \Vert \bar{x}_{n} \Vert ^{2\alpha } \biggl[- \bigl(k_{1}(c_{10}+c_{11})+2c_{9}\bigr) - \Vert \bar{x}_{n} \Vert ^{-2\alpha } \int ^{1}_{0}V(t,\bar{x}_{n})\,dt \biggr] \\ &{}-(k_{2}+1) (c_{10}+c_{11}). \end{aligned}

Taking $$c_{0}>k_{1}(c_{10}+c_{11})+2c_{9}$$, by this inequality and (1.4) of condition (A7) we also obtain that $$\{\|\bar{x} _{n}\|\}$$ is bounded. Hence $$\{\|x_{n}\|\}$$ is bounded by (3.9). Arguing then as in PropositionÂ 4.1 of [10], we easily conclude that the (PS)-condition is satisfied.

Next, we will check that

$$-I(x)\to +\infty$$
(3.11)

as $$\|x\|\to +\infty$$ in $$Z^{+}(B_{1})$$. In fact, by the proof of (3.10) we have

\begin{aligned}& \biggl\vert \int ^{1}_{0}\bigl(V(t,x)-V(t,\theta )\bigr)\,dt \biggr\vert \\& \quad \leq \frac{1}{1+\alpha } \int ^{1}_{0}f(t) \vert x \vert ^{1+\alpha } \,dt+ \int ^{1} _{0}g(t) \vert x \vert \,dt \\& \quad \leq 2 \biggl( \int ^{1}_{0}f^{2}(t)\,dt \biggr)^{\frac{1}{2}} \biggl( \int ^{1}_{0} \vert x \vert ^{2(1+\alpha )}\,dt \biggr)^{\frac{1}{2}} + \Vert x \Vert \int ^{1}_{0}g(t)\,dt \\& \quad \leq \frac{\varepsilon _{0}\beta _{0}}{2} \int ^{1}_{0} \vert x \vert ^{2}\,dt+ \frac{2 \beta _{0}}{\varepsilon _{0}} \Vert x \Vert _{\infty }^{2\alpha } + \Vert x \Vert \int ^{1} _{0}g(t)\,dt \end{aligned}

for all $$x\in Z^{+}(B_{1})$$. It follows that

\begin{aligned} -I(x) =& \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x' \bigr\vert ^{2}-\bigl(B_{1}(t)x,x\bigr) \bigr]\,dt \\ &{}- \int ^{1}_{0}\bigl(V(t,x)-V(t,\theta )\bigr)\,dt- \int ^{1}_{0}V(t,\theta )\,dt \\ \geq & \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x' \bigr\vert ^{2}-\bigl(B_{1}(t)x,x\bigr)- \varepsilon _{0}\beta _{0} \vert x \vert ^{2} \bigr] \,dt-\frac{2\beta _{0}}{ \varepsilon _{0}} \Vert x \Vert _{\infty }^{2\alpha } \\ &{}- \Vert x \Vert \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,\theta )\,dt \\ \geq &c_{8} \Vert x \Vert ^{2}-\frac{2\beta _{0}}{\varepsilon _{0}} \Vert x \Vert ^{2 \alpha } - \Vert x \Vert \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,\theta )\,dt \\ \to &+\infty \end{aligned}

as $$\|x\|\to +\infty$$ in $$Z^{+}(B_{1})$$, which shows (3.11).

On the other hand, if (1.3) of condition (A7) holds, then we clearly have

$$-I(x)\to -\infty$$
(3.12)

as $$\|x\|\to +\infty$$ in $$Z^{0}(B_{1})$$. Thus by (3.11), (3.12), and the saddle point theorem (see TheoremÂ 4.6 in [12] or [10]) we obtain that problem (1.1) has at least one solution in Z. If (1.4) of condition (A7) holds, then we have

$$-I(x)\to +\infty$$

as $$\|x\|\to +\infty$$ in $$Z^{0}(B_{1})$$. Thus by (3.11) we can see that $$-I(x)\to +\infty$$ as $$\|x\|\to +\infty$$ in Z. From TheoremÂ 1.1 and CorollaryÂ 1.1 in [10] we know that problem (1.1) also has at least one solution in Z.

Further, if condition ($$\mathrm{A}'_{6}$$) holds, then $$Z=Z^{-}(B_{01})\oplus Z ^{0}(B_{01})\oplus Z^{0}(B_{02})\oplus Z^{+}(B_{02})$$ via (5) of PropositionÂ 2.5. Let $$X_{1}=Z^{-}(B_{01})\oplus Z^{0}(B_{01})$$ and $$X_{2}=Z^{-}(B_{01})$$. Then $$X_{1}^{\perp }=Z^{0}(B_{02})\oplus Z^{+}(B _{02})$$ and $$X_{2}^{\perp }= Z^{0}(B_{01})\oplus Z^{0}(B_{02})\oplus Z^{+}(B_{02})$$. Note that $$I\in C^{1}(Z,\mathbf{R})$$ satisfies the (PS)-condition. By TheoremÂ 5.29 and ExampleÂ 5.26 in [12] we only need to verify that

($$I_{1}$$):

$$\liminf \|x\|^{-2}I(x)>0$$ as $$\|x\|\to 0$$ in $$X_{1}$$,

($$I_{2}$$):

$$I(x)\leq 0$$ for all $$x\in X^{\perp }_{1}$$, and

($$I_{3}$$):

$$I(x)\to -\infty$$ as $$\|x\|\to +\infty$$ in $$X_{2}^{\perp}$$.

By condition ($$\mathrm{A}'_{6}$$) we can see that $$V(t,\theta )=0$$. Since

$$V(t,x)-V(t,\theta )= \int _{0}^{1}\bigl(\nabla _{x} V(t,sx),x \bigr)\,ds$$

for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$, from condition ($$\mathrm{A}'_{4}$$) we obtain

$$\bigl\vert V(t,x) \bigr\vert \leq \frac{1}{1+\alpha }f(t) \vert x \vert ^{1+\alpha }+g(t) \vert x \vert$$

for all $$x\in \mathbf{R}^{n}$$ and a.e. $$t\in [0,1]$$, and there exist $$c_{17},c_{18}>0$$ such that

\begin{aligned} \biggl\vert \int ^{1}_{0}V(t,x)\,dt \biggr\vert \leq & \frac{1}{1+\alpha } \int ^{1} _{0}f(t) \vert x \vert ^{1+\alpha } \,dt+ \int ^{1}_{0}g(t) \vert x \vert \,dt \\ \leq & \frac{1}{1+\alpha } \Vert x \Vert _{\infty }^{1+\alpha } \int ^{1}_{0}f(t)\,dt+ \Vert x \Vert _{\infty } \int ^{1}_{0}g(t)\,dt \\ \leq &c_{17} \Vert x \Vert ^{1+\alpha }+c_{18} \Vert x \Vert \leq k_{3} \Vert x \Vert ^{3} \end{aligned}

for all $$\|x\|\geq r$$ and $$k_{3}>0$$ given by $$k_{3}=c_{17}r^{\alpha -2}+c _{18}r^{-2}$$. Now it follows from condition ($$\mathrm{A}'_{6}$$) that

$$\int ^{1}_{0}V(t,x)\,dt \geq \int ^{1}_{0} \frac{1}{2}\bigl(\bigl(\epsilon I_{n}+B _{01}(t)-B_{1}(t)\bigr)x,x\bigr)\,dt- k_{3} \Vert x \Vert ^{3}$$

for all $$x\in Z$$. Hence by (3.1) we have

\begin{aligned} I(x) \geq &- \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x' \bigr\vert ^{2}-\bigl(B_{01}(t)x,x\bigr) \bigr] \,dt+ \frac{1}{2}\epsilon \int _{0}^{1} \vert x \vert ^{2}\,dt- k_{3} \Vert x \Vert ^{3} \\ =&-\frac{1}{2}q_{B_{01}}(x,x)+\frac{1}{2}\epsilon \Vert x \Vert _{L^{2}}^{2}- k_{3} \Vert x \Vert ^{3}. \end{aligned}

Noting that $$X_{1}=Z^{-}(B_{01})\oplus Z^{0}(B_{01})$$ is finite-dimensional, we can see that there exists $$k_{4}>0$$ such that

$$I(x) \geq \frac{1}{2}\epsilon k_{4} \Vert x \Vert ^{2}- k_{3} \Vert x \Vert ^{3}$$

for all $$x\in X_{1}$$, from which ($$I_{1}$$) follows.

For $$x\in X^{\perp }_{1}$$, again by condition ($$\mathrm{A}'_{6}$$) we have

$$I(x) \leq - \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x' \bigr\vert ^{2}-\bigl(B_{02}(t)x,x\bigr) \bigr] \,dt \leq 0$$

via $$X_{1}^{\perp }=Z^{0}(B_{02})\oplus Z^{+}(B_{02})$$ and PropositionÂ 2.1, which shows that ($$I_{2}$$) holds.

Since $$B_{02}>B_{01}>B_{1}$$, by (2.3) we have

$$q_{B_{1}}(x,x)>q_{B_{01}}(x,x)>q_{B_{02}}(x,x), \quad x\in Z \setminus \{\theta \}.$$

Noticing that $$X_{2}^{\perp }= Z^{0}(B_{01})\oplus Z^{0}(B_{02}) \oplus Z^{+}(B_{02})$$, we have $$X_{2}^{\perp }\subset Z^{+}(B_{1})$$. Finally, ($$I_{3}$$) follows from (3.11). Hence the proof is completed.â€ƒâ–¡

### 3.4 Proof of Theorem 1.4

In the section, we first use the saddle point theorem (see TheoremÂ 4.6, [12] or [10]) to prove that problem (1.1) has at least one solution. Then, to prove that problem (1.1) has multiple periodic solutions, we need the following abstract critical point theorem developed recently in [3].

### Lemma 3.3

([3], TheoremÂ 5.2.23)

Let X be a Banach space, and let $$\varphi \in C^{1}(X,\mathbf{R})$$ be an even function satisfying the (PS)-condition. Assume that $$a< b$$ and $$\varphi (\theta )\geq b$$. Further, suppose that

1. (1)

there are an m-dimensional linear subspace G and $$\rho >0$$ such that

$$\sup_{x\in G\cap \partial B_{\rho }(\theta )}\varphi (x)< b,$$

where $$\partial B_{\rho }(\theta )=\{x\in X|\|x\|=\rho \}$$;

2. (2)

there is a j-dimensional linear subspace F such that

$$\inf_{x\in F^{\perp }}\varphi (x)> a,$$

where $$F^{\perp }$$ is the orthogonal complementary space of F;

3. (3)

$$m>j$$.

Then Ï† has at least $$m-j$$ pairs of distinct critical points.

### Proof of TheoremÂ 1.4

By assumption ($$\mathrm{A}'_{1}$$), Propositions 2.1â€“2.4 and DefinitionÂ 2.3 we have $$Z=Z^{-}(B_{1})\oplus Z^{0}(B_{1})\oplus Z^{+}(B_{1})$$. Set $$X_{0}=Z^{-}(B_{1})$$, $$X_{1}=Z^{0}(B_{1})$$, $$X_{2}=Z^{+}(B_{1})$$, $$x\in Z$$, $$x=x_{0}+x_{1}+x_{2}$$ with $$x_{0}\in X_{0}$$, $$x_{1}\in X_{1}$$, $$x_{2}\in X_{2}$$. Next, we divide the proof into four steps.

Step 1. We verify that I satisfies the (PS)-condition. Suppose that $$I'(x_{n})\to 0$$ as $$n\to \infty$$ and $$I(x_{n})$$ is bounded. Let $$x_{n}=x_{n0}+x_{n1}+x_{n2}$$ with $$x_{n0}\in Z^{-}(B_{1})$$, $$x_{n1}\in Z^{0}(B_{1})$$ and $$x_{n2}\in Z^{+}(B_{1})$$. In a way similar to (3.8), by assumption ($$\mathrm{A}'_{4}$$) we have

\begin{aligned}& \biggl\vert \int ^{1}_{0}\bigl(\nabla _{x}V(t,x_{n}),x_{n2}-x_{n0} \bigr)\,dt \biggr\vert \\& \quad \leq \int ^{1}_{0}f(t) \vert x_{n2}+x_{n1}+x_{n0} \vert ^{\alpha } \vert x_{n2}-x_{n0} \vert \,dt+ \int ^{1}_{0}g(t) \vert x_{n2}-x_{n0} \vert \,dt \\& \quad \leq \int ^{1}_{0}2f(t) \bigl( \vert x_{n2}+x_{n0} \vert ^{\alpha }+ \vert x_{n1} \vert ^{\alpha }\bigr) \vert x _{n2}-x_{n0} \vert \,dt+ \int ^{1}_{0}g(t) \vert x_{n2}-x_{n0} \vert \,dt \\& \quad \leq \int ^{1}_{0}2f(t) \vert x_{n1} \vert ^{\alpha } \vert x_{n2} \vert \,dt+ \int ^{1}_{0}2f(t) \vert x _{n1} \vert ^{\alpha } \vert x_{n0} \vert \,dt \\& \qquad {} +2\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr)^{1+\alpha } \int ^{1} _{0}f(t)\,dt +\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \int ^{1}_{0}g(t)\,dt \\& \quad \leq 2 \biggl( \int ^{1}_{0}f^{2}(t)\,dt \biggr)^{\frac{1}{2}} \biggl( \int _{0}^{1} \vert x_{n1} \vert ^{2\alpha } \vert x_{n2} \vert ^{2}\,dt \biggr)^{\frac{1}{2}} \\& \qquad {} +2 \Vert x_{n1} \Vert _{\infty }^{\alpha } \bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \int ^{1}_{0}f(t)\,dt \\& \qquad {} +2\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr)^{1+\alpha } \int ^{1} _{0}f(t)\,dt +\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \int ^{1}_{0}g(t)\,dt \\& \quad \leq 2\beta _{0} \Vert x_{n1} \Vert _{\infty }^{\alpha } \biggl( \int _{0}^{1} \vert x _{n2} \vert ^{2}\,dt \biggr)^{\frac{1}{2}}+ 2 \Vert x_{n1} \Vert _{\infty }^{\alpha }\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \int ^{1}_{0}f(t)\,dt \\& \qquad {} +2\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr)^{1+\alpha } \int ^{1} _{0}f(t)\,dt +\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \int ^{1}_{0}g(t)\,dt \\& \quad \leq \varepsilon \beta _{0} \int _{0}^{1} \vert x_{n2} \vert ^{2}\,dt+\frac{\beta _{0}}{\varepsilon } \Vert x_{n1} \Vert _{\infty }^{2\alpha }+ 2 \Vert x_{n1} \Vert _{ \infty }^{\alpha }\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \\& \qquad {} \cdot \int ^{1}_{0}f(t)\,dt+2\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr)^{1+\alpha } \int ^{1}_{0}f(t)\,dt \\& \qquad {} +\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \int ^{1}_{0}g(t)\,dt \end{aligned}
(3.13)

for all n, where $$\beta _{0}=(\int ^{1}_{0}f^{2}(t)\,dt)^{\frac{1}{2}}$$ and $$\varepsilon >0$$. Thus from $$x_{n0}\in Z^{-}(B_{1})$$, $$x_{n1}\in Z ^{0}(B_{1})$$, $$x_{n2}\in Z^{+}(B_{1})$$, (3.13), (2.3), and PropositionÂ 2.1 we have

\begin{aligned} \Vert x_{n2} \Vert + \Vert x_{n0} \Vert \geq& \Vert x_{n2}-x_{n0} \Vert \geq \bigl\langle -I'(x_{n}),x _{n2}-x_{n0}\bigr\rangle \\ =& \int _{0}^{1} \bigl[\bigl(x'_{n},x'_{n2}-x'_{n0} \bigr)-\bigl(B_{1}(t)x_{n},x_{n2}-x _{n0}\bigr) \bigr]\,dt \\ &{}- \int ^{1}_{0}\bigl(\nabla _{x}V(t,x_{n}),x_{n2}-x_{n0} \bigr)\,dt \\ \geq & \int _{0}^{1} \bigl[ \bigl\vert x'_{n2} \bigr\vert ^{2}-\bigl(B_{1}(t)x_{n2},x_{n2} \bigr)- \varepsilon \beta _{0} \vert x_{n2} \vert ^{2} \bigr]\,dt \\ &{}- \int _{0}^{1} \bigl[ \bigl\vert x'_{n0} \bigr\vert ^{2}-\bigl(B_{1}(t)x_{n0},x_{n0} \bigr) \bigr]\,dt-\frac{ \beta _{0}}{\varepsilon } \Vert x_{n1} \Vert _{\infty }^{2\alpha } \\ &{}-2 \Vert x_{n1} \Vert _{\infty }^{\alpha }\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \int ^{1}_{0}f(t)\,dt \\ &{}-2\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr)^{1+\alpha } \int ^{1} _{0}f(t)\,dt -\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \int ^{1}_{0}g(t)\,dt \\ =&q_{B_{1}+\varepsilon \beta _{0}I_{n}}(x_{n2},x_{n2})-q_{B_{1}}(x _{n0},x_{n0})-\frac{\beta _{0}}{\varepsilon } \Vert x_{n1} \Vert _{\infty }^{2 \alpha } -2 \Vert x_{n1} \Vert _{\infty }^{\alpha } \\ &{}\cdot \bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \int ^{1}_{0}f(t)\,dt -2\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr)^{1+\alpha } \int ^{1}_{0}f(t)\,dt \\ &{}-\bigl( \Vert x_{n2} \Vert _{\infty }+ \Vert x_{n0} \Vert _{\infty }\bigr) \int ^{1}_{0}g(t)\,dt \end{aligned}

for n large enough. By PropositionÂ 2.7 we can choose $$\varepsilon _{0}>0$$ such that $$\nu _{M}^{s}(B_{1}+\varepsilon _{0}\beta _{0}I_{n})=0$$ and $$i_{M}^{s}(B_{1}+\varepsilon _{0}\beta _{0}I_{n})=i _{M}^{s}(B_{1})+\nu _{M}^{s}(B_{1})$$. From (5) of PropositionÂ 2.5 we know that $$(-q_{B_{1}}(x_{0},x_{0}))^{\frac{1}{2}}+(q _{B_{1}+\varepsilon _{0}\beta _{0}I_{n}}(x_{2},x_{2}))^{\frac{1}{2}}$$ is an equivalent norm on Z for $$x=x_{0}+x_{2}$$ with $$x_{0}\in Z^{-}(B _{1})$$ and $$x_{2}\in Z^{+}(B_{1})=Z^{+}(B_{1}+\varepsilon _{0}\beta _{0}I_{n})$$. Hence there exist $$c_{19}, c_{20},c_{21},c_{22}>0$$ such that

\begin{aligned}& \bigl( \Vert x_{n2} \Vert + \Vert x_{n0} \Vert \bigr)^{2} \\& \quad \leq c_{19} \Vert x_{n1} \Vert ^{2\alpha }+c_{20} \Vert x_{n1} \Vert ^{\alpha }\bigl( \Vert x _{n2} \Vert + \Vert x_{n0} \Vert \bigr)+c_{21}\bigl( \Vert x_{n2} \Vert + \Vert x_{n0} \Vert \bigr)^{1+\alpha } \\& \qquad {} +c_{22}\bigl( \Vert x_{n2} \Vert + \Vert x_{n0} \Vert \bigr). \end{aligned}
(3.14)

From (3.14) we claim that there exist n large enough and $$k_{5},k_{6}>0$$ such that

$$k_{5} \Vert x_{n1} \Vert ^{2\alpha }+k_{6} \geq \bigl( \Vert x_{n2} \Vert + \Vert x_{n0} \Vert \bigr)^{2}.$$
(3.15)

In fact, we only need to consider two cases: $$\|x_{n2}\|+\|x_{n0}\|$$ is bounded, or $$\|x_{n2}\|+\|x_{n0}\|$$ is unbounded.

(i) If $$\|x_{n2}\|+\|x_{n0}\|$$ is bounded, then $$\|x_{n2}\|+\|x_{n0} \|\leq c_{23}$$. By (3.14) we have

$$(c_{19}+c_{20}+c_{23}) \Vert x_{n1} \Vert ^{2\alpha }+c_{21}c_{23}^{1+\alpha }+2c _{22}c_{23}\geq \bigl( \Vert x_{n2} \Vert + \Vert x_{n0} \Vert \bigr)^{2}.$$

Thus (3.15) follows.

(ii) If $$\|x_{n2}\|+\|x_{n0}\|$$ is unbounded, then there is n large enough such that

\begin{aligned}& c_{19}\frac{ \Vert x_{n1} \Vert ^{2\alpha }}{( \Vert x_{n2} \Vert + \Vert x_{n0} \Vert )^{2}}+c _{20}\frac{ \Vert x_{n1} \Vert ^{\alpha }}{ \Vert x_{n2} \Vert + \Vert x_{n0} \Vert } \\& \quad \geq 1- c_{21}\frac{1}{( \Vert x_{n2} \Vert + \Vert x_{n0} \Vert )^{1-\alpha }}-c_{22} \frac{1}{ \Vert x_{n2} \Vert + \Vert x_{n0} \Vert }\geq \frac{1}{2}, \end{aligned}

which implies that there is $$c_{24}>0$$ such that $$\frac{\|x_{n1}\| ^{\alpha }}{\|x_{n2}\|+\|x_{n0}\|}\geq c_{24}$$. From (i) and (ii) we get that (3.15) holds.

To prove the boundedness of $$\{x_{n}\}$$, by (3.15) it suffices to prove that $$\{x_{n1}\}$$ is bounded. In a way similar to (3.8), for all n, we have

\begin{aligned}& \biggl\vert \int ^{1}_{0}\bigl(V(t,x_{n})-V(t,x_{n1}) \bigr)\,dt \biggr\vert \\& \quad = \biggl\vert \int ^{1}_{0} \int ^{1}_{0}\bigl(\nabla _{x}V \bigl(t,x_{n1}+s(x_{n0}+x _{n2}) \bigr),x_{n0}+x_{n2}\bigr)\,ds\,dt \biggr\vert \\& \quad \leq \int ^{1}_{0} \int ^{1}_{0}f(t) \bigl\vert x_{n1}+s(x_{n0}+x_{n2}) \bigr\vert ^{\alpha } \vert x_{n0}+x_{n2} \vert \,ds\,dt \\& \qquad {} + \int ^{1}_{0} \int ^{1}_{0}g(t) \vert x_{n0}+x_{n2} \vert \,ds\,dt \\& \quad \leq \int ^{1}_{0}2f(t) \biggl( \vert x_{n1} \vert ^{\alpha }+\frac{1}{1+\alpha }\bigl( \vert x _{n0} \vert + \vert x_{n2} \vert \bigr)^{\alpha }\biggr) \bigl( \vert x_{n0} \vert + \vert x_{n2} \vert \bigr)\,dt \\& \qquad {} + \int ^{1}_{0}g(t) \bigl( \vert x_{n0} \vert + \vert x_{n2} \vert \bigr)\,dt \\& \quad \leq 2\beta _{0} \Vert x_{n1} \Vert ^{\alpha }_{\infty } \biggl( \int ^{1}_{0}\bigl( \vert x _{n0} \vert + \vert x_{n2} \vert \bigr)^{2}\,dt \biggr)^{\frac{1}{2}} +2\bigl( \Vert x_{n0} \Vert _{\infty }+ \Vert x_{n2} \Vert _{\infty }\bigr)^{1+\alpha } \\& \qquad {} \cdot \int ^{1}_{0}f(t)\,dt +\bigl( \Vert x_{n0} \Vert _{\infty }+ \Vert x_{n2} \Vert _{\infty }\bigr) \int ^{1}_{0}g(t)\,dt \\& \quad \leq \frac{\varepsilon _{0}\beta _{0}}{4} \int ^{1}_{0}\bigl( \vert x_{n0} \vert + \vert x_{n2} \vert \bigr)^{2}\,dt+ \frac{4\beta _{0}}{\varepsilon _{0}} \Vert x_{n1} \Vert ^{2\alpha }_{\infty }+2 \int ^{1}_{0}f(t)\,dt \\& \qquad {} \cdot \bigl( \Vert x_{n0} \Vert _{\infty }+ \Vert x_{n2} \Vert _{\infty }\bigr)^{1+\alpha } +\bigl( \Vert x _{n0} \Vert _{\infty }+ \Vert x_{n2} \Vert _{\infty }\bigr) \int ^{1}_{0}g(t)\,dt \\& \quad \leq -\frac{\varepsilon _{0}\beta _{0}}{2} \int ^{1}_{0} \vert x_{n2} \vert ^{2}\,dt+ \varepsilon _{0}\beta _{0}\bigl( \Vert x_{n0} \Vert _{\infty }+ \Vert x_{n2} \Vert _{\infty }\bigr)^{2}+ \frac{4\beta _{0}}{\varepsilon _{0}} \Vert x_{n1} \Vert ^{2\alpha }_{\infty } \\& \qquad {} +2\bigl( \Vert x_{n0} \Vert _{\infty }+ \Vert x_{n2} \Vert _{\infty }\bigr)^{1+\alpha } \int ^{1} _{0}f(t)\,dt \\& \qquad {} +\bigl( \Vert x_{n0} \Vert _{\infty }+ \Vert x_{n2} \Vert _{\infty }\bigr) \int ^{1}_{0}g(t)\,dt. \end{aligned}
(3.16)

Notice that by the boundedness of $$\{I(x_{n})\}$$ and $$\alpha \in [0,1)$$, the equivalence of the norm $$(q_{B_{1}+\varepsilon _{0}\beta _{0}I_{n}}(x _{2},x_{2}))^{\frac{1}{2}}$$ on Z for $$x=x_{2}$$ with $$x_{2}\in Z ^{+}(B_{1})=Z^{+}(B_{1}+\varepsilon _{0}\beta _{0}I_{n})$$, $$q_{B_{1}}(x _{1},x_{1})<0$$ on Z and for $$x=x_{0}$$ with $$x_{0}\in Z^{-}(B_{1})$$, (3.15), and (3.16) we obtain that there exist $$c_{25}\in \mathbf{R}$$ and $$c_{26},c_{27}>0$$ such that

\begin{aligned} c_{25} \leq& -I(x_{n}) \\ =& \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x'_{n} \bigr\vert ^{2}-\bigl(B_{1}(t)x_{n},x_{n} \bigr) \bigr] \,dt- \int ^{1}_{0}\bigl(V(t,x_{n})-V(t,x_{n1}) \bigr)\,dt \\ &{}- \int ^{1}_{0}V(t,x_{n1})\,dt \\ \leq & \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x'_{n2} \bigr\vert ^{2}-\bigl(B_{1}(t)x_{n2},x _{n2}\bigr)-\varepsilon _{0}\beta _{0} \vert x_{n2} \vert ^{2} \bigr] \,dt \\ &{}+ \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x'_{n1} \bigr\vert ^{2}-\bigl(B_{1}(t)x_{n1},x_{n1} \bigr) \bigr] \,dt+\varepsilon _{0}\beta _{0}\bigl( \Vert x_{n0} \Vert _{\infty }+ \Vert x_{n2} \Vert _{\infty }\bigr)^{2} \\ &{}+\frac{4\beta _{0}}{\varepsilon _{0}} \Vert x_{n1} \Vert _{\infty }^{2\alpha } +2\bigl( \Vert x_{n0} \Vert _{\infty }+ \Vert x_{n2} \Vert _{\infty }\bigr)^{1+\alpha } \int ^{1}_{0}f(t)\,dt \\ &{}+\bigl( \Vert x_{n0} \Vert _{\infty }+ \Vert x_{n2} \Vert _{\infty }\bigr) \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,x_{n1})\,dt \\ \leq &c_{26} \Vert x_{n2} \Vert ^{2}+c_{26} \bigl( \Vert x_{n0} \Vert + \Vert x_{n2} \Vert \bigr)^{2}+c_{26} \Vert x_{n1} \Vert ^{2\alpha } \\ &{}+c_{26}\bigl( \Vert x_{n0} \Vert + \Vert x_{n2} \Vert \bigr)^{1+\alpha }+c_{26}\bigl( \Vert x_{n0} \Vert + \Vert x _{n2} \Vert \bigr)- \int ^{1}_{0}V(t,x_{n1})\,dt \\ \leq &4c_{26}\bigl( \Vert x_{n0} \Vert + \Vert x_{n2} \Vert \bigr)^{2}+c_{26} \Vert x_{n1} \Vert ^{2\alpha }+c _{27}- \int ^{1}_{0}V(t,x_{n1})\,dt \\ \leq &(4c_{26}k_{5}+c_{26}) \Vert x_{n1} \Vert ^{2\alpha }+4c_{26}k_{6}+c_{27}- \int ^{1}_{0}V(t,x_{n1})\,dt \\ \leq & \Vert x_{n1} \Vert ^{2\alpha } \biggl((4c_{26}k_{5}+c_{26})- \Vert x_{n1} \Vert ^{-2\alpha } \int ^{1}_{0}V(t,x_{n1})\,dt \biggr)+4c_{26}k_{6}+c_{27} \end{aligned}

for n large enough. Taking $$c_{0}>4c_{26}k_{5}+c_{26}$$ in this inequality, by (1.3) of condition (A7) we get that $$\{\|x_{n1}\|\}$$ is bounded. If (1.4) of condition (A7) holds, then similarly to the proof of TheoremÂ 1.3, by this inequality, (3.15), and (3.16) we also get that $$\{\|x_{n1}\|\}$$ is bounded. Hence $$\{\|x_{n}\|\}$$ is bounded by (3.15). Arguing then as in PropositionÂ 4.1 in [10], we easily conclude that the (PS)-condition is satisfied.

Step 2. We prove that $$-I(x_{2})\to +\infty$$ as $$\|x_{2}\| \to +\infty$$ with $$x_{2}\in X_{2}=Z^{+}(B_{1})$$ and $$-I(x_{0})\to - \infty$$ as $$\|x_{0}\|\to +\infty$$ with $$x_{0}\in X_{2}=Z^{-}(B_{1})$$.

For $$x_{2}\in Z^{+}(B_{1})$$, from condition ($$\mathrm{A}'_{4}$$) we have

\begin{aligned}& \biggl\vert \int ^{1}_{0}\bigl(V(t,x_{2})-V(t,\theta ) \bigr)\,dt \biggr\vert \\ & \quad \leq \frac{1}{1+\alpha } \int ^{1}_{0}f(t) \vert x_{2} \vert ^{1+\alpha }\,dt+ \int ^{1}_{0}g(t) \vert x_{2} \vert \,dt \\ & \quad \leq 2\beta _{0} \biggl( \int ^{1}_{0} \vert x_{2} \vert ^{2(1+\alpha )}\,dt \biggr) ^{\frac{1}{2}} + \Vert x_{2} \Vert \int ^{1}_{0}g(t)\,dt \\& \quad \leq \frac{\varepsilon _{0}\beta _{0}}{2} \int ^{1}_{0} \vert x_{2} \vert ^{2}\,dt+\frac{2 \beta _{0}}{\varepsilon _{0}} \Vert x_{2} \Vert ^{2\alpha } + \Vert x_{2} \Vert \int ^{1} _{0}g(t)\,dt. \end{aligned}

It follows that

\begin{aligned} -I(x_{2}) =& \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x'_{2} \bigr\vert ^{2}-\bigl(B_{1}(t)x_{2},x _{2}\bigr) \bigr] \,dt- \int ^{1}_{0}\bigl(V(t,x_{2})-V(t,\theta ) \bigr)\,dt \\ &{}- \int ^{1}_{0}V(t,\theta )\,dt \\ \geq & \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x_{2}' \bigr\vert ^{2}-\bigl(B_{1}(t)x_{2},x _{2}\bigr)-\varepsilon _{0}\beta _{0} \vert x_{2} \vert ^{2} \bigr]\,dt \\ &{}-\frac{2\beta _{0}}{\varepsilon _{0}} \Vert x_{2} \Vert ^{2\alpha } - \Vert x_{2} \Vert \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,\theta )\,dt \\ \geq &c_{28} \Vert x_{2} \Vert ^{2}- \frac{2\beta _{0}}{\varepsilon _{0}} \Vert x_{2} \Vert ^{2\alpha } - \Vert x_{2} \Vert \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,\theta )\,dt \\ \to &+\infty \end{aligned}
(3.17)

as $$\|x\|\to +\infty$$ in $$Z^{+}(B_{1})$$, where $$c_{28}>0$$.

Similarly, for $$x_{0}\in Z^{-}(B_{1})$$, from condition ($$\mathrm{A}'_{4}$$) we have

\begin{aligned}& \biggl\vert \int ^{1}_{0}\bigl(V(t,x_{0})-V(t,\theta ) \bigr)\,dt \biggr\vert \\& \quad \leq \frac{1}{1+\alpha } \int ^{1}_{0}f(t) \vert x_{0} \vert ^{1+\alpha }\,dt+ \int ^{1}_{0}g(t) \vert x_{0} \vert \,dt \\& \quad \leq \frac{1}{1+\alpha } \Vert x_{0} \Vert ^{1+\alpha } \int ^{1}_{0}f(t)\,dt + \Vert x _{0} \Vert \int ^{1}_{0}g(t)\,dt. \end{aligned}

It follows that

\begin{aligned} -I(x_{0}) \leq & \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x'_{0} \bigr\vert ^{2}-\bigl(B_{1}(t)x _{0},x_{0}\bigr) \bigr] \,dt+\frac{1}{1+\alpha } \Vert x_{0} \Vert ^{1+\alpha } \int ^{1}_{0}f(t)\,dt \\ &{}+ \Vert x_{0} \Vert \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,\theta )\,dt \\ =&-\frac{1}{2}\bigl(-q_{B_{1}}(x_{0},x_{0}) \bigr)+\frac{1}{1+\alpha } \Vert x_{0} \Vert ^{1+\alpha } \int ^{1}_{0}f(t)\,dt \\ &{}+ \Vert x_{0} \Vert \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,\theta )\,dt \\ \leq &-\frac{c_{29}}{2} \Vert x_{0} \Vert ^{2}+ \frac{1}{1+\alpha } \Vert x_{0} \Vert ^{1+\alpha } \int ^{1}_{0}f(t)\,dt \\ &{}+ \Vert x_{0} \Vert \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,\theta )\,dt \\ \to &-\infty \end{aligned}
(3.18)

as $$\|x_{0}\|\to +\infty$$ in $$Z^{-}(B_{1})$$, where $$c_{29}>0$$.

Step 3. Next, we prove that problem (1.1) has at least one solution in Z. If (1.3) of condition (A7) holds, then we let $$X^{-}= Z^{-}(B_{1})\oplus Z^{0}(B_{1})$$ and $$X^{+}=Z^{+}(B _{1})$$. For $$x=x_{0}+x_{1}\in X^{-}$$ with $$x_{0}\in Z^{-}(B_{1})$$ and $$x_{1}\in Z^{0}(B_{1})$$, by condition ($$\mathrm{A}'_{4}$$) we have

\begin{aligned}& \biggl\vert \int ^{1}_{0}\bigl(V(t,x)-V(t,x_{1})\bigr) \,dt \biggr\vert \\ & \quad = \biggl\vert \int ^{1}_{0} \int ^{1}_{0}\bigl(\nabla _{x}V(t,sx_{0}+x_{1}),x_{0} \bigr)\,ds\,dt \biggr\vert \\ & \quad \leq \int ^{1}_{0} \int ^{1}_{0}f(t) \vert sx_{0}+x_{1} \vert ^{\alpha } \vert x_{0} \vert \,ds\,dt+ \int ^{1}_{0} \int ^{1}_{0}g(t) \vert x_{0} \vert \,ds\,dt \\ & \quad \leq 2 \int ^{1}_{0}f(t) \vert x_{1} \vert ^{\alpha } \vert x_{0} \vert \,dt+2 \Vert x_{0} \Vert ^{1+ \alpha } \int ^{1}_{0}f(t)\,dt+ \Vert x_{0} \Vert \int ^{1}_{0}g(t)\,dt. \end{aligned}

Thus, there is $$\varepsilon >0$$ with $$2\varepsilon \int ^{1}_{0}f(t)\,dt< c _{29}$$ such that

\begin{aligned} -I(x) \leq& \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x'_{0} \bigr\vert ^{2}-\bigl(B_{1}(t)x_{0},x _{0}\bigr) \bigr] \,dt+2 \int ^{1}_{0}f(t) \vert x_{1} \vert ^{\alpha } \vert x_{0} \vert \,dt \\ &{} +2 \Vert x_{0} \Vert ^{1+\alpha } \int ^{1}_{0}f(t)\,dt+ \Vert x_{0} \Vert \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,x_{1})\,dt \\ \leq& -\frac{c_{29}}{2} \Vert x_{0} \Vert ^{2}+ \varepsilon \Vert x_{0} \Vert ^{2} \int ^{1}_{0}f(t)\,dt+\frac{1}{\varepsilon } \Vert x_{1} \Vert ^{2\alpha } \int ^{1}_{0}f(t)\,dt \\ & {} +2 \Vert x_{0} \Vert ^{1+\alpha } \int ^{1}_{0}f(t)\,dt+ \Vert x_{0} \Vert \int ^{1}_{0}g(t)\,dt- \int ^{1}_{0}V(t,x_{1})\,dt \\ \leq& \biggl[- \biggl(\frac{c_{29}}{2}-\varepsilon \int ^{1}_{0}f(t)\,dt \biggr) \Vert x_{0} \Vert ^{2} +2 \Vert x_{0} \Vert ^{1+\alpha } \int ^{1}_{0}f(t)\,dt+ \Vert x_{0} \Vert \int ^{1}_{0}g(t)\,dt \biggr] \\ & {} + \Vert x_{1} \Vert ^{2\alpha } \biggl( \frac{1}{\varepsilon } \int ^{1}_{0}f(t)\,dt - \Vert x_{1} \Vert ^{-2\alpha } \int ^{1}_{0}V(t,x_{1})\,dt \biggr). \end{aligned}

Taking $$c_{0}>\frac{1}{\varepsilon }\int ^{1}_{0}f(t)\,dt$$, by (1.3) of condition (A7) we see that $$-I(x)\to -\infty$$ as $$\|x\| \to +\infty$$ in $$X^{-}$$.

If (1.4) of condition (A7) holds, then we let $$X^{-}= Z^{-}(B _{1})$$ and $$X^{+}= Z^{0}(B_{1})\oplus Z^{+}(B_{1})$$. As before, we easily get that $$-I(x)\to +\infty$$ as $$\|x\|\to +\infty$$ in $$X^{+}$$. Together, from (3.17) and (3.18) we have $$-I(x)\to +\infty$$ as $$\|x\|\to +\infty$$ in $$X^{+}$$ and $$-I(x) \to -\infty$$ as $$\|x\|\to +\infty$$ in $$X^{-}$$. By the saddle point theorem (see TheoremÂ 4.6 in [12] or [10]) we see that problem (1.1) has at least one solution in Z.

Step 4. Finally, we prove that problem (1.1) has at least $$i_{M}^{s}(B_{2})-i_{M}^{s}(B_{1})-\nu _{M}^{s}(B_{1})$$ pairs of solutions in Z. Since $$B_{2}> B_{1}$$ and $$\nu _{M}^{s}(B_{2}) \neq 0$$, we have $$Z=Z^{-}(B_{1})\oplus Z^{0}(B_{1})\oplus Z^{+}(B_{1})=Z ^{-}(B_{2})\oplus Z^{0}(B_{2})\oplus Z^{+}(B_{2})$$ and $$Z^{+}(B_{2}) \subset Z^{+}(B_{1})$$. Set $$G=Z^{-}(B_{2})$$, $$F=Z^{-}(B_{1})\oplus Z^{0}(B _{1})$$, $$b=0$$. We define $$\varphi (x)=-I(x)+\int ^{1}_{0}V(t,\theta )\,dt$$ for all $$x\in Z$$. Then $$\varphi (\theta )=0\geq b$$. Noting that $$F^{\perp }=Z^{+}(B_{1})$$, we get that (2) of LemmaÂ 3.3 holds. By condition (A8) and the proof of Step 1 it suffices to show that (1) of LemmaÂ 3.3 holds.

By condition (A9), for any $$x\in G\cap B_{r}(\theta )$$, we have

\begin{aligned} \varphi (x) =& \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x' \bigr\vert ^{2}-\bigl(B_{1}(t)x,x\bigr) \bigr] \,dt- \int ^{1}_{0}V(t,x)\,dt+ \int ^{1}_{0}V(t,\theta )\,dt \\ \leq & \int _{0}^{1}\frac{1}{2} \bigl[ \bigl\vert x' \bigr\vert ^{2}-\bigl(B_{2}(t)x,x\bigr) \bigr] \,dt-\frac{ \epsilon }{2} \int _{0}^{1} \vert x \vert ^{2}\,dt =\frac{1}{2}q_{B_{2}}(x,x)-\frac{\epsilon }{2} \Vert x \Vert ^{2}_{L^{2}}. \end{aligned}

Noticing that $$E=Z^{-}(B_{2})$$ is finite-dimensional, we get that there exists $$k_{7}>0$$ such that

$$\varphi (x)\leq -\frac{k_{7}\epsilon }{2} \Vert x \Vert ^{2},$$

which implies that, for all $$x\in G\cap \partial B_{r}(\theta )$$,

$$\varphi (x)\leq -\frac{rk_{7}\epsilon }{2}< 0.$$

Thus Ï† has at least $$i_{M}^{s}(B_{2})-i_{M}^{s}(B_{1})-\nu _{M}^{s}(B_{1})$$ pairs of distinct critical points, which implies that problem (1.1) has at least $$i_{M}^{s}(B_{2})-i_{M}^{s}(B _{1})-\nu _{M}^{s}(B_{1})$$ pairs of solutions in Z.â€ƒâ–¡

### 3.5 Proof of the corollaries

In the section, we use Theorems 1.1â€“1.4 to prove that the corollaries.

### Proof of CorollaryÂ 1.5

Letting $$M=N=I_{n}$$ and $$B_{1}(t)\equiv 0$$, from the index theory of Sect.Â 2 we easily see that $$Z=H^{1}_{0}$$, $$\nu _{I_{n}}^{s}(0)\neq 0$$, $$\operatorname{ker}(\varLambda )=\mathbf{R}^{n}$$, and $$i_{I_{n}}^{s}(0)=0$$, that is, (A1) holds. By $$\operatorname{ker}(\varLambda )=\mathbf{R}^{n}$$ we know that $$|x|=\|x\|$$ for all $$x\in \operatorname{ker}(\varLambda )$$. Again setting $$B_{2}(t)\equiv (2 \pi )^{2}$$, from (2.4) and (2.5) of RemarkÂ 2.8 we have $$\nu _{I_{n}}^{s}((2\pi )^{2})\neq 0$$ and $$i_{I_{n}}^{s}((2\pi )^{2})=i _{I_{n}}^{s}(0)+\nu _{I_{n}}^{s}(0)$$. Thus (A2) and (A3) follow from (H1) and (H2). The proof is complete.â€ƒâ–¡

### Proof of CorollaryÂ 1.7

Letting $$M=N=I_{n}$$ and $$B_{1}(t)\equiv 0$$, we have $$Z=H^{1}_{0}$$, $$\nu _{I_{n}}^{s}(0)\neq 0$$, $$\operatorname{ker}(\varLambda )=\mathbf{R}^{n}$$, and $$i_{I_{n}}^{s}(0)=0$$, that is, (A1) holds. We need only to show that (A4) follows from (H3) and (A6) follows from (H4). In fact, from (2.4) and (2.5) of RemarkÂ 2.8 we obtain that $$\nu _{I_{n}}^{s}(f(t)I_{n})=0$$ and $$i_{I_{n}}^{s}(f(t)I _{n})=i_{I_{n}}^{s}(0)+\nu _{I_{n}}^{s}(0)$$, which shows that (A4) holds.

Moreover, setting $$B_{01}(t)\equiv (2k\pi )^{2}$$ and $$B_{02}(t)\equiv (2(k+1)\pi )^{2}$$, from (2.4) and (2.5) of RemarkÂ 2.8 we see that $$\nu _{I_{n}}^{s}(B_{0i})\neq 0$$ ($$i=1,2$$) and $$i_{I_{n}} ^{s}(B_{02})=i_{I_{n}}^{s}(B_{01})+\nu _{I_{n}}^{s}(B_{01})$$. Noting that $$|x|\leq \|x\|_{\infty }\leq \|x\|$$ for $$x\in H_{0}^{1}$$, we have $$|x|\leq \delta$$ as $$\|x\|\leq \delta$$, which implies that (A6) holds. The proof is complete.â€ƒâ–¡

### Proof of CorollaryÂ 1.9

Letting $$M=N=I_{n}$$ and $$B_{1}(t)\equiv 0$$, we have $$Z=H^{1}_{0}$$, $$\nu _{I_{n}}^{s}(0)\neq 0$$, $$\operatorname{ker}(\varLambda )=\mathbf{R}^{n}$$, and $$i_{I_{n}}^{s}(0)=0$$, that is, (A1) holds. Similarly to the proof of CorollaryÂ 1.7, ($$\mathrm{A}'_{6}$$) follows from ($$\mathrm{H}'_{4}$$). Since $$\operatorname{ker}(\varLambda )=\mathbf{R}^{n}$$, we have that $$|x|=\|x\|$$ for all $$x\in \operatorname{ker}(\varLambda )$$. So (A7) follows from (H5). The proof is complete.â€ƒâ–¡

### Proof of CorollaryÂ 1.12

Letting $$M=N=I_{n}$$, $$B_{1}(t) \equiv (2k\pi )^{2}$$, and $$B_{2}(t)\equiv (2(k+m)\pi )^{2}$$, from (2.4) and (2.5) of RemarkÂ 2.8 we obtain $$Z=H^{1}_{0}$$, $$\nu _{I_{n}}^{s}(B_{i})\neq 0$$ ($$i=1,2$$), $$\nu _{I_{n}}^{s}(B_{1}+f(t)I _{n})=0$$, $$i_{I_{n}}^{s}(B_{1}+f(t)I_{n})=i_{I_{n}}^{s}(B_{1})+\nu _{I_{n}}^{s}(B_{1})$$, and $$i_{I_{n}}^{s}(B_{2})-i_{I_{n}}^{s}(B_{1})- \nu _{I_{n}}^{s}(B_{1})=2nm-2n>0$$ via some simple calculation. Similarly to the proof of CorollaryÂ 1.7, we see that the conditions of TheoremÂ 1.4 hold. The proof is complete.â€ƒâ–¡

### Proof of CorollaryÂ 1.14

Letting $$M=N=I_{n}$$ and $$B_{1}(t)=A(t)$$, from (2.4) and (2.5) of RemarkÂ 2.8 we obtain that $$Z=H^{1}_{0}$$, $$\nu _{I_{n}}^{s}(B_{1})\neq 0$$, and $$i_{I_{n}}^{s}(A(t))$$ is at most finite-dimensional. If $$i_{I_{n}} ^{s}(A(t))=0$$, then we see that the conditions of TheoremÂ 1.3 hold, and if $$i_{I_{n}}^{s}(A(t))\neq 0$$, then we also see that the conditions of TheoremÂ 1.4 hold. The proof is complete.â€ƒâ–¡

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The author thanks the referees for the valuable comments and suggestions.

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## Funding

This research was supported by the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (No.Â 18KJB110006).

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Song, M. Existence and multiplicity of solutions for second-order Hamiltonian systems satisfying generalized periodic boundary value conditions at resonance. Bound Value Probl 2019, 118 (2019). https://doi.org/10.1186/s13661-019-1233-x