Existence of periodic solutions for a class of second order Hamiltonian systems

Abstract

By using the least action principle and the minimax methods, the existence of periodic solutions for a class of second order Hamiltonian systems is considered. The results obtained in this paper extend some previous results.

1 Introduction and main results

Consider the second order Hamiltonian system

$$ \textstyle\begin{cases} \ddot{u}(t)=\nabla F(t,u(t)),\\ u(T)-u(0)=\dot{u}(T)-\dot{u}(0)=0, \end{cases} $$

(1.1)

where $T>0$ and $F:[0,T]\times\mathbb{R}^{N}\rightarrow\mathbb{R}$ satisfies the following assumption:

(A) $F(t,x)$ is measurable in t for every $x\in\mathbb{R}^{N}$, continuously differentiable in x for a.e. $t\in[0,T]$, and there exist $a\in C(\mathbb{R}^{+},\mathbb{R}^{+})$, $b\in L^{1}([0,T];\mathbb {R}^{+})$ such that

$$\bigl\vert F(t,x)\bigr\vert \leq a\bigl(\vert x\vert \bigr)b(t), \qquad \bigl\vert \nabla F(t,x)\bigr\vert \leq a\bigl(\vert x\vert \bigr)b(t) $$

for all $x\in\mathbb{R}^{N}$ and a.e. $t\in[0,T]$.

The corresponding functional $\varphi:H_{T}^{1}\rightarrow\mathbb {R}$,

$$\varphi(u)=\frac{1}{2}\int_{0}^{T} \bigl\vert \dot{u}(t)\bigr\vert ^{2}\,dt+\int_{0}^{T}F \bigl(t,u(t) \bigr)\,dt, $$

is continuously differentiable and weakly lower semi-continuous on $H_{T}^{1}$, where $H_{T}^{1}$ is the usual Sobolev space with the norm

$$\Vert u\Vert = \biggl(\int_{0}^{T} \bigl\vert u(t)\bigr\vert ^{2} \,dt+\int_{0}^{T} \bigl\vert \dot {u}(t)\bigr\vert ^{2}\,dt \biggr)^{1/2} $$

for $u\in H^{1}_{T}$, and

$$\bigl\langle \varphi'(u),v \bigr\rangle =\int ^{T}_{0} \bigl[ \bigl(\dot{u}(t),\dot{v} \bigr)+ \bigl( \nabla F \bigl(t,u(t) \bigr),v(t) \bigr) \bigr]\,dt $$

for all $u, v\in H^{1}_{T}$, It is well known that the solutions of problem (1.1) correspond to the critical points of φ.

The existence of periodic solutions for problem (1.1) is obtained in [1–22] with many solvability conditions by using the least action principle and the minimax methods, such as the coercive type potential condition (see [2]), the convex type potential condition (see [5]), the periodic type potential conditions (see [16]), the even type potential condition (see [4]), the subquadratic potential condition in Rabinowitz’s sense (see [9]), the bounded nonlinearity condition (see [6]), the subadditive condition (see [11]), the sublinear nonlinearity condition (see [3, 13]), and the linear nonlinearity condition (see [7, 15, 19, 20]).

In particular, when the nonlinearity $\triangledown F(t,x)$ is bounded, that is, there exists $g(t)\in L^{1}([0,T],\mathbb{R}^{+})$ such that $\vert \triangledown F(t,x)\vert \leq g(t)$ for all $x\in\mathbb{R}^{N}$ and a.e. $t\in[0,T]$, and that

$$\int_{0}^{T}F(t,x)\,dt\rightarrow\pm \infty \quad \mbox{as } \vert x\vert \rightarrow\infty, $$

Mawhin and Willem [6] proved that problem (1.1) has at least one periodic solution.

In [3, 13], Han and Tang generalized these results to the sublinear case:

$$ \bigl\vert \triangledown F(t,x)\bigr\vert \leq f(t)\vert x\vert ^{\alpha}+g(t) \quad \text{for all } x\in \mathbb{R}^{N} \text{ and a.e. } t\in[0,T] $$

(1.2)

and

$$ \vert x\vert ^{-2\alpha}\int_{0}^{T}F(t,x) \,dt \rightarrow\pm\infty \quad \text{as } \vert x\vert \rightarrow\infty, $$

(1.3)

where $f(t),g(t)\in L^{1}([0,T],\mathbb{R}^{+})$ and $\alpha\in[0,1)$.

Subsequently, when $\alpha=1$ Zhao and Wu [19, 20] and Meng and Tang [7, 15] also proved the existence of periodic solutions for problem (1.1), i.e. $\nabla F(t,x)$ was linear:

$$\bigl\vert \triangledown F(t,x)\bigr\vert \leq f(t)\vert x \vert +g(t) \quad \mbox{for all }x\in \mathbb{R}^{N} \text{ and a.e. } t\in[0,T], $$

where $f(t),g(t)\in L^{1}([0,T],\mathbb{R}^{+})$.

Recently, Wang and Zhang [21] used a control function $h(\vert x\vert )$ instead of $\vert x\vert ^{\alpha}$ in (1.2) and (1.3) and got some new results, where h satisfied the following conditions:

(B) $h\in C([0,\infty),[0,\infty))$ and there exist constants $C_{0}>0$, $K_{1}>0$, $K_{2}>0$, $\alpha\in[0,1)$ such that

• (i) $h(s)\leq h(t)$ $\forall s\leq t$, $s, t\in[0,\infty)$,

• (ii) $h(s+t)\leq C_{0}(h(s)+h(t))$ $\forall s, t\in[0,\infty)$,

• (iii) $0\leq h(s)\leq K_{1}s^{\alpha}+K_{2}$ $\forall s\in[0,\infty )$,

• (iv) $h(s)\rightarrow\infty$ as $s\rightarrow\infty$.

Motivated by the results mentioned above, we will consider the periodic solutions for problem (1.1). The following are our main results.

Theorem 1.1

Suppose that $F(t,x)=F_{1}(t,x)+F_{2}(x)$, where $F_{1}$ and $F_{2}$ satisfy assumption (A) and the following conditions:

• (1) there exist $f,g\in L^{1}([0,T];\mathbb{R}^{+})$ such that

  $$\bigl\vert \nabla F_{1}(t,x)\bigr\vert \leq f(t)h\bigl( \vert x\vert \bigr)+g(t), $$

  for all $x\in\mathbb{R}^{N}$ and a.e. $t\in[0,T]$, here h satisfies (B);

• (2) there exist constants $r>0$ and $\gamma\in[0,2)$ such that

  $$\bigl(\nabla F_{2}(x)-\nabla F_{2}(y),x-y \bigr) \geq-r\vert x-y\vert ^{\gamma}, $$

  for all $x,y \in\mathbb{R}^{N}$;

• (3)
  $$\liminf_{\vert x\vert \rightarrow\infty}h^{-2}\bigl(\vert x\vert \bigr)\int_{0}^{T}F(t,x)\,dt> \frac {T^{2}C_{0}^{2}}{8\pi^{2}}\int _{0}^{T}f^{2}(t)\,dt. $$

Then problem (1.1) has at least one periodic solution which minimizes φ on $H_{T}^{1}$.

Theorem 1.2

Suppose that $F(t,x)=F_{1}(t,x)+F_{2}(x)$, where $F_{1}$ and $F_{2}$ satisfy assumption (A), (1), (2), and the following conditions:

• (4) there exist $\delta\in[0,2)$ and $\mu>0$ such that

  $$\bigl(\nabla F_{2}(x)-\nabla F_{2}(y),x-y \bigr) \leq \mu \vert x-y\vert ^{\delta}, $$

  for all $x,y \in R^{N}$;

• (5)
  $$\limsup_{\vert x\vert \rightarrow\infty}h^{-2}\bigl(\vert x\vert \bigr)\int_{0}^{T}F(t,x)\,dt< -\frac {3T^{2}C_{0}^{2}}{8\pi^{2}} \int_{0}^{T}f^{2}(t)\,dt. $$

Then problem (1.1) has at least one periodic solution which minimizes φ on $H_{T}^{1}$.

Theorem 1.3

Suppose that $F(t,x)=F_{1}(t,x)+F_{2}(x)$, where $F_{1}$ and $F_{2}$ satisfy assumption (A), (1), and the following conditions:

• (6) there exists a constant $0< r<4\pi^{2}/T^{2}$, such that

  $$\bigl(\nabla F_{2}(x)-\nabla F_{2}(y),x-y \bigr) \geq-r\vert x-y\vert ^{2}, $$

  for all $x,y \in R^{N}$;

• (7)
  $$\liminf_{\vert x\vert \rightarrow\infty}h^{-2}\bigl(\vert x\vert \bigr)\int_{0}^{T}F(t,x)\,dt> \frac {T^{2}}{2(4\pi^{2}-rT^{2})}\int _{0}^{T}f^{2}(t)\,dt. $$

Then problem (1.1) has at least one periodic solution which minimizes φ on $H_{T}^{1}$.

Theorem 1.4

Suppose that $F=F_{1}+F_{2}$, where $F_{1}$ and $F_{2}$ satisfy assumption (A), (1), and the following conditions:

• (8) there exist $k\in L^{1}([0,T];\mathbb{R}^{+})$ and ($\lambda,\mu$)-subconvex potential $G: \mathbb{R}^{N}\rightarrow\mathbb{R}$ with $\lambda>1/2$ and $0<\mu<2\lambda^{2}$, such that

  $$\bigl(\nabla F_{2}(t,x),y \bigr)\geq-k(t)G(x-y), $$

  for all $x,y \in\mathbb{R}^{N}$;

• (9)
  $$\begin{aligned}& \limsup_{\vert x\vert \rightarrow\infty} h^{-2}\bigl(\vert x\vert \bigr)\int_{0}^{T}F_{1}(t,x)\,dt < - \frac{3T^{2}C_{0}^{2}}{8\pi^{2}}\int_{0}^{T}f^{2}(t) \,dt, \\& \limsup_{\vert x\vert \rightarrow\infty} \vert x\vert ^{-\beta}\int _{0}^{T}F_{2}(t,x) \,dt \leq-8\mu\max _{\vert s\vert \leq1}G(s)\int_{0}^{T}k(t) \,dt, \end{aligned}$$

  where $\beta=\log_{2\lambda}(2\mu)$.

Then problem (1.1) has at least one periodic solution which minimizes φ on $H_{T}^{1}$.

Remark 1.5

Theorems 1.1-1.4 extend some existing results: (i) [22], Theorems 1.1-1.4, are special cases of Theorems 1.1-1.4 with control function $h(t)=t^{\alpha}, \alpha\in[0,1)$, $t\in[0,+\infty)$; (ii) if $F_{2}=0$, [15], Theorems 1 and 2, are special cases of Theorem 1.1 and Theorem 1.2, respectively; (iii) If $F_{2}=0$, Theorem 1.1 and Theorem 1.2 extend [21], Theorems 1.1 and 1.2, since we weaken the so-called Ahmad-Lazer-Paul type conditions with the control function $h(t)$.

2 Proof of theorems

For $u\in H_{T}^{1}$, let $\bar{u}=\frac{1}{T}\int_{0}^{T}\vert \dot {u}(t)\vert \,dt$ and $\tilde{u}(t)=u(t)-\bar{u}$. Then one has

$$\begin{aligned}& \Vert \tilde{u}\Vert _{\infty}^{2} \leq\frac{T}{12}\int _{0}^{T}\bigl\vert \dot{u}(t)\bigr\vert ^{2}\,dt \quad \text{(Sobolev's inequality),} \\& \Vert \tilde{u}\Vert _{L^{2}}^{2} \leq\frac{T^{2}}{4\pi^{2}}\int _{0}^{T}\bigl\vert \dot{u}(t)\bigr\vert ^{2}\,dt \quad \text{(Wirtinger's inequality).} \end{aligned}$$

For the sake of convenience, we denote $M_{1}=(\int_{0}^{T}f^{2}(t)\,dt)^{1/2}$, $M_{2}=\int_{0}^{T}f(t)\,dt$, $M_{3}=\int_{0}^{T}g(t)\,dt$.

Proof of Theorem 1.1

Due to (3), we can choose an $a_{1}>T^{2}/(4\pi^{2})$ such that

$$ \liminf_{\vert x\vert \rightarrow\infty}h^{-2}\bigl(\vert x\vert \bigr)\int _{0}^{T}F(t,x) \,dt>\frac {a_{1}C_{0}^{2}}{2}M_{1}^{2}. $$

(2.1)

For (B) and the Sobolev inequality, for any $u\in H_{T}^{1}$ we have

$$\begin{aligned} & \biggl\vert \int_{0}^{T} \bigl[F_{1} \bigl(t,u(t) \bigr)-F_{1}(t,\bar{u}) \bigr]\,dt \biggr\vert \\ &\quad = \biggl\vert \int_{0}^{T}\int _{0}^{1} \bigl(\nabla F_{1} \bigl(t,\bar {u}+s\tilde{u}(t) \bigr),\tilde{u}(t) \bigr)\,ds\,dt \biggr\vert \\ &\quad \leq\int_{0}^{T}\int_{0}^{1}f(t)h \bigl(\bigl\vert \bar{u}+s\tilde {u}(t)\bigr\vert \bigr)\bigl\vert \tilde{u}(t)\bigr\vert \,ds\,dt +\int_{0}^{T}\int _{0}^{1}g(t)\bigl\vert \tilde {u}(t)\bigr\vert \,ds\,dt \\ &\quad \leq\int_{0}^{T}\int _{0}^{1}C_{0}f(t) \bigl(h\bigl(\vert \bar {u}\vert \bigr)+h \bigl(\bigl\vert \tilde{u}(t)\bigr\vert \bigr) \bigr) \bigl\vert \widetilde{u}(t)\bigr\vert \,ds\,dt + M_{3}\Vert \tilde{u}\Vert _{\infty} \\ &\quad \leq C_{0}h\bigl(\vert \bar{u}\vert \bigr) \biggl(\int _{0}^{T}f^{2}(t) \,dt \biggr)^{1/2} \biggl(\int_{0}^{T}\bigl\vert \tilde{u}(t)\bigr\vert ^{2}\,dt \biggr)^{1/2} \\ &\qquad {}+ C_{0}\int_{0}^{T}f(t)h \bigl(\bigl\vert \tilde{u}(t)\bigr\vert \bigr)\bigl\vert \tilde {u}(t)\bigr\vert \,dt+M_{3}\Vert \tilde{u} \Vert _{\infty} \\ &\quad \leq C_{0}M_{1}h\bigl(\vert \bar{u}\vert \bigr)\Vert \tilde{u} \Vert _{L^{2}}+C_{0}\int_{0}^{T}f(t) \bigl(K_{1}\bigl\vert \tilde{u}(t)\bigr\vert ^{\alpha}+K_{2} \bigr)\bigl\vert \tilde{u}(t)\bigr\vert \,dt +M_{3}\Vert \tilde{u}\Vert _{\infty} \\ &\quad \leq C_{0}M_{1}h\bigl(\vert \bar{u}\vert \bigr)\Vert \tilde{u}\Vert _{L^{2}}+C_{0}M_{2}K_{1} \Vert \tilde{u} \Vert _{\infty}^{1+\alpha} +C_{0}M_{2}K_{2} \bigl\Vert \tilde{u}(t)\bigr\Vert _{\infty}+M_{3}\bigl\Vert \tilde{u}(t)\bigr\Vert _{\infty} \\ &\quad \leq\frac{1}{2a_{1}}\Vert \widetilde{u}\Vert ^{2}_{L^{2}}+ \frac {a_{1}(C_{0}M_{1})^{2}}{2}h^{2}\bigl(\vert \bar{u}\vert \bigr)+ C_{0}M_{2}K_{1} \Vert \tilde{u}\Vert _{\infty}^{1+\alpha} \\ &\qquad {}+C_{0}M_{2}K_{2} \bigl\Vert \tilde{u}(t)\bigr\Vert _{\infty}+M_{3}\bigl\Vert \tilde{u}(t)\bigr\Vert _{\infty} \\ &\quad \leq\frac{T^{2}}{8\pi^{2}a_{1}}\Vert \dot{u}\Vert _{L^{2}}^{2}+ \frac{a_{1}(C_{0}M_{1})^{2}}{2}h^{2}\bigl(\vert \bar{u}\vert \bigr) + \biggl( \frac{T}{12} \biggr)^{(1+\alpha)/2}C_{0}M_{2}K_{1} \Vert \dot{u}\Vert _{L^{2}}^{1+\alpha} \\ &\qquad {}+ \biggl(\frac{T}{12} \biggr)^{1/2}C_{0}M_{2}K_{2} \Vert \dot{u}\Vert _{L^{2}}+ \biggl(\frac{T}{12} \biggr)^{1/2}M_{3}\Vert \dot{u} \Vert _{L^{2}.} \end{aligned}$$

(2.2)

Similarly, from (2) and the Sobolev inequality, for any $u\in H_{T}^{1}$ we get

$$\begin{aligned} &\int^{T}_{0} \bigl[F_{2} \bigl(u(t) \bigr)-F_{2}(\bar{u}) \bigr]\,dt \\ &\quad =\int_{0}^{T}\int_{0}^{1} \frac{1}{s} \bigl(\nabla F_{2} \bigl(\bar{u}+s\tilde {u}(t) \bigr)- \nabla F_{2}(\bar{u}),s\tilde{u}(t) \bigr)\,ds\,dt \\ &\quad \geq-\int_{0}^{T}\int _{0}^{1}rs^{\gamma-1}\bigl\vert \tilde {u}(t) \bigr\vert ^{\gamma}\,ds\,dt \\ &\quad \geq-\frac{rT}{\gamma} \Vert \widetilde{u}\Vert _{\infty}^{\gamma } \\ &\quad \geq-\frac{rT}{\gamma} \biggl(\frac{T}{12} \biggr)^{\gamma /2}\Vert \dot{u}\Vert _{L^{2}}^{\gamma}. \end{aligned}$$

(2.3)

From (2.2) and (2.3) we have

$$\begin{aligned} \varphi(u)={}&\frac{1}{2}\Vert \dot{u}\Vert _{L^{2}}^{2}+ \int_{0}^{T} \bigl[F_{1} \bigl(t,u(t) \bigr)-F_{1}(t,\bar{u}) \bigr]\,dt \\ &{} +\int^{T}_{0} \bigl[F_{2} \bigl(u(t) \bigr)-F_{2}(\bar{u}) \bigr]\,dt+\int_{0}^{T}F(t, \bar{u})\,dt \\ \geq {}&\biggl(\frac{1}{2}-\frac{T^{2}}{8\pi^{2}a_{1}} \biggr)\Vert \dot{u} \Vert _{L^{2}}^{2}-\frac{a_{1}(C_{0}M_{1})^{2}}{2}h^{2}\bigl(\vert \bar{u}\vert \bigr) - \biggl(\frac{T}{12} \biggr)^{\frac{1+\alpha}{2}}C_{0}M_{2}K_{1} \Vert \dot {u}\Vert _{L^{2}}^{1+\alpha} \\ &{} - \biggl(\frac{T}{12} \biggr)^{1/2}C_{0}M_{2}K_{2} \Vert \dot{u}\Vert _{L^{2}}- \biggl(\frac{T}{12} \biggr)^{1/2}M_{3} \Vert \dot{u}\Vert _{L^{2}} \\ &{}- \frac{rT}{\gamma} \biggl(\frac{T}{12} \biggr)^{\alpha/2}\Vert \dot{u} \Vert _{L^{2}}^{\gamma}+\int_{0}^{T}F(t, \bar{u})\,dt \\ \geq {}&\biggl(\frac{1}{2}-\frac{T^{2}}{8\pi^{2}a_{1}} \biggr)\Vert \dot{u} \Vert _{L^{2}}^{2} +h^{2}\bigl(\vert \bar{u}\vert \bigr) \biggl(h^{-2}\bigl(\vert \bar{u}\vert \bigr)\int _{0}^{T}F(t,\bar{u})\,dt- \frac {a_{1}(C_{0}M_{1})^{2}}{2} \biggr) \\ &{} - \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3}) \Vert \dot{u}\Vert _{L^{2}}\\ &{} - \biggl(\frac{T}{12} \biggr)^{\frac{1+\alpha}{2}}C_{0}M_{2}K_{1} \Vert \dot {u}\Vert _{L^{2}}^{1+\alpha} -\frac{rT}{\gamma} \biggl( \frac{T}{12} \biggr)^{\gamma/2}\Vert \dot{u}\Vert _{L^{2}}^{\gamma}, \end{aligned}$$

for all $u\in H_{T}^{1}$. So, by (2.1) we get $\varphi(u)\rightarrow \infty$ as $\Vert u\Vert \rightarrow\infty$.

Hence, applying the least action principle (see [6], Theorem 1.1 and Corollary 1.1), the proof is complete. □

Proof of Theorem 1.2

Step 1. First, we assert that φ satisfies the (PS) condition. Suppose that $\{u_{n}\}$ is a (PS) sequence, that is, $\varphi '(u_{n})\rightarrow0$ as $n\rightarrow\infty$ and $\{\varphi(u_{n})\}$ is bounded. For (5), we can choose an $a_{2}>T^{2}/(4\pi^{2})$ such that

$$ \limsup_{\vert x\vert \rightarrow\infty} h^{-2}\bigl(\vert x\vert \bigr) \int_{0}^{T}F(t,x) \,dt < - \biggl( \frac{a_{2}}{2}+\frac{\sqrt{a_{2}}T}{2\pi} \biggr)C_{0}^{2} \int _{0}^{T}f^{2}(t)\,dt. $$

(2.4)

Similar to the proof of Theorem 1.1, we have

$$\begin{aligned} & \biggl\vert \int_{0}^{T} \bigl(\nabla F_{1} \bigl(t,u_{n}(t) \bigr),\tilde {u}_{n}(t) \bigr)\,dt \biggr\vert \\ & \quad \leq\frac{T^{2}}{8\pi^{2}a_{2}}\Vert \dot{u}_{n}\Vert _{L^{2}}^{2}+ \frac {a_{2}(C_{0}M_{1})^{2}}{2}h^{2}\bigl(\vert \bar{u}_{n}\vert \bigr) + \biggl( \frac{T}{12} \biggr)^{(1+\alpha)/2}C_{0}M_{2}K_{1} \Vert \dot{u}_{n}\Vert _{L^{2}}^{1+\alpha} \\ &\qquad {} + \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3}) \Vert \dot{u}_{n}\Vert _{L^{2}} \end{aligned}$$

(2.5)

and

$$\begin{aligned} \int_{0}^{T} \bigl(\nabla F_{2} \bigl(u_{n}(t) \bigr),\tilde{u}_{n}(t) \bigr)\,dt \geq - \frac {rT}{\gamma} \biggl(\frac{T}{12} \biggr)^{\gamma/2}\Vert \dot{u} \Vert _{L^{2}}^{\gamma}, \end{aligned}$$

for all n. Hence we have

$$\begin{aligned} \Vert \tilde{u}_{n}\Vert \geq {}&\bigl\langle \varphi'(u_{n}), \tilde {u}_{n} \bigr\rangle \\ ={}&\Vert \dot{u}_{n}\Vert _{L^{2}}^{2}+\int _{0}^{T} \bigl(\nabla F \bigl(t,u_{n}(t) \bigr),\tilde{u}_{n}(t) \bigr)\,dt \\ \geq{}& \biggl(1- \frac{T^{2}}{8\pi^{2}a_{2}} \biggr)\Vert \dot {u}_{n} \Vert _{L^{2}}^{2}-\frac{a_{2}(C_{0}M_{1})^{2}}{2}h^{2}\bigl( \vert \bar{u}_{n}\vert \bigr) \\ &{}- \biggl(\frac{T}{12} \biggr)^{\frac{1+\alpha}{2}}C_{0}M_{2}K_{1} \Vert \dot {u}_{n}\Vert _{L^{2}}^{1+\alpha} \\ &{}- \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3}) \Vert \dot{u}_{n}\Vert _{L^{2}} -\frac{rT}{\gamma} \biggl( \frac{T}{12} \biggr)^{\gamma/2}\Vert \dot{u}_{n}\Vert _{L^{2}}^{\gamma}, \end{aligned}$$

(2.6)

for large n. So, by Wirtinger’s inequality we get

$$ \bigl\Vert (\tilde{u}_{n})\bigr\Vert \leq\frac{ (T^{2}+4\pi^{2} )^{1/2}}{2\pi} \Vert \dot{u}_{n} \Vert _{L^{2}}. $$

(2.7)

From (2.6) and (2.7),

$$\begin{aligned} &\frac{a_{2}(C_{0}M_{1})^{2}}{2}h^{2}\bigl(\vert \bar{u}_{n}\vert \bigr) \\ &\quad \geq \biggl(1-\frac{T^{2}}{8\pi^{2}a_{2}} \biggr)\Vert \dot{u}_{n} \Vert _{L^{2}}^{2}- C_{0}M_{2}K_{1} \biggl(\frac{T}{12} \biggr)^{\frac{1+\alpha}{2}}\Vert \dot {u}_{n} \Vert _{L^{2}}^{1+\alpha} \\ &\qquad {}- \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3}) \Vert \dot{u}_{n}\Vert _{L^{2}} \\ &\qquad {}-\Vert \tilde{u}_{n}\Vert - \frac{rT}{\gamma} \biggl( \frac{T}{12} \biggr)^{\gamma/2}\Vert \dot{u}_{n}\Vert _{L^{2}}^{\gamma} \geq\frac{1}{2}\Vert \dot{u}_{n} \Vert _{L^{2}}^{2}+C_{1}, \end{aligned}$$

(2.8)

where

$$\begin{aligned} C_{1}={}& \min_{s\in[0,+\infty]} \biggl\{ \frac{4\pi ^{2}a_{2}-T^{2}}{8\pi^{2}a_{2}}s^{2}- \biggl(\frac{T}{12} \biggr)^{\frac {1+\alpha}{2}}C_{0}M_{2}K_{1}s^{1+\alpha} \\ &{}- \biggl[\frac{ (T^{2}+4\pi^{2} )^{1/2}}{2\pi }+C_{0}M_{2}K_{2} \biggl(\frac{T}{12} \biggr)^{1/2}+ \biggl(\frac {T}{12} \biggr)^{1/2}M_{3} \biggr]s- \frac{rT}{\gamma} \biggl( \frac{T}{12} \biggr)^{\gamma/2}s^{\gamma} \biggr\} . \end{aligned}$$

Note that $a_{2}>T^{2}/(4\pi^{2})$ implies $-\infty < C_{1} <0$. Hence, it follows from (2.8) that

$$ \Vert \dot{u}_{n}\Vert _{L^{2}}^{2}\leq a_{2}C_{0}^{2}M_{1}^{2}h^{2} \bigl(\vert \overline {u}_{n}\vert \bigr)-2C_{1}, $$

(2.9)

and then

$$ \Vert \dot{u}_{n}\Vert _{L^{2}}\leq\sqrt{a_{2}}C_{0}M_{1}h \bigl(\vert \overline{u}_{n}\vert \bigr)+C_{2}, $$

(2.10)

where $0 < C_{2} < +\infty$. Similar to the proof of Theorem 1.1, we have

$$\begin{aligned} & \biggl\vert \int_{0}^{T} \bigl[F_{1} \bigl(t,u_{n}(t) \bigr)-F_{1}(t,\bar {u}_{n}) \bigr] \,dt \biggr\vert \\ &\quad \leq C_{0}M_{1}h\bigl(\vert \bar{u}_{n} \vert \bigr)\Vert \tilde{u}_{n}\Vert _{L^{2}}+C_{0}M_{2}K_{1} \Vert \tilde{u}_{n}\Vert _{\infty}^{1+\alpha}+ (C_{0}M_{2}K_{2}+M_{3}) \Vert \tilde{u}_{n}\Vert _{\infty} \\ &\quad \leq\frac{\pi}{\sqrt{a_{2}}T}\Vert \tilde{u}_{n}\Vert _{L^{2}}^{2}+ \frac{\sqrt{a_{2}}TC_{0}^{2}}{4\pi}{M_{1}^{2}}h^{2} \bigl(\vert \bar{u}_{n}\vert \bigr) \\ &\qquad {} +C_{0}M_{2}K_{1} \Vert \tilde{u}_{n}\Vert _{\infty}^{1+\alpha }+(C_{0}M_{2}K_{2}+M_{3}) \Vert \tilde{u}_{n}\Vert _{\infty} \\ &\quad \leq\frac{T}{4\pi\sqrt{a_{2}}}\Vert \dot{u}_{n}\Vert _{L^{2}}^{2}+ \frac{\sqrt{a_{2}}TC_{0}^{2}}{4\pi}{M_{1}^{2}}h^{2} \bigl(\vert \bar{u}_{n}\vert \bigr) + \biggl(\frac{T}{12} \biggr)^{(1+\alpha)/2}C_{0}M_{2}K_{1} \Vert \dot{u}_{n}\Vert _{L^{2}}^{1+\alpha} \\ &\qquad {}+ \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3}) \Vert \dot{u}_{n}\Vert _{L^{2}}. \end{aligned}$$

(2.11)

By (4), we obtain

$$\begin{aligned} &\int_{0}^{T} \bigl[F_{2} \bigl(u_{n}(t) \bigr)-F_{2}(\bar{u}_{n}) \bigr] \,dt \\ &\quad =\int_{0}^{T}\int_{0}^{1} \frac{1}{s} \bigl(\nabla F_{2} \bigl(\bar {u}_{n}+s \tilde{u}_{n}(t) \bigr)-\nabla F_{2}(\bar{u}_{n}),s \tilde{u}_{n}(t) \bigr)\,ds\,dt \\ &\quad \leq\int_{0}^{T}\int_{0}^{1} \mu s^{\delta-1}\bigl\vert \tilde {u}_{n}(t)\bigr\vert ^{\delta}\,ds\,dt \leq \frac{\mu T}{\delta} \Vert \tilde{u}_{n} \Vert _{\infty}^{\delta} \\ &\quad \leq\frac{\mu T}{\delta} \biggl(\frac{T}{12} \biggr)^{\delta/2}\Vert \dot{u}_{n}\Vert _{L^{2}}^{\delta}. \end{aligned}$$

From the boundedness of ${\varphi(u_{n})}$ and (2.9)-(2.11), we have

$$\begin{aligned} C_{3}\leq{}&\varphi(u_{n}) \\ ={}&\frac{1}{2}\Vert \dot {u}_{n}\Vert _{L^{2}}^{2} + \int_{0}^{T} \bigl[F_{1} \bigl(t,u_{n}(t) \bigr)-F_{1}(t,\bar{u}_{n}) \bigr] \,dt+ \int_{0}^{T} \bigl[F_{2} \bigl(u_{n}(t) \bigr)-F_{2}(\bar{u}_{n}) \bigr] \,dt \\ &{}+\int_{0}^{T}F \bigl(t,\bar{u}_{n}(t) \bigr)\,dt \\ \leq {}&\biggl(\frac{1}{2}+\frac{T}{4\pi\sqrt{a_{2}}} \biggr)\Vert \dot{u} \Vert _{L^{2}}^{2}+\frac{\sqrt{a_{2}}TC_{0}^{2}}{4\pi }{M_{1}^{2}}h^{2} \bigl(\vert \bar{u}\vert \bigr) +C_{0}M_{2}K_{1} \biggl(\frac{T}{12} \biggr)^{(1+\alpha)/2}\Vert \dot{u}_{n} \Vert _{L^{2}}^{1+\alpha} \\ &{}+ \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3}) \Vert \dot{u}_{n}\Vert _{L^{2}} +\int_{0}^{T}F(t, \bar{u}_{n})\,dt+\frac{\mu T}{\delta} \biggl(\frac {T}{12} \biggr)^{\delta/2}\Vert \dot{u}_{n}\Vert _{L^{2}}^{\delta} \\ \leq {}&\biggl(\frac{1}{2}+\frac{T}{4\pi\sqrt{a_{2}}} \biggr) \bigl(a_{2}C_{0}^{2}M_{1}^{2}h^{2} \bigl(\vert \bar{u}_{n}\vert \bigr)-2C_{1} \bigr) + \frac{\sqrt{a_{2}}TC_{0}^{2}}{4\pi}{M_{1}^{2}}h^{2}\bigl(\vert \bar{u}\vert \bigr) \\ &{}+C_{0}M_{2}K_{1} \biggl(\frac{T}{12} \biggr)^{(1+\alpha )/2} \bigl(\sqrt{a_{2}}C_{0}M_{1}h \bigl(\vert \bar{u}_{n}\vert \bigr)+C_{2} \bigr)^{1+\alpha} \\ &{}+ \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3}) \bigl(\sqrt{a_{2}}C_{0}M_{1}h\bigl(\vert \bar {u}_{n}\vert \bigr)+C_{2} \bigr) \\ &{}+ \int_{0}^{T}F(t,\bar{u}_{n})\,dt+ \frac{\mu T}{\delta} \biggl(\frac {T}{12} \biggr)^{\delta/2}\Vert \dot{u}_{n}\Vert _{L^{2}}^{\delta} \\ \leq {}&\biggl(\frac{a_{2}}{2}+\frac{\sqrt{a_{2}}T}{2\pi } \biggr)C_{0}^{2}M_{1}^{2}h^{2} \bigl(\vert \bar{u}_{n}\vert \bigr) - \biggl(1+\frac{T}{2\pi\sqrt{a_{2}}} \biggr)C_{1} \\ &{}+ \biggl(\frac{T}{12} \biggr)^{(1+\alpha)/2}2^{\alpha}C_{0}M_{2}K_{1} \bigl[ (\sqrt{a_{2}}C_{0}M_{1} )^{1+\alpha}h \bigl(\vert \bar {u}_{n}\vert \bigr)^{1+\alpha}+C_{2}^{1+\alpha} \bigr] \\ &{}+ \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3}) \bigl(\sqrt{a_{2}}C_{0}M_{1}h\bigl(\vert \bar {u}_{n}\vert \bigr)+C_{2} \bigr) \\ &{}+\int_{0}^{T}F(t,\bar{u}_{n})\,dt + \frac{\mu T}{\delta} \biggl(\frac{T}{12} \biggr)^{\delta/2}2^{\delta -1} \bigl((\sqrt{a_{2}}M_{1})^{\delta}h^{\delta} \bigl(\vert \overline {u}_{n}\vert \bigr)+C_{2}^{\delta} \bigr) \\ ={}&h^{2}\bigl(\vert \bar{u}_{n}\vert \bigr) \biggl[h^{-2}\bigl(\vert \bar{u}_{n}\vert \bigr)\int _{0}^{T}F(t, \bar{u}_{n})\,dt + \biggl( \frac{a_{2}}{2}+\frac{\sqrt{a_{2}}T}{2\pi} \biggr)C_{0}^{2}M_{1}^{2} \\ &{}+ \biggl(\frac{T}{12} \biggr)^{(1+\alpha)/2}2^{\alpha }C_{0}^{2+\alpha}M_{2}M_{1}^{1+\alpha}K_{1}h^{\alpha-1} \bigl(\vert \bar{u}_{n}\vert \bigr) + \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2} \\ &{}+M_{3}) \bigl(\sqrt{a_{2}}C_{0}M_{1}h^{-1} \bigl(\vert \bar{u}_{n}\vert \bigr) \bigr) +\frac{\mu T}{\delta} \biggl(\frac{T}{12} \biggr)^{\delta/2}2^{\delta -1}( \sqrt{a_{2}}M_{1})^{\delta}h^{\delta-2}\bigl(\vert \overline{u}_{n}\vert \bigr) \biggr] \\ &{}+ \biggl(\frac{T}{12} \biggr)^{(1+\alpha)/2}2^{\alpha }C_{0}M_{2}K_{1}C_{2}^{1+\alpha}+ \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2} \\ &{}+M_{3})C_{2} - \biggl(1+\frac{T}{2\pi\sqrt{a_{2}}} \biggr)C_{1}+ \frac{\mu T}{\delta} \biggl(\frac{T}{12} \biggr)^{\delta/2}2^{\delta -1}C_{2}^{\delta}, \end{aligned}$$

for large n. So, by (2.4) we see that $\vert \bar{u}\vert$ is bounded. Hence $\{u_{n}\}$ is bounded by (2.9). Arguing as in the proof of Proposition 4.1 of [6], we conclude that the (PS) condition is satisfied.

Step 2. Let $\tilde{H}_{T}^{1}=\{u\in H_{T}^{1}:\bar{u}=0\}$. We assert that for $u\in\tilde{H}_{T}^{1}$,

$$ \varphi(u)\rightarrow+\infty,\qquad \Vert u\Vert \rightarrow\infty. $$

(2.12)

In fact, from (1) and Sobolev’s inequality, we get

$$\begin{aligned} & \biggl\vert \int_{0}^{T} \bigl[F_{1} \bigl(t,u(t) \bigr)-F_{1}(t,0) \bigr]\,dt \biggr\vert \\ &\quad = \biggl\vert \int_{0}^{T}\int _{0}^{1} \bigl(\nabla F \bigl(t,su(t) \bigr),u(t) \bigr)\,ds\,dt \biggr\vert \\ &\quad \leq\int_{0}^{T}f(t)h \bigl(\bigl\vert u(t) \bigr\vert \bigr)\bigl\vert u(t)\bigr\vert \,dt+\int_{0}^{T}g(t) \bigl\vert u(t)\bigr\vert \,dt \\ &\quad \leq\int_{0}^{T} f(t) \bigl(K_{1}\bigl\vert u(t)\bigr\vert ^{\alpha }+K_{2} \bigr)\bigl\vert u(t) \bigr\vert \,dt+M_{3}\Vert u\Vert _{\infty} \\ &\quad \leq M_{2}K_{1}\Vert u\Vert _{\infty}^{1+\alpha}+M_{2}K_{2} \Vert u\Vert _{\infty}+M_{3}\Vert u\Vert _{\infty} \\ &\quad \leq \biggl(\frac{T}{12} \biggr)^{\frac{1+\alpha }{2}}M_{2}K_{1} \Vert \dot{u}\Vert _{L^{2}}^{1+\alpha} + \biggl(\frac{T}{12} \biggr)^{1/2}(M_{2}K_{2}+M_{3})\Vert \dot{u} \Vert _{L^{2}}, \end{aligned}$$

for all $u\in\tilde{H}_{T}^{1}$. It follows from (2) that

$$\begin{aligned} &\int^{T}_{0} \bigl[F_{2} \bigl(u(t) \bigr)-F_{2}(0) \bigr]\,dt \\ &\quad =\int_{0}^{T}\int_{0}^{1} \frac{1}{s} \bigl(\nabla F_{2} \bigl(s\tilde{u}(t) \bigr)-\nabla F_{2}(0),su(t) \bigr)\,ds\,dt \\ &\quad \geq-\int_{0}^{T}\int_{0}^{1}rs^{\gamma-1} \vert u\vert ^{\gamma } \,ds\,dt \\ &\quad \geq-\frac{rT}{\gamma} \Vert \dot{u}\Vert _{\infty}^{\gamma} \\ &\quad \geq-\frac{rT}{\gamma} \biggl(\frac{T}{12} \biggr)^{\gamma /2}\Vert \dot{u}\Vert _{L^{2}}^{\gamma}. \end{aligned}$$

So, we get

$$\begin{aligned} \varphi(u)={}&\frac{1}{2}\Vert \dot{u}\Vert _{L^{2}}^{2} +\int_{0}^{T} \bigl[F \bigl(t,u(t) \bigr)-F(t,0) \bigr]\,dt+ \int_{0}^{T}F(t,0)\,dt \\ \geq{}&\frac{1}{2}\Vert \dot{u}\Vert _{L^{2}}^{2}- \biggl( \frac {T}{12} \biggr)^{\frac{1+\alpha}{2}}M_{2}K_{1} \Vert \dot{u}\Vert _{L^{2}}^{1+\alpha} - \biggl(\frac{T}{12} \biggr)^{1/2}(M_{2}K_{2}+M_{3}) \Vert \dot{u}\Vert _{L^{2}} \\ &{}-\frac{rT}{\gamma} \biggl(\frac{T}{12} \biggr)^{\gamma/2}\Vert \dot{u} \Vert _{L^{2}}^{\gamma} +\int_{0}^{T}F(t,0) \,dt. \end{aligned}$$

By Wirtinger’s inequality, $\Vert u\Vert \rightarrow \infty$ if and only if $\Vert \dot{u}\Vert _{L^{2}} \rightarrow \infty$ in $\tilde{H}_{T}^{1}$. Hence (2.12) holds.

Step 3. By (5), we can easily see that$\int_{0}^{T}F(t,x)\,dt \rightarrow -\infty$ as $\vert x\vert \rightarrow \infty$ for all $x\in\mathbb{R}^{N}$. Thus, for all $u\in (\tilde {H}_{T}^{1})^{\perp} =\mathbb{R}^{N}$,

$$\begin{aligned} \varphi(u)=\int_{0}^{T}F(t,u)\,dt \rightarrow- \infty \quad \text{as } \vert u\vert \rightarrow\infty. \end{aligned}$$

Now, by saddle point theorem (see, [10], Theorem 4.6), the proof is completed. □

Proof of Theorem 1.3

By (7), we can choose an $a_{3}>\frac{T^{2}}{4\pi^{2}-rT^{2}}$ such that

$$ \liminf_{\vert x\vert \rightarrow\infty} h^{-2}\bigl(\vert x\vert \bigr) \int_{0}^{T}F(t,x) \,dt>\frac {a_{3}}{2}M_{1}^{2}C_{0}^{2}. $$

(2.13)

By (6) and the Sobolev inequality, we have

$$\begin{aligned} &\int_{0}^{T} \bigl[F_{2} \bigl(u(t) \bigr)-F_{2}(\bar{u}) \bigr]\,dt \\ &\quad =\int_{0}^{T}\int_{0}^{1} \frac{1}{s} \bigl(\nabla F_{2} \bigl(\bar{u}+s\tilde {u}(t) \bigr)- \nabla F_{2}(\bar{u}),s\tilde{u}(t) \bigr)\,ds\,dt \\ &\quad \geq -\int_{0}^{T}\int_{0}^{1}rs \bigl\vert \tilde{u}(t)\bigr\vert ^{2}\,ds\,dt \geq- \frac{rT^{2}}{8\pi^{2}}\Vert \dot{u} \Vert _{L^{2}}^{2}. \end{aligned}$$

By a similar method to that of the proof of Theorem 1.1, we get

$$\begin{aligned} \varphi(u)={}&\frac{1}{2}\Vert \dot{u}\Vert _{L^{2}}^{2} +\int_{0}^{T}F \bigl(t,u(t) \bigr)\,dt \\ ={}&\frac{1}{2}\Vert \dot{u}\Vert _{L^{2}}^{2} +\int _{0}^{T} \bigl[F_{1} \bigl(t,u(t) \bigr)-F_{1}(t,\overline{u}) \bigr]\,dt \\ &{}+\int_{0}^{T} \bigl[F_{2} \bigl(u(t) \bigr)-F_{2}(\overline{u}) \bigr]\,dt +\int_{0}^{T}F(t, \overline{u})\,dt \\ \geq {}&\biggl(\frac{1}{2}-\frac{T^{2}}{8\pi^{2}a_{3}}-\frac{rT^{2}}{8\pi ^{2}} \biggr) \Vert \dot{u}\Vert _{L^{2}}^{2} - \biggl(\frac{T}{12} \biggr)^{(1+\alpha)/2}C_{0}M_{2}K_{1}\Vert \dot{u}\Vert ^{1+\alpha}_{L_{2}} \\ &{}- \biggl(\frac{T}{12} \biggr)^{1/2} \biggl(M_{3}+ \frac {C_{0}M_{2}K_{2}}{2} \biggr)\Vert \dot{u}\Vert _{L^{2}} - \frac{a_{3}C_{0}^{2}M_{1}^{2}}{2}h^{2}\bigl(\vert \bar{u}\vert \bigr)+\int _{0}^{T}F(t,\overline{u})\,dt \\ ={}& \biggl(\frac{1}{2}-\frac{T^{2}}{8\pi^{2}a_{3}}-\frac {rT^{2}}{8\pi^{2}} \biggr)\Vert \dot{u}\Vert _{L^{2}}^{2} - \biggl(\frac{T}{12} \biggr)^{1/2} \biggl(M_{3}+\frac {C_{0}M_{2}K_{2}}{2} \biggr)\Vert \dot{u} \Vert _{L^{2}} \\ &{}- \biggl(\frac{T}{12} \biggr)^{(1+\alpha )/2}C_{0}M_{2}K_{1} \Vert \dot{u}\Vert ^{1+\alpha}_{L_{2}}+h^{2}\bigl( \vert \bar{u}\vert \bigr) \biggl(h^{-2}\bigl(\vert \bar{u}\vert \bigr)\int_{0}^{T}F(t, \bar{u})\,dt- \frac {a_{3}C_{0}^{2}M_{1}^{2}}{2} \biggr), \end{aligned}$$

for all $u\in{H}_{T}^{1}$, which implies that $\varphi(u)\rightarrow\infty$ as $\Vert u\Vert \rightarrow\infty$ by (2.13), due to the facts that $r<\frac{4\pi^{2}}{T^{2}}$ and $\Vert u\Vert \rightarrow\infty$ if and only if $(\vert \bar{u}\vert ^{2}+\Vert \dot{u}\Vert _{L^{2}}^{2})^{1/2}\rightarrow\infty$. So, applying the least action principle, Theorem 1.3 holds. □

Proof of Theorem 1.4

First, we assert that φ satisfies the (PS) condition. Suppose that $\{u_{n}\}$ satisfies $\varphi'(u_{n})\rightarrow0$ as $n\rightarrow\infty$ and $\{\varphi (u_{n})\}$ is bounded. By (9), we can choose an $a_{4}>\frac{T^{2}}{4\pi^{2}}$ such that

$$ \limsup_{\vert x\vert \rightarrow\infty} h^{-2}\bigl(\vert x\vert \bigr) \int_{0}^{T}F_{1}(t,x) \,dt< - \biggl( \frac{a_{4}}{2}+\frac{\sqrt{a_{4}}T}{2\pi } \biggr)C_{0}^{2}M_{1}^{2}. $$

(2.14)

By the ($\lambda,\mu$)-subconvexity of $G(x)$, we have

$$ G(x)\leq \bigl(2\mu \vert x\vert ^{\beta}+1 \bigr)G_{0} $$

(2.15)

for all $x\in\mathbb{R}^{N}$, and a.e. $t\in[0,T]$, where $G_{0}=\max_{\vert s\vert \leq1}G(s)$, $\beta=\log_{2\lambda}(2\mu)<2$ Then

$$\begin{aligned} &\int_{0}^{T} \bigl(\nabla F_{2} \bigl(t,u_{n}(t) \bigr),\tilde{u}_{n}(t) \bigr)\,dt \\ &\quad \geq-\int_{0}^{T}k(t)G(\bar{u}_{n}) \,dt \\ &\quad \geq-\int_{0}^{T}k(t) \bigl(2\mu \vert \bar{u}_{n}\vert ^{\beta}+1 \bigr)G_{0}\,dt \\ &\quad =-2\mu M_{4}\vert \bar{u}_{n}\vert ^{\beta}- M_{4}, \end{aligned}$$

(2.16)

where $M_{4}=G_{0}\int_{0}^{T}k(t)\,dt$. From (2.5) and (2.16), for large n, we have

$$\begin{aligned} \Vert \bar{u}_{n}\Vert \geq {}&\bigl\langle \varphi(u_{n}), \tilde {u}_{n} \bigr\rangle \\ ={}&\Vert \dot{u}_{n}\Vert _{L^{2}}^{2}+\int _{0}^{T} \bigl(\nabla F \bigl(t,u_{n}(t) \bigr),\tilde{u}_{n}(t) \bigr) \\ \geq {}&\biggl(1-\frac{T^{2}}{8\pi^{2}a_{4}}\biggr)\Vert \dot{u}_{n} \Vert _{L^{2}}^{2}- \biggl( \frac{T}{12} \biggr)^{(1+\alpha )/2}C_{0}M_{2}K_{1} \Vert \dot{u}_{n}\Vert ^{1+\alpha}_{L^{2}} - \frac{(C_{0}M_{1})^{2}a_{4}}{2}h^{2}\bigl(\vert \bar{u}_{n}\vert \bigr) \\ &{}- \biggl(\frac{T}{12} \biggr)^{1/2} \biggl(M_{3}+ \frac {C_{0}M_{2}K_{2}}{2} \biggr)\Vert \dot{u}\Vert _{L^{2}} -2\mu M_{4}\vert \bar{u}_{n}\vert ^{\beta}- M_{4}. \end{aligned}$$

(2.17)

So, from (2.7) and (2.17) we have

$$\begin{aligned} &\frac{(C_{0}M_{1})^{2}a_{4}}{2}h^{2}\bigl(\vert \bar{u}_{n}\vert \bigr) +2\mu M_{4}\vert \bar{u}_{n}\vert ^{\beta} \\ &\quad \geq \biggl(1-\frac{T^{2}}{8\pi^{2}a_{4}} \biggr)\Vert \dot{u}_{n} \Vert _{L^{2}}^{2} - \biggl(\frac{T}{12} \biggr)^{(1+\alpha)/2}C_{0}M_{2}K_{1}\Vert \dot{u}_{n}\Vert ^{1+\alpha}_{L^{2}} \\ &\qquad {}- \biggl(\frac{T}{12} \biggr)^{1/2} \biggl(M_{3}+ \frac {C_{0}M_{2}K_{2}}{2} \biggr)\Vert \dot{u}\Vert _{L^{2}} - \frac{(T^{2}+4\pi^{2})^{1/2}}{2\pi} \Vert \dot{u}_{n}\Vert _{L^{2}}-M_{4} \\ &\quad \geq\frac{1}{2}\Vert \dot{u}_{n}\Vert ^{2}_{L^{2}}+C_{4}, \end{aligned}$$

(2.18)

where

$$\begin{aligned} C_{4}={}&\min \biggl\{ \biggl(\frac{1}{2}-\frac{T^{2}}{8\pi ^{2}a_{4}} \biggr)s^{2}- \biggl(\frac{T}{12} \biggr)^{(1+\alpha )/2}C_{0}M_{2}K_{1}s^{1+\alpha} - \biggl[\frac{ (T^{2}+4\pi^{2} )^{1/2}}{2\pi} \\ &{}+ \biggl(\frac{T}{12} \biggr)^{1/2} \biggl(M_{3}+ \frac{C_{0}M_{2}K_{2}}{2} \biggr) \biggr]s -M_{4} \biggr\} . \end{aligned}$$

Note that $-\infty< C_{4}<0$ due to $a_{4}>\frac{T^{2}}{4\pi^{2}}$, by (2.18), one has

$$ \Vert \dot{u}_{n}\Vert _{L^{2}}^{2}\leq a_{4}(C_{0}M_{1})^{2}h^{2} \bigl(\vert \bar{u}_{n}\vert \bigr)+4\mu M_{4}\vert \overline {u}_{n}\vert ^{\beta}-2C_{4}, $$

(2.19)

and then

$$ \Vert \dot{u}_{n}\Vert _{L^{2}}\leq \sqrt{a_{4}}C_{0}M_{1}h \bigl(\vert \bar{u}_{n}\vert \bigr)+2\sqrt{\mu M_{4}} \vert \overline {u}_{n}\vert ^{\beta/2}+C_{5}, $$

(2.20)

where $C_{5} >0$. From (8) and (2.15), we have

$$\begin{aligned} & \biggl\vert \int_{0}^{T} \bigl[F_{2} \bigl(t,u(t) \bigr)-F_{2}(t,\bar{u}) \bigr]\,dt \biggr\vert \\ &\quad =\int_{0}^{T}\int_{0}^{1} \bigl(\nabla F_{2} \bigl(t,\bar {u}_{n}+s \tilde{u}_{n}(t) \bigr),\tilde{u}_{n}(t) \bigr)\,ds\,dt \\ &\quad \leq\int_{0}^{T}\int_{0}^{1}k(t)G \bigl(\bar {u}_{n}+(s+1)\tilde{u}_{n} \bigr)\,ds\,dt \\ &\quad \leq\int_{0}^{T}\int_{0}^{1}k(t) \bigl(2\mu\bigl\vert \bar {u}_{n}+(s+1)\tilde{u}_{n}(t) \bigr\vert ^{\beta}+1 \bigr) \\ &\quad \leq4\mu\int_{0}^{T}k(t) \bigl(\vert \bar{u}_{n}\vert ^{\beta}+2^{\beta }\vert \tilde{u}_{n}\vert ^{\beta} \bigr)G_{0}\int _{0}^{T}k(t)\,dt \\ &\quad \leq \biggl(\frac{T}{12} \biggr)^{\beta/2}2^{\beta+2}\mu M_{4}\Vert \dot{u}_{n}\Vert _{L^{2}}^{\beta} +4\mu M_{4}\vert \bar{u}_{n}\vert ^{\beta}+M_{4}, \end{aligned}$$

(2.21)

for all $u\in H_{T}^{1}$. By the boundedness of $\{\varphi(u_{n})\}$ and the inequalities (2.19)-(2.21), we get

$$\begin{aligned} C_{6}\leq{}&\varphi(u_{n}) \\ ={}&\frac{1}{2}\Vert \dot{u}_{n}\Vert _{L^{2}}^{2} +\int_{0}^{T} \bigl[F_{1} \bigl(t,u_{n}(t) \bigr)-F_{1}(t,\bar{u}_{n}) \bigr]\,dt \\ &{}+\int_{0}^{T} \bigl[F_{2} \bigl(t,u_{n}(t) \bigr)-F_{2}(t,\bar{u}_{n}) \bigr] \,dt +\int_{0}^{T}F(t,\bar{u}_{n}) \,dt \\ \leq {}&\biggl(\frac{1}{2}+\frac{T}{4\pi\sqrt{a_{4}}} \biggr)\Vert \dot{u}_{n} \Vert _{L^{2}}^{2}+\frac{\sqrt{a_{4}}TC_{0}^{2}}{4\pi}{M_{1}^{2}}h^{2} \bigl(\vert \bar{u}_{n}\vert \bigr) + \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3}) \Vert \dot{u}_{n}\Vert _{L^{2}} \\ &{}+ \biggl(\frac{T}{12} \biggr)^{(1+\alpha )/2}C_{0}M_{2}K_{1} \Vert \dot{u}_{n}\Vert _{L^{2}}^{1+\alpha} + \biggl( \frac{T}{12} \biggr)^{\beta/2}2^{\beta+2}\mu M_{4} \Vert \dot{u}\Vert _{L^{2}}^{\beta}\\ &{}+4\mu M_{4}\vert \bar{u}_{n}\vert ^{\beta}+M_{4}+\int _{0}^{T}F(t, \bar{u}_{n})\,dt \\ \leq {}&\biggl(\frac{1}{2}+\frac{T}{4\pi\sqrt{a_{4}}} \biggr) \bigl(a_{4}(C_{0}M_{1})^{2}h^{2} \bigl(\vert \bar{u}_{n}\vert \bigr)+4\mu M_{4}\vert \overline {u}_{n}\vert ^{\beta}-2C_{4} \bigr)\\ &{}+ \frac{\sqrt{a_{4}}TC_{0}^{2}}{4\pi }{M_{1}^{2}}h^{2}\bigl(\vert \bar{u}_{n}\vert \bigr) \\ &{}+ \biggl(\frac{T}{12} \biggr)^{(1+\alpha )/2}C_{0}M_{2}K_{1} \bigl(\sqrt{a_{4}}C_{0}M_{1}h\bigl(\vert \bar{u}_{n}\vert \bigr)+2\sqrt {\mu M_{4}}\vert \overline{u}_{n}\vert ^{\beta/2}+C_{5} \bigr)^{1+\alpha} \\ &{}+ \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3}) \bigl(\sqrt{a_{4}}C_{0}M_{1}h\bigl(\vert \bar {u}_{n}\vert \bigr)+2\sqrt{\mu M_{4}}\vert \overline{u}_{n}\vert ^{\beta/2}+C_{5} \bigr) \\ &{}+ \biggl(\frac{T}{12} \biggr)^{\beta/2}2^{\beta+2}\mu M_{4} \bigl(\sqrt{a_{4}}C_{0}M_{1}h \bigl(\vert \bar{u}_{n}\vert \bigr)+2\sqrt{\mu M_{4}} \vert \overline{u}_{n}\vert ^{\beta/2}+C_{5} \bigr)^{\beta}\\ &{}+4\mu M_{4}\vert \bar {u}_{n}\vert ^{\beta}+M_{4}+ \int_{0}^{T}F(t, \bar{u}_{n})\,dt \\ \leq {}&\biggl(\frac{a_{4}}{2}+\frac{\sqrt{a_{4}}T}{2\pi } \biggr) \bigl((C_{0}M_{1})^{2}h^{2} \bigl(\vert \bar{u}_{n}\vert \bigr) \bigr) \\ &{}+ \biggl(6+ \frac{T}{\pi\sqrt{a_{4}}} \biggr)\mu M_{4}\vert \overline {u}_{n} \vert ^{\beta}- \biggl(1+ \frac{T}{2\pi\sqrt{a_{4}}} \biggr)C_{4} \\ &{} + \biggl(\frac{T}{12} \biggr)^{(1+\alpha)/2}C_{0}M_{2}K_{1} \bigl(2^{\alpha }a_{4}^{(1+\alpha)/2}(C_{0}M_{1})^{1+\alpha}h^{1+\alpha} \bigl(\vert \overline{u}_{n}\vert \bigr) \\ &{}+2^{3\alpha+1} \mu^{\frac{1+\alpha}{2}}M_{4}^{\frac{1+\alpha }{2}}\vert \overline{u}_{n} \vert ^{\beta(1+\alpha)} +2^{2\alpha}C_{5}^{1+\alpha} \bigr) \\ &{}+ \biggl(\frac{T}{12} \biggr)^{\beta/2}2^{2+\beta}\mu M_{4} \bigl(2^{\beta-1}a_{4}^{\beta/2} \bigl(C_{0}M_{1}h\bigl(\vert \overline {u}_{n} \vert \bigr) \bigr)^{\beta}\\ &{}+2^{3\beta-2}\mu^{\beta/2}M_{4}^{\beta/2} \vert \overline{u}_{n}\vert ^{\beta^{2}/2}+2^{2(\beta-1)}C_{5}^{\beta} \bigr) \\ &{}+ \biggl(\frac{T}{12} \biggr)^{1/2} (C_{0}M_{2}K_{2}+M_{3} ) \bigl(\sqrt{a_{4}}C_{0}M_{1}h\bigl(\vert \bar{u}_{n}\vert \bigr)+2\sqrt{\mu M_{4}}\vert \overline {u}_{n}\vert ^{\beta/2}+C_{5} \bigr) \\ &{}+M_{4}+\int_{0}^{T}F_{1}(t, \bar{u}_{n})\,dt+\int_{0}^{T}F_{2}( \bar{u}_{n})\,dt \\ ={}&h^{2}\bigl(\vert \bar{u}_{n}\vert \bigr) \biggl[h^{-2}\bigl(\vert \bar{u}_{n}\vert \bigr)\int _{0}^{T}F_{1}(t, \bar{u}_{n}) \,dt+ \biggl(\frac{a_{4}}{2}+\frac{\sqrt{a_{4}}T}{2\pi} \biggr) (C_{0}M_{1})^{2} \\ &{}+ \biggl(\frac{T}{12} \biggr)^{(1+\alpha)/2}C_{0}^{2+\alpha }M_{1}^{1+\alpha}M_{2}K_{1}2^{\alpha}a_{4}^{(1+\alpha)/2}h^{\alpha -1} \bigl(\vert \overline{u}_{n}\vert \bigr) \\ &{}+ \biggl(\frac{T}{12} \biggr)^{\beta/2}2^{1+2\beta}\mu (C_{0}M_{1})^{\beta}M_{4}a_{4}^{\beta/2}h^{\beta-2} \bigl(\vert \overline {u}_{n}\vert \bigr)\\ &{}+ \biggl(\frac{T}{12} \biggr)^{1/2} (C_{0}M_{2}K_{2}+M_{3} ) \sqrt{a_{4}}C_{0}M_{1}h^{-1}\bigl( \vert \bar{u}_{n}\vert \bigr) \biggr] \\ &{}+\vert \overline{u}_{n}\vert ^{\beta} \biggl[\vert \overline{u}_{n}\vert ^{-\beta } \int_{0}^{T}F_{2}( \bar{u}_{n}) \,dt + \biggl(6+\frac{T}{\pi\sqrt{a_{4}}} \biggr)\mu M_{4}\\ &{}+ \biggl( \frac {T}{12} \biggr)^{(1+\alpha)/2}2^{3\alpha+1}C_{0}M_{2}K_{1} \mu^{(1+\alpha )/2}M_{4}^{(1+\alpha)/2}\vert \overline{u}_{n} \vert ^{\alpha\beta} \\ &{}+ \biggl(\frac{T}{12} \biggr)^{\beta/2}2^{4\beta}\mu ^{(\beta+2)/2}M_{4}^{(\beta+2)/2}\vert \overline{u}_{n} \vert ^{\frac{\beta ^{2}}{2}-\beta} \\ &{}+ \biggl(\frac{T}{12} \biggr)^{1/2}2 (C_{0}M_{2}K_{2}+M_{3} )\sqrt {\mu M_{4}}\vert \overline{u}_{n}\vert ^{-\beta/2} \biggr] \\ &{}- \biggl(1+\frac {T}{2\pi\sqrt{a_{4}}} \biggr)C_{4}+ \biggl(\frac{T}{12} \biggr)^{(1+\alpha )/2}2^{2\alpha}C_{0}M_{2}K_{1}C^{1+\alpha}_{5}\\ &{} + \biggl(\frac{T}{12} \biggr)^{1/2}(C_{0}M_{2}K_{2}+M_{3})C_{5} + \biggl(\frac{T}{12} \biggr)^{\beta/2}2^{3\beta}\mu M_{4}C^{\beta}_{5}+M_{4}, \end{aligned}$$

for large n. The above inequality and (2.14) imply that $\{\vert \overline {u}\vert \}$ is bounded. Hence $\{u_{n}\}$ is bounded by (2.19). By using the standard method, the (PS) condition holds.

Since the rest of the proof is similar to that of Theorem 1.2, we omit the details here. □