RETRACTED ARTICLE: Landesman-Lazer type condition for second-order differential equations at resonance with impulsive effects

Abstract

In this paper, we study the existence of periodic solutions of second-order impulsive differential equations at resonance. We prove the existence of periodic solutions under a generalized Landesman-Lazer type condition by using the variational method. The impulses can generate a periodic solution.

1 Introduction

We are concerned with periodic boundary value problem of second-order impulsive differential equations at resonance

$$\{\begin{array}{c}{x}^{″}(t)+m^{2}x(t)+f(t,x(t))=e(t),\text{a.e. }t\in [0,2\pi ],\hfill \\ x(0)-x(2\pi )=x^{\prime }(0)-x^{\prime }(2\pi )=0,\hfill \\ x(t_j^{+})=x(t_j^{-}),\hfill \\ \mathrm{\Delta }{x}^{\prime }(t_j):=x^{\prime }(t_j^{+})-x^{\prime }(t_j^{-})=I_j(t_j,x(t_j)),j=1,2,\dots ,p,\hfill \end{array}$$

(1.1)

where $m\in \mathbb{N}$, $f:[0,2\pi ]\times \mathbb{R}\to \mathbb{R}$ is a Carathéodory function, $e\in L^1(0,2\pi )$, $0<t_1<t_2<\cdots <t_p<2\pi$, and $I_j:[0,2\pi ]\times \mathbb{R}\to \mathbb{R}$ is continuous for every j.

When $\mathrm{\Delta }{x}^{\prime }(t_j)\equiv 0$, problem (1.1) becomes to the well-known periodic boundary value problem at resonance

$$\{\begin{array}{c}{x}^{″}(t)+m^{2}x(t)+f(t,x(t))=e(t),\text{a.e. }t\in [0,2\pi ],\hfill \\ x(0)-x(2\pi )=x^{\prime }(0)-x^{\prime }(2\pi )=0.\hfill \end{array}$$

(1.2)

There are many existence results for problem (1.2) in the literature. Let us mention some pioneering works by Lazer [1], Lazer and Leach [2], and Landesman and Lazer [3]. In [3], a key sufficient condition for the existence of solutions of problem (1.2) is the so-called Landesman-Lazer condition,

$$\begin{array}{rl}{\int }_0^{2\pi }e(t)sin(mt+\theta )dt<& {\int }_0^{2\pi }[(\underset{x\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}f(t,x)){sin}^{+}(mt+\theta )\\ -(\underset{x\to -\mathrm{\infty }}{lim\hspace{0.17em}sup}f(t,x)){sin}^{-}(mt+\theta )]dt,\mathrm{\forall }\theta \in \mathbb{R},\end{array}$$

(1.3)

where ${sin}^{\pm }(mt+\theta )=max\{\pm sin(mt+\theta ),0\}$.

It is well known that the theory of impulsive differential equations has been recognized to not only be richer than that of differential equations without impulses, but also to provide a more adequate mathematical model for numerous processes and phenomena studied in physics, biology, engineering, etc. We refer the reader to the book [4]. Recently, the Dirichlet and periodic boundary conditions problems for second-order differential equations with impulses in the derivative and without impulses are studied by some authors via variational method [5–11]. In this paper, we will investigate problem (1.1) under a more general Landesman-Lazer type condition. Define

$$F(t,x)={\int }_0^{x}f(t,s)ds,F_{+}(t)=\underset{x\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{F(t,x)}{x},F_{-}(t)=\underset{x\to -\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{F(t,x)}{x}$$

and for $j=1,2,\dots ,p$,

$$J_j(t,x)={\int }_0^x{I}_j(t,s)ds,J_j^{+}(t)=\underset{x\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{J_j(t,x)}{x},J_j^{-}(t)=\underset{x\to -\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{J_j(t,x)}{x}.$$

Throughout this paper, we give the following fundamental assumptions.

(${\mathrm{H}}_1$) There exists $p\in L^1([0,2\pi ],[0,+\mathrm{\infty }))$ such that $|f(t,x)|⩽p(t)$, for a.e. $t\in [0,2\pi ]$ and for all $x\in \mathbb{R}$.

(${\mathrm{H}}_2$) There exist positive constants $c_1,c_2,\dots ,c_p$ such that for all $t,x\in \mathbb{R}$,

$$|I_j(t,x)|⩽c_j,j=1,2,\dots ,p.$$

(${\mathrm{H}}_3$) For all $\theta \in \mathbb{R}$,

$$\begin{array}{r}\sum _{j=1}^p{J}_j^{+}(t_j){sin}^{+}(m{t}_j+\theta )-\sum _{j=1}^p{J}_j^{-}(t_j){sin}^{-}(m{t}_j+\theta )+{\int }_0^{2\pi }e(t)sin(mt+\theta )dt\\ <{\int }_0^{2\pi }(F_{+}(t){sin}^{+}(mt+\theta )-F_{-}(t){sin}^{-}(mt+\theta ))dt.\end{array}$$

We now can state the main theorem of this paper.

Theorem 1.1 Assume that the conditions (${\mathrm{H}}_1$), (${\mathrm{H}}_2$), and (${\mathrm{H}}_3$) hold. Then problem (1.1) has at least one 2π-periodic solution.

To demonstrate the impulsive effects clearly, we can take

$$I_j(t,x)\equiv d_j,j=1,2,\dots ,p,$$

(1.4)

where $d_1,d_2,\dots ,d_p$ are constants. Hence, $J_j^{\pm }(t)=d_j$.

From Theorem 1.1, we obtain the following result.

Corollary 1.2 Assume that we have the conditions (${\mathrm{H}}_1$), (1.4), and the following.

(${\mathrm{H}}_3^{\prime }$) For all $\theta \in \mathbb{R}$,

$$\begin{array}{r}\sum _{j=1}^p{d}_{j}sin(m{t}_j+\theta )+{\int }_0^{2\pi }e(t)sin(mt+\theta )dt\\ <{\int }_0^{2\pi }(F_{+}(t){sin}^{+}(mt+\theta )-F_{-}(t){sin}^{-}(mt+\theta ))dt\end{array}$$

hold. Then problem (1.1) has at least one 2π-periodic solution.

Moreover, we have the following corollary.

Corollary 1.3 Assume that we have the conditions (${\mathrm{H}}_1$) and the following.

(${\mathrm{H}}_3^{″}$) For all $\theta \in \mathbb{R}$,

$${\int }_0^{2\pi }e(t)sin(mt+\theta )dt<{\int }_0^{2\pi }(F_{+}(t){sin}^{+}(mt+\theta )-F_{-}(t){sin}^{-}(mt+\theta ))dt$$

(1.5)

holds. Then problem (1.2) has at least one 2π-periodic solution.

Remark 1.4 By a simple calculation, one can easily derive

$$F_{+}(t)=\underset{x\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{F(t,x)}{x}⩾\underset{x\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}f(t,x),F_{-}(t)=\underset{x\to -\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{F(t,x)}{x}⩽\underset{x\to -\mathrm{\infty }}{lim\hspace{0.17em}sup}f(t,x).$$

A simple example $f(t,x)=sint+cosx$ illustrates it. Thus condition (${\mathrm{H}}_3^{″}$) generalizes condition (1.3). Hence, our results improve the related results in the literature mentioned above. Moreover, since we consider the problem with impulses, Theorem 1.1 is also a complement of the pioneering works.

Remark 1.5 It is remarkable that Landesman-Lazer condition (${\mathrm{H}}_3^{″}$) is an ‘almost’ necessary and sufficient condition when $F_{+}$ and $F_{-}$ are replaced by $f_{+}$ and $f_{-}$, where $f_{+}={lim}_{x\to +\mathrm{\infty }}f(t,x)$, $f_{-}={lim}_{x\to -\mathrm{\infty }}f(t,x)$, and $f_{-}(t)⩽f(t,x)⩽f_{+}(t)$ (see [[12], p.70]). If the condition (1.5) is not satisfied, i.e., $\mathrm{\exists }\theta \in \mathbb{R}$,

$${\int }_0^{2\pi }e(t)sin(mt+\theta )dt⩾{\int }_0^{2\pi }(F_{+}(t){sin}^{+}(mt+\theta )-F_{-}(t){sin}^{-}(mt+\theta ))dt,$$

problem (1.2) cannot be guaranteed to have periodic solution. For example, we consider resonant differential equation

$$x^{″}+m^{2}x+(1+sinmt)arctanx=8sinmt.$$

(1.6)

Obviously, $f(t,x)=(1+sinmt)arctanx$, $e(t)=8sinmt$, and $F_{+}(t)=\frac{\pi }{2}(1+sinmt)$, $F_{-}(t)=-\frac{\pi }{2}(1+sinmt)$. Taking $\theta =0$, we have

$$\begin{array}{r}{\int }_0^{2\pi }e(t)sinmtdt-{\int }_0^{2\pi }(F_{+}(t){sin}^{+}mt-F_{-}(t){sin}^{-}mt)dt\\ =8\pi -\frac{\pi }{2}{\int }_0^{2\pi }(1+sinmt)|sinmt|dt\\ ⩾8\pi -2{\pi }^2>0.\end{array}$$

Then (${\mathrm{H}}_3^{″}$) is not satisfied. From now on, we prove that (1.6) has not 2π-periodic solution by contradiction. Assume that (1.6) has 2π-periodic solution. Multiplying both sides of (1.6) by $sinmt$ and integrating over $[0,2\pi ]$, we get

$$\begin{array}{rl}8\pi & ={\int }_0^{2\pi }(1+sinmt)arctanxsinmtdt\\ ⩽{\int }_0^{2\pi }|(1+sinmt)arctanxcosmt|dt\\ ⩽\pi {\int }_0^{2\pi }dt=2{\pi }^2,\end{array}$$

which is impossible. Hence, problem (1.2) may have no solution if the condition (${\mathrm{H}}_3^{″}$) is not satisfied. However, as long as (${\mathrm{H}}_3$) holds, problem (1.1) will have at least one periodic solution. Therefore, the impulses can generate a periodic solution.

The rest of the paper is organized as follows. In Section 2, we shall state some notations, some necessary definitions, and a saddle theorem due to Rabinowitz. In Section 3, we shall prove Theorem 1.1.

2 Preliminaries

In the following, we introduce some notations and some necessary definitions.

Define

$$H=\{x\in H^1(0,2\pi ):x(0)=x(2\pi )\},$$

with the norm

$$\parallel x\parallel ={\left({\int }_0^{2\pi }(x^{\prime 2}(t)+x^2(t))dt\right)}^{\frac{1}{2}}.$$

Consider the functional $\phi (x)$ defined on H by

$$\begin{array}{rl}\phi (x)=& \frac{1}{2}{\int }_0^{2\pi }{x}^{\prime 2}(t)dt-\frac{m^2}{2}{\int }_0^{2\pi }{x}^2(t)dt-{\int }_0^{2\pi }F(t,x(t))dt\\ +{\int }_0^{2\pi }e(t)x(t)dt+\sum _{j=1}^p{J}_j(t_j,x(t_j)).\end{array}$$

(2.1)

Similarly as in [7], $\phi (x)$ is continuously differentiable on H, and

$$\begin{array}{rl}{\phi }^{\prime }(x)v(t)=& {\int }_0^{2\pi }{x}^{\prime }(t)v^{\prime }(t)dt-m^2{\int }_0^{2\pi }x(t)v(t)dt-{\int }_0^{2\pi }f(t,x(t))v(t)dt\\ +{\int }_0^{2\pi }e(t)v(t)dt+\sum _{j=1}^p{I}_j(t_j,x(t_j))v(t_j),\text{for }\mathrm{\forall }v(t)\in H.\end{array}$$

(2.2)

Now, we have the following lemma.

Lemma 2.1 If $x\in H$ is a critical point of φ, then x is a 2π-periodic solution of (1.1).

The proof of Lemma 2.1 is similar to Lemma 2.1 in [6], so we omit it.

We say that φ satisfies (PS) if every sequence $(x_n)$ for which $\phi (x_n)$ is bounded in ℝ and ${\phi }^{\prime }(x_n)\to 0$ (as $n\to \mathrm{\infty }$) possesses a convergent subsequence.

To prove the main result, we will use the following saddle point theorem due to Rabinowitz [13] (or see [12]).

Theorem 2.2 Let $\phi \in C^1(H,\mathbb{R})$ and $H=H^{-}\oplus H^{+}$, $dim(H^{-})<\mathrm{\infty }$, $dim(H^{+})=\mathrm{\infty }$. We suppose that:

• (a) There exists a bounded neighborhood D of 0 in $H^{-}$ and a constant α such that $\phi {|}_{\partial D}⩽\alpha$;

• (b) there exists a constant $\beta >\alpha$ such that $\phi {|}_{H^{+}}⩾\beta$;

• (c) φ satisfies (PS).

Then the functional φ has a critical point in H.

3 The proof of Theorem 1.1

In this section, we first show that the functional φ satisfies the Palais-Smale condition.

Lemma 3.1 Assume that the conditions (${\mathrm{H}}_1$), (${\mathrm{H}}_2$), and (${\mathrm{H}}_3$) hold. Then φ defined by (2.1) satisfies (PS).

Proof Let $M>0$ be a constant and $\{x_n\}\subset H$ be a sequence satisfying

$$\begin{array}{rl}|\phi (x_n)|=& |\frac{1}{2}{\int }_0^{2\pi }{x}_n^{\prime 2}dt-\frac{m^2}{2}{\int }_0^{2\pi }{x}_n^{2}dt-{\int }_0^{2\pi }F(t,x_n)dt\\ +{\int }_0^{2\pi }e(t)x_n(t)dt+\sum _{j=1}^p{J}_j(t_j,x_n(t_j))|\\ ⩽& M\end{array}$$

(3.1)

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel {\phi }^{\prime }(x_n)\parallel =0.$$

(3.2)

We first prove that $\{x_n\}$ is bounded in H by contradiction. Assume that $\{x_n\}$ is unbounded. Let $\{z_k\}$ be an arbitrary sequence bounded in H. It follows from (3.2) that, for any $k\in \mathbb{N}$,

$$\underset{n\to \mathrm{\infty }}{lim}|{\phi }^{\prime }(x_n)z_k|⩽\underset{n\to \mathrm{\infty }}{lim}\parallel {\phi }^{\prime }(x_n)\parallel \parallel z_k\parallel =0.$$

Thus

$$\underset{n\to \mathrm{\infty }}{lim}{\phi }^{\prime }(x_n)z_k=0\text{uniformly for }k\in \mathbb{N}.$$

Hence,

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}({\int }_0^{2\pi }(x_n^{\prime }{z}_k^{\prime }-m^2{x}_n{z}_k)dt-{\int }_0^{2\pi }(f(t,x_n)z_k-e(t)z_k)dt\\ +\sum _{j=1}^p{I}_j(t_j,x_n(t_j))z_k(t_j))=0.\end{array}$$

(3.3)

By (${\mathrm{H}}_1$) and (${\mathrm{H}}_2$), we have

$$\underset{n\to \mathrm{\infty }}{lim}({\int }_0^{2\pi }\frac{f(t,x_n)z_k-e(t)z_k}{\parallel x_n\parallel }dt-\frac{{\sum }_{j=1}^p{I}_j(t_j,x_n(t_j))z_k(t_j)}{\parallel x_n\parallel })=0.$$

(3.4)

From (3.3) and (3.4), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^{2\pi }(\frac{x_n^{\prime }}{\parallel x_n\parallel }{z}_k^{\prime }-m^2\frac{x_n}{\parallel x_n\parallel }{z}_k)dt=0.$$

(3.5)

Set

$$y_n=\frac{x_n}{\parallel x_n\parallel }.$$

Then we have

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^{2\pi }(y_n^{\prime }{z}_k^{\prime }-m^2{y}_n{z}_k)dt=0,$$

and furthermore,

$$\underset{i\to \mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{lim}}{\int }_0^{2\pi }[{(y_n-y_i)}^{\prime }{z}_k^{\prime }-m^2(y_n-y_i)z_k]dt=0.$$

(3.6)

Replacing $z_k$ in (3.6) by $(y_n-y_i)$, we get

$$\underset{i\to \mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{lim}}({\parallel y_n-y_i\parallel }^2-(m^2+1){\parallel y_n-y_i\parallel }_2^2)=0.$$

Due to the compact embedding $H\hookrightarrow L^2(0,2\pi )$, going to a subsequence,

$$y_n\rightharpoonup y_0\text{weakly in }H,y_n\to y_0\text{in }{L}^2(0,2\pi ).$$

Therefore,

$$\underset{i\to \mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{lim}}{\parallel y_n-y_i\parallel }_2^2=0.$$

Furthermore, we have

$$\underset{i\to \mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{lim}}{\parallel y_n-y_i\parallel }^2=0,$$

which implies $(y_n)$ is Cauchy sequence in H. Thus, $y_n\to y_0$ in H. It follows from (3.5) and the usual regularity argument for ordinary differential equations (see [14]) that

$$y_0=k_{1}sinmt+k_{2}cosmt,$$

(3.7)

where $k_1^2+k_2^2=\frac{1}{(m^2+1)\pi }$ ($\parallel y_0\parallel =1$). (Different subsequences of $\{y_n\}$ correspond to different $k_1$ and $k_2$.)

Write (3.7) as

$$y_0=\frac{1}{\sqrt{(m^2+1)\pi }}sin(mt+\theta ),$$

where θ satisfies $sin\theta =\frac{k_2}{\sqrt{k_1^2+k_2^2}}$ and $cos\theta =\frac{k_1}{\sqrt{k_1^2+k_2^2}}$.

Taking $z_k=\frac{1}{\sqrt{(m^2+1)\pi }}sin(mt+\theta )$, we get, for any $n\in \mathbb{N}$,

$${\int }_0^{2\pi }(x_n^{\prime }{z}_k^{\prime }-m^2{x}_n{z}_k)dt=0.$$

(3.8)

Thus, it follows from (3.3) and (3.8) that

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}[{\int }_0^{2\pi }(f(t,x_n)-e(t))\frac{1}{\sqrt{(m^2+1)\pi }}sin(mt+\theta )dt\\ -\sum _{j=1}^p{I}_j(t_j,x_n(t_j))\frac{1}{\sqrt{(m^2+1)\pi }}sin(m{t}_j+\theta )]=0.\end{array}$$

(3.9)

By (${\mathrm{H}}_1$) and (${\mathrm{H}}_2$), we obtain

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}[{\int }_0^{2\pi }(f(t,x_n)-e(t))(\frac{1}{\sqrt{(m^2+1)\pi }}sin(mt+\theta )-y_n)dt\\ -\sum _{j=1}^p{I}_j(t_j,x_n(t_j))(\frac{1}{\sqrt{(m^2+1)\pi }}sin(m{t}_j+\theta )-y_n(t_j))]=0.\end{array}$$

(3.10)

It follows from (3.9) and (3.10) that

$$\underset{n\to \mathrm{\infty }}{lim}[{\int }_0^{2\pi }(f(t,x_n)-e(t))y_{n}dt-\sum _{j=1}^p{I}_j(t_j,x_n(t_j))y_n(t_j)]=0.$$

Hence, replacing $z_k$ in (3.3) by $y_n$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^{2\pi }(x_n^{\prime }\frac{x_n^{\prime }}{\parallel x_n\parallel }-m^2{x}_n\frac{x_n}{\parallel x_n\parallel })dt=0.$$

(3.11)

Now, dividing (3.1) by $\parallel x_n\parallel$, we get

$$\begin{array}{rl}\frac{-M}{\parallel x_n\parallel }& ⩽\frac{1}{2}{\int }_0^{2\pi }(\frac{x_n^{\prime 2}}{\parallel x_n\parallel }-\frac{m^2{x}_n^2}{\parallel x_n\parallel })dt-{\int }_0^{2\pi }\frac{F(t,x_n)-e(t)x_n}{\parallel x_n\parallel }+\frac{{\sum }_{j=1}^p{J}_j(t_j,x_n(t_j))}{\parallel x_n\parallel }\\ ⩽\frac{M}{\parallel x_n\parallel },\end{array}$$

which yields

$$\begin{array}{r}{\int }_0^{2\pi }\frac{F(t,x_n)-e(t)x_n}{\parallel x_n\parallel }⩽\frac{M}{\parallel x_n\parallel }+\frac{1}{2}{\int }_0^{2\pi }(\frac{x_n^{\prime 2}}{\parallel x_n\parallel }-\frac{m^2{x}_n^2}{\parallel x_n\parallel })dt+\frac{{\sum }_{j=1}^p{J}_j(t_j,x_n(t_j))}{\parallel x_n\parallel }.\end{array}$$

(3.12)

Note that $\frac{x_n}{\parallel x_n\parallel }\to \frac{1}{\sqrt{(m^2+1)\pi }}sin(mt+\theta )$ in H. Due to the compact embedding $H\hookrightarrow C(0,2\pi )$ and $|x_n(t)|\to +\mathrm{\infty }$, we have $\frac{x_n}{\parallel x_n\parallel }\to \frac{1}{\sqrt{(m^2+1)\pi }}sin(mt+\theta )$ in $C(0,2\pi )$. Furthermore,

$$\underset{n\to \mathrm{\infty }}{lim}{x}_n(t)=\{\begin{array}{cc}+\mathrm{\infty },\hfill & \mathrm{\forall }t\in I_{+}:=\{t\in [0,2\pi ]|sin(mt+\theta )>0\},\hfill \\ -\mathrm{\infty },\hfill & \mathrm{\forall }t\in I_{-}:=\{t\in [0,2\pi ]|sin(mt+\theta )<0\}.\hfill \end{array}$$

Hence, from (3.11) and (3.12), we have

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\int }_0^{2\pi }\frac{F(t,x_n)-e(t)x_n}{\parallel x_n\parallel }dt⩽& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\sum _{j=1}^p\frac{J_j(t_j,x_n(t_j))}{x_n(t_j)}\cdot \frac{x_n^{+}(t_j)-x_n^{-}(t_j)}{\parallel x_n\parallel }\\ ⩽& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\sum _{j=1}^p\frac{J_j(t_j,x_n(t_j))}{x_n(t_j)}\cdot \frac{x_n^{+}(t_j)}{\parallel x_n\parallel }\\ -\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\sum _{j=1}^p\frac{J_j(t_j,x_n(t_j))}{x_n(t_j)}\cdot \frac{x_n^{-}(t_j)}{\parallel x_n\parallel }\\ =& \frac{1}{\sqrt{(m^2+1)\pi }}\sum _{j=1}^p{J}_j^{+}(t_j){sin}^{+}(m{t}_j+\theta )\\ -\frac{1}{\sqrt{(m^2+1)\pi }}\sum _{j=1}^p{J}_j^{-}(t_j){sin}^{-}(m{t}_j+\theta ).\end{array}$$

(3.13)

Using Fatou’s lemma, we get

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\int }_0^{2\pi }\frac{F(t,x_n)}{\parallel x_n\parallel }dt& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}[{\int }_{I_{+}}\frac{F(t,x_n)}{x_n}\frac{x_n}{\parallel x_n\parallel }dt-{\int }_{I_{-}}\frac{F(t,x_n)}{x_n}\frac{-x_n}{\parallel x_n\parallel }dt]\\ ⩾{\int }_{I_{+}}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{F(t,x_n)}{x_n}\frac{x_n}{\parallel x_n\parallel }dt-{\int }_{I_{-}}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{F(t,x_n)}{x_n}\frac{-x_n}{\parallel x_n\parallel }dt.\end{array}$$

Thus, by a simple computation, we have

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\int }_0^{2\pi }\frac{F(t,x_n)}{\parallel x_n\parallel }dt\\ ⩾\frac{1}{\sqrt{(m^2+1)\pi }}{\int }_0^{2\pi }[F_{+}(t){sin}^{+}(mt+\theta )-F_{-}(t){sin}^{-}(mt+\theta )]dt.\end{array}$$

(3.14)

Hence, it follows from (3.13) and (3.14) that

$$\begin{array}{r}\sum _{j=1}^p{J}_j^{+}(t_j){sin}^{+}(m{t}_j+\theta )-\sum _{j=1}^p{J}_j^{-}(t_j){sin}^{-}(m{t}_j+\theta )+{\int }_0^{2\pi }e(t)sin(mt+\theta )dt\\ ⩾{\int }_0^{2\pi }[F_{+}(t){sin}^{+}(mt+\theta )-F_{-}(t){sin}^{-}(mt+\theta )]dt.\end{array}$$

This contradicts (${\mathrm{H}}_3$). It implies that the sequence $(x_n)$ is bounded. Thus, there exists $x_0\in H$ such that $x_n\rightharpoonup x_0$ weakly in H. Due to the compact embedding $H\hookrightarrow L^2(0,2\pi )$ and $H\hookrightarrow C(0,2\pi )$, going to a subsequence,

$$x_n\to x_0\text{in }{L}^2(0,2\pi ),x_n\to x_0\text{in }C(0,2\pi ).$$

From (3.3), we obtain

$$\begin{array}{r}\underset{i\to \mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{lim}}({\int }_0^{2\pi }((x_n^{\prime }-x_i^{\prime })z_k^{\prime }-m^2(x_n-x_i)z_k)dt-{\int }_0^{2\pi }(f(t,x_n)-f(t,x_i))z_{k}dt\\ +\sum _{j=1}^p(I_j(t_j,x_n(t_j))-I_j(t_j,x_i(t_j)))z_k(t_j))=0.\end{array}$$

Replacing $z_k$ by $x_n-x_i$ in the above equality, we get

$$\begin{array}{r}\underset{i\to \mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{lim}}({\int }_0^{2\pi }({(x_n^{\prime }-x_i^{\prime })}^2-m^2{(x_n-x_i)}^2)dt-{\int }_0^{2\pi }(f(t,x_n)-f(t,x_i))(x_n-x_i)dt\\ +\sum _{j=1}^p(I_j(t_j,x_n(t_j))-I_j(t_j,x_i(t_j)))(x_n(t_j)-x_i(t_j)))=0.\end{array}$$

(3.15)

By (${\mathrm{H}}_1$) and (${\mathrm{H}}_2$), we have

$$\underset{i\to \mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{lim}}{\int }_0^{2\pi }(f(t,x_n)-f(t,x_i))(x_n-x_i)dt=0$$

(3.16)

and

$$\underset{i\to \mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{lim}}\sum _{j=1}^p(I_j(t_j,x_n(t_j))-I_j(t_j,x_i(t_j)))(x_n(t_j)-x_i(t_j))=0.$$

(3.17)

Thus, it follows from (3.15), (3.16), and (3.17) that

$$\underset{i\to \mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{lim}}{\int }_0^{2\pi }[{(x_n^{\prime }-x_i^{\prime })}^2-m^2{(x_n-x_i)}^2]dt=0.$$

Therefore,

$$\underset{i\to \mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{lim}}{\parallel x_n-x_i\parallel }^2=0,$$

which implies $x_n\to x_0$ in H. It shows that φ satisfies (PS). □

Now, we can give the proof of Theorem 1.1.

Proof of Theorem 1.1 Denote

$$H^{-}=\mathbb{R}\oplus span\{sint,cost,sin2t,cos2t,\dots ,sinmt,cosmt\}$$

and

$$H^{+}=span\{sin(m+1)t,cos(m+1)t,\dots \}.$$

We first prove that

$$\underset{\parallel x\parallel \to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (x)=-\mathrm{\infty },\text{for }x\in H^{-},$$

(3.18)

by contradiction. Assume that there exists a sequence $(x_n)\subset H^{-}$ such that $\parallel x_n\parallel \to \mathrm{\infty }$ (as $n\to \mathrm{\infty }$) and there exists a constant $c_{-}$ satisfying

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (x_n)⩾c_{-}.$$

(3.19)

By (${\mathrm{H}}_1$), we have

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^{2\pi }\frac{F(t,x_n)-e(t)x_n}{{\parallel x_n\parallel }^2}dt=0.$$

(3.20)

By (${\mathrm{H}}_2$), we get

$$\underset{n\to \mathrm{\infty }}{lim}\sum _{j=1}^p\frac{J_j(t_j,x_n(t_j))}{{\parallel x_n\parallel }^2}=0.$$

(3.21)

From (3.19) and the definition of φ, we obtain

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}[\frac{1}{2}{\int }_0^{2\pi }\frac{x_n^{\prime 2}-m^2{x}_n^2}{{\parallel x_n\parallel }^2}dt-{\int }_0^{2\pi }\frac{F(t,x_n)-e(t)x_n}{{\parallel x_n\parallel }^2}dt+\sum _{j=1}^p\frac{J_j(t_j,x_n(t_j))}{{\parallel x_n\parallel }^2}]\\ ⩾0.\end{array}$$

(3.22)

For $x\in H^{-}$, we have

$${\int }_0^{2\pi }(x^{\prime 2}-m^2{x}^2)dt={\parallel x\parallel }^2-(m^2+1){\parallel x\parallel }_2^2⩽0.$$

(3.23)

The equality in (3.23) holds only for

$$x=\frac{1}{\sqrt{(m^2+1)\pi }}sin(mt+\theta ),\theta \in \mathbb{R}.$$

Set $y_n=\frac{x_n}{\parallel x_n\parallel }$. Since $dim{H}^{-}<\mathrm{\infty }$, going to a subsequence, there exists $y_0\in H^{-}$ such that $y_n\to y_0$ in H and $y_n\to y_0$ in $L^2(0,2\pi )$. Then (3.20), (3.21), (3.22), and (3.23) imply that

$$y_0=\frac{1}{\sqrt{(m^2+1)\pi }}sin(mt+\theta ),\theta \in \mathbb{R}.$$

By (3.19), we have, for n large enough,

$$\frac{1}{2}{\int }_0^{2\pi }\frac{x_n^{\prime 2}-m^2{x}_n^2}{\parallel x_n\parallel }dt-{\int }_0^{2\pi }\frac{F(t,x_n)-e(t)x_n}{\parallel x_n\parallel }dt+\sum _{j=1}^p\frac{J_j(t_j,x_n(t_j))}{\parallel x_n\parallel }⩾\frac{c_{-}}{\parallel x_n\parallel }.$$

(3.24)

It follows from $x_n\in H^{-}$ that

$${\int }_0^{2\pi }\frac{x_n^{\prime 2}-m^2{x}_n^2}{\parallel x_n\parallel }⩽0.$$

(3.25)

From (3.24) and (3.25), we get, for n large enough,

$$\frac{c_{-}}{\parallel x_n\parallel }⩽-{\int }_0^{2\pi }\frac{F(t,x_n)-e(t)x_n}{\parallel x_n\parallel }dt+\sum _{j=1}^p\frac{J_j(t_j,x_n(t_j))}{\parallel x_n\parallel }.$$

Thus,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\int }_0^{2\pi }(\frac{F(t,x_n)}{x_n}-e(t))\frac{x_n}{\parallel x_n\parallel }dt⩽\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\sum _{j=1}^p\frac{J_j(t_j,x_n(t_j))}{\parallel x_n\parallel }.$$

Using an argument similar to the proof of Lemma 3.1, we get

$$\begin{array}{r}\sum _{j=1}^p{J}_j^{+}(t_j){sin}^{+}(m{t}_j+\theta )-\sum _{j=1}^p{J}_j^{-}(t_j){sin}^{-}(m{t}_j+\theta )+{\int }_0^{2\pi }e(t)sin(mt+\theta )dt\\ ⩾{\int }_0^{2\pi }(F_{+}(t){sin}^{+}(mt+\theta )-F_{-}(t){sin}^{-}(mt+\theta ))dt,\end{array}$$

which is a contradiction to (${\mathrm{H}}_3$).

Then (3.18) holds.

Next, we prove that

$$\underset{\parallel x\parallel \to \mathrm{\infty }}{lim}\phi (x)=\mathrm{\infty },\text{for all }x\in H^{+},$$

and φ is bounded on bounded sets.

Because of the compact embedding of $H\hookrightarrow C(0,2\pi )$ and $H\hookrightarrow L^2(0,2\pi )$, there exists constants $m_1$, $m_2$ such that

$${\parallel x\parallel }_{\mathrm{\infty }}⩽m_1\parallel x\parallel ,{\parallel x\parallel }_2⩽m_2\parallel x\parallel .$$

Then by (${\mathrm{H}}_1$) and (${\mathrm{H}}_2$), one has

$$\begin{array}{rl}|\phi (x)|=& |\frac{1}{2}{\int }_0^{2\pi }{x}^{\prime 2}dt-\frac{m^2}{2}{\int }_0^{2\pi }{x}^{2}dt-{\int }_0^{2\pi }[F(t,x)-e(t)x]dt\\ +\sum _{j=1}^p{J}_j(t_j,x(t_j))|\\ ⩽& \frac{1}{2}{\parallel x\parallel }^2+\frac{m^2}{2}{m}_2^2{\parallel x\parallel }^2+{\int }_0^{2\pi }(|p(t)||x|+|e(t)||x|)dt\\ +\sum _{j=1}^p{c}_j|x(t_j)|\\ ⩽& \frac{1+m^2{m}_2^2}{2}{\parallel x\parallel }^2+m_1({\parallel p\parallel }_1+{\parallel e\parallel }_1)\parallel x\parallel +\sum _{j=1}^p{c}_j{m}_1\parallel x\parallel .\end{array}$$

(3.26)

Hence, φ is bounded on bounded sets of H.

Since $x\in H^{+}$, we have

$${\parallel x\parallel }^2⩾({(m+1)}^2+1){\parallel x\parallel }_2^2.$$

(3.27)

Thus, from (3.26) and (3.27), we obtain

$$\begin{array}{rl}\phi (x)& =\frac{1}{2}{\int }_0^{2\pi }{x}^{\prime 2}dt-\frac{m^2}{2}{\int }_0^{2\pi }{x}^{2}dt-{\int }_0^{2\pi }[F(t,x)-e(t)x]dt+\sum _{j=1}^p{J}_j(t_j,x(t_j))\\ ⩾\frac{2m+1}{2({(m+1)}^2+1)}{\parallel x\parallel }^2-m_1({\parallel p\parallel }_1+{\parallel e\parallel }_1+\sum _{j=1}^p{c}_j)\parallel x\parallel ,\end{array}$$

which implies

$$\underset{\parallel x\parallel \to \mathrm{\infty }}{lim}\phi (x)=\mathrm{\infty },\text{for all }x\in H^{+}.$$

Up to now, the conditions (a) and (b) of Theorem 2.2 are satisfied. According to Lemma 3.1, (c) is also satisfied. Hence, by Theorem 2.2, (1.1) has at least one solution. This completes the proof. □