Notes on Multiple Periodic Solutions for Second Order Hamiltonian Systems

Abstract

In this paper, we study the multiplicity of periodic solutions for the second order Hamiltonian systems $\ddot{u}+\nabla F(t,u)=0$ with the boundary condition $u(0)-u(T)=\dot{u}(0)-\dot{u}(T)=0$, where the potential F is either subquadratic k(t)-concave or subquadratic $\mu (t)$-convex. Based on the reduction method and a three-critical-point theorem due to Brezis and Nirenberg, we obtain the multiplicity results, which complement and sharply improve some related results in the literature.

1 Introduction and Main Results

Consider the second order Hamiltonian systems

$$\begin{aligned} \left\{ \begin{array}{ll} \ddot{u}(t)+\nabla F(t,u(t))=0\ \ \ \ \ \ \hbox {a.e. }t\in [0,T],\\ u(0)-u(T)=\dot{u}(0)-\dot{u}(T)=0, \end{array} \right. \end{aligned}$$

(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 $x\in {\mathbb {R}}^N$ and 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

  $$\begin{aligned} |F(t,x)|\le a(|x|)b(t),\ \ \ \ \ \ |\nabla F(t,x)|\le a(|x|)b(t) \end{aligned}$$

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

With the aid of variational methods, the existence and multiplicity of periodic solutions of problem (1.1) have been extensively investigated in the literature during the past two decades, see [3,4,5, 7,8,9,10,11,12,13,14,15,16, 18,19,26] and references therein. Many solvability conditions are obtained, such as: the convexity conditions (see [14, 17, 19] and the references therein); the subquadratic conditions (including the sublinear nonlinearity case, see [18, 19] the references therein); the superquadratic conditions (see [6, 26] and the references therein) and the asymptotically linear conditions (see [20, 26, 29] and the references therein).

Lazer, Landesman and Meyers in [8] consider a minimax theorem for a $C^2$ functional $\varphi$ defined on a Hilbert space H. They prove the existence of a critical point characterized by $\displaystyle {\varphi (u_0)=\max _{v\in V} \min _{w\in W} \varphi (v+w)}$, where V, W are subspaces of H, V is finite-dimensional and $H=V\bigoplus W$. This result has been generalized by Castro and Lazer [4], supposing a weaker condition that $\varphi (v)\rightarrow -\infty$ as $\Vert v\Vert \rightarrow \infty$, $v\in V$. Manasevich [12] extend the result to infinite-dimensional cases by using a global inversion theorem. See also Bates and Ekeland [2] and Manasevich [13] for related results. Recently, by means of the reduction method, a perturbation argument and the least action principle, Tang and Wu [19] greatly improve the result of Lazer et al. [8] in three aspects: requiring the spaces being reflexive Banach space instead of Hilbert space; requiring the functionals being $C^1$ instead of $C^2$; and using much weaker convexity of the functionals. As a main application, they successively studied the existence of periodic solutions of problem (1.1) with subquadratic convex potential, with subquadratic $\mu (t)$-convex potential and with subquadratic k(t)-concave potential, which unifies and significantly improves the recent results in [14, 24, 28]. Here we say that the potential F is $\mu (t)$-convex, if there exists $\mu \in L^1(0,T;{\mathbb {R}})$ such that $F(t,x)-\frac{1}{2}\mu (t)|x|^2$ in convex in x for a.e. $t\in [0,T]$. It is worth pointing out that the reduction technique, which has been motivated in earlier papers by Amann [1] and Thews [21] to discuss the existence and multiplicity of solutions for nonlinear equations, is rather powerful. For its applications on semilinear elliptic equations, we refer the readers to [24].

In the present paper, based on the reduction method and some abstract critical point theorem, i.e., the three-critical-point theorem proposed by Brezis and Nirenberg, we shall study the multiplicity of periodic solutions of (1.1) with subquadratic k(t)-concave or subquadratic $\mu (t)$-convex potential, which complements the results mentioned above and improves the corresponding results in Wu [24] and Zhao and Wu [27].

1.1 The Subquadratic k(t)-Concave Case

For the subquadratic k(t)-concave case, we make the following hypotheses:

($A_1$):

There exists $k\in L^1(0,T;{\mathbb {R}}^+)$ with $\int _0^Tk(t)dt<12/T$ such that $-F(t,x)+\frac{1}{2}k(t)|x|^2$ is convex in x for a.e. $t\in [0,T]$.

$(A_2)$:

${\int _0^TF(t,x)dt\rightarrow +\infty }$ as $|x|\rightarrow +\infty$, $x\in {\mathbb {R}}^N$.

$(A_3)$:

There exists $\delta >0$ such that $F(t,x)-F(t,0)\le 0$ for all $|x|\le \delta \hbox { and a.e. }t\in [0,T]$.

$(A_4)$:

There exists $k\in L^1(0,T;{\mathbb {R}}^+)$ with $k(t)\le \omega ^2$ for a.e. $t\in [0,T]$ and

$$\begin{aligned} \text {meas}\left\{ t\in [0,T]:k(t)<\omega ^2\right\} >0, \end{aligned}$$

such that $-F(t,x)+\frac{1}{2}k(t)|x|^2$ is convex in x for a.e. $t\in [0,T]$, where $\omega =2\pi \backslash T$.

Theorem 1.1

Assume that the potential F satisfies assumptions (A) and $(A_1)$-$(A_3)$. Then problem (1.1) possesses at least three distinct solutions.

Remark 1.1

Theorem 1.1 generalizes [27, Theorem 1], where the existence of one nonzero solution is obtained under the assumptions (A), $(A_2)$ and

$(A_1')$:

There exists $k\in L^1(0,T;{\mathbb {R}}^+)$ with $0<\int _0^Tk(t)dt<12/T$ such that $|\nabla F(t,x_1)-\nabla F(t,x_2)|\le k(t)|x_1-x_2|$ for all $x_1$, $x_2\in {\mathbb {R}}^N$ and a.e. $t\in [0,T]$.

$(A_3')$:

For a.e. $t\in [0,T]$, $\nabla F(t,0)=0$, and there exists $\delta >0$ such that $F(t,x)-F(t,0)<0$ for all $x\in {\mathbb {R}}^N$ with $0<|x|\le \delta$ and for a.e. $t\in [0,T]$.

Since, by condition $(A_1')$, we have

$$\begin{aligned} (\nabla (-F(t,x))-\nabla (-F(t,y)),x-y)&\ge -|\nabla F(t,x)-\nabla F(t,y)||x-y|\\&\ge -k(t)|x-y|^2 \end{aligned}$$

for all $x,y\in {\mathbb {R}}^N$ and a.e. $t\in [0,T]$, it follows that $-F(t,x)+\frac{1}{2}k(t)|x|^2$ is convex in x for a.e. $t\in [0,T]$. Hence $(A_1')$ implies $(A_1)$. In addition, the condition $\nabla F(t,0)=0$ for a.e. $t\in [0,T]$ (see assumption $(A_3')$) is deleted and our conclusion is better. There are functions F satisfying Theorem 1.1 but not satisfying the results in [14, 24, 27, 28]. For example, let

$$\begin{aligned} F(t,x)=\frac{1}{2}\alpha (t)|x|^2-|x|^\frac{3}{2},\ \ \ \ \ \ \ \forall x\in {\mathbb {R}}^N \text { and a.e. }t\in [0,T], \end{aligned}$$

where $\alpha \in L^\infty (0,T;{\mathbb {R}})$ with

$$\begin{aligned} \frac{6}{T}\le \int _0^T\alpha ^-(t) dt<\int _0^T\alpha ^+(t)dt<\frac{12}{T}, \end{aligned}$$

and $\alpha ^\pm (t)$=max$\left\{ \pm \alpha (t),0\right\}$. Then F satisfies Theorem 1.1 with $k=\alpha ^+$. But it does not satisfy the corresponding results in [14, 24, 27, 28], for $\int _0^T|\alpha (t)|dt>12/T$ and F is not convex in x for $t\in [0,T]$ with $\alpha ^-(t)>0$.

Theorem 1.2

The conclusion of Theorem 1.1 holds if we replace ($A_1$) by ($A_4$).

Remark 1.2

Theorem 1.2 complements [19, Theorem 5.1], where the existence of one solution was obtained under the assumptions (A), ($A_2$) and $(A_4)$. There are functions F satisfying Theorem 1.2 but not satisfying Theorem 1.1. For example, let

$$\begin{aligned} F(t,x)=\frac{2}{3T}\omega ^2\left( \frac{3T}{4}-t\right) |x|^2-|x|^\frac{3}{2},\ \ \ \ \ \forall x\in {\mathbb {R}}^N \text { and a.e. }t\in [0,T]. \end{aligned}$$

Then $\alpha (t)=\frac{4}{3T}\omega ^2\left( \frac{3T}{4}-t\right)$ for $t\in [0,T]$,

$$\begin{aligned} \alpha ^+(t)={\left\{ \begin{array}{ll} \alpha (t), &{} t\in \big [0,\frac{3T}{4}\big ],\\ 0, &{} t\in \big [\frac{3T}{4},T\big ], \end{array}\right. }\ \ \ \ \ \ \ \ \ \alpha ^-(t)={\left\{ \begin{array}{ll} 0, &{} t\in \big [0,\frac{3T}{4}\big ],\\ -\alpha (t), &{} t\in \big [\frac{3T}{4},T\big ], \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} -F(t,x)+\frac{1}{2}\alpha ^+(t)|x|^2=\frac{1}{2}\alpha ^-(t)|x|^2+|x|^\frac{3}{2} \end{aligned}$$

for $x\in {\mathbb {R}}^N$ and a.e. $t\in [0,T]$. It is easy to check that F satisfies the conditions of Theorem 1.2 with $k=\alpha ^+$. However, since

$$\begin{aligned} \int _0^T\alpha ^+ (t)dt=\int _0^\frac{3T}{4} \frac{4}{3T}\omega ^2\left( \frac{3T}{4}-t\right) dt=\frac{3\pi ^2}{2T}>\frac{12}{T}, \end{aligned}$$

it does not satisfy condition $(A_1)$ and hence not satisfy Theorem 1.1.

1.2 The Subquadratic $\mu (t)$-Convex Case

For the subquadratic $\mu (t)$-convex case, we make the following hypotheses:

$(A_5)$:

There exists $\mu \in L^1(0,T;{\mathbb {R}})$ with $\int _0^T\mu (t)dt>0$ such that $F(t,x)-\frac{1}{2}\mu (t)|x|^2$ is convex in x for a.e. $t\in [0,T]$.

$(A_6)$:

There exist $\alpha \in L^1(0,T;{\mathbb {R}}^+)$ with $\int _0^T\alpha (t)dt<12/T$ and $\gamma \in L^1(0,T;{\mathbb {R}}^+)$ such that

$$\begin{aligned} F(t,x)\le \frac{1}{2}\alpha (t)|x|^2+\gamma (t),\ \ \ \ \ \ \ \ \ \forall x\in {\mathbb {R}}^N \text { and a.e. }t\in [0,T]. \end{aligned}$$

(1.2)

$(A_7)$:

There exist $\delta >0$ and an integer $k\ge 1$ such that

$$\begin{aligned} \frac{1}{2}k^2\omega ^2|x|^2\le F(t,x)-F(t,0) \le \frac{1}{2}(k+1)^2\omega ^2|x|^2 \end{aligned}$$

for all $|x|\le \delta$ and a.e. $t\in [0,T]$.

$(A_8)$:

There exists $\gamma \in L^1(0,T;{\mathbb {R}}^+)$ such that $F(t,x)\le \frac{1}{2}\omega ^2|x|^2+\gamma (t)$ for all $x\in {\mathbb {R}}^N$ and a.e. $t\in [0,T]$, and $\hbox {meas}\left\{ t\in [0,T]:F(t,x)-\frac{1}{2}\omega ^2|x|^2\rightarrow -\infty \hbox { as} |x|\rightarrow \infty \right\} >0.$

$(A_9)$:

There exists $\alpha \in L^\infty (0,T;{\mathbb {R}}^+)$ with $\alpha (t)\le \omega ^2$ for a.e. $t\in [0,T]$ and

$$\begin{aligned} \text {meas}\left\{ t\in [0,T]:\alpha (t)<\omega ^2\right\} >0, \end{aligned}$$

such that $\displaystyle {\limsup _{|x|\rightarrow \infty }|x|^{-2}F(t,x)}\le \frac{1}{2}\alpha (t)$ $\text {uniformly for a.e. }t\in [0,T]$.

Theorem 1.3

Assume that assumptions (A) and $(A_5)$-$(A_7)$ are satisfied. Then problem (1.1) possesses at least three distinct solutions.

Remark 1.3

Theorem 1.3 extends [24, Theorem 2.2]; there Wu consider problem (1.1) under the hypothesis (A), $(A_5)$ and

$(A_6')$:

There exist f, $g\in L^1(0,T;{\mathbb {R}}^+)$ with $\int _0^Tf(t)dt<12/T$ such that $|\nabla F(t,x)|\le f(t)|x|+g(t)$ for all $x\in {\mathbb {R}}^N \text { and a.e. }t\in [0,T]$.

$(A_7')$:

There exist $\delta >0$ and an integer $k\ge 1$ such that $\frac{1}{2}k^2\omega ^2|x|^2\le F(t,x) \le \frac{1}{2}(k+1)^2\omega ^2|x|^2$ for all $|x|\le \delta$ and a.e. $t\in [0,T]$.

We point out that $(A_6')$ is stronger than $(A_6)$. This is because by (A) and ($A_6'$), we deduce that

$$\begin{aligned} F(t,x)&= \int _0^1(\nabla F(t,sx),x)ds+F(t,0)\\&\le \int _0^1(f(t)|sx|+g(t))|x|ds+a(0)b(t)\\&\le \frac{1}{2}f(t)|x|^2+g(t)|x|+a(0)b(t)\\&\le \frac{1}{2}f(t)|x|^2+\frac{1}{2}g(t) \left( \frac{12-T|f|_1}{T(|g|_1+1)}|x|^2+\frac{T(|g|_1+1)}{12-T|f|_1}\right) +a(0)b(t), \end{aligned}$$

which is just (1.2) with

$$\begin{aligned} \alpha (t)=f(t)+\frac{12-T|f|_1}{T(|g|_1+1)}g(t) \text { and } \gamma (t)=\frac{T(|g|_1+1)}{2(12-T|f|_1)}g(t)+a(0)b(t), \end{aligned}$$

where $|\cdot |_1$ denotes the usual norm of $L^1(0,T)$. Moreover, $(A_7')$ implies that $F(t,0)=0$ for a.e. $t\in [0,T]$, which gives $(A_7)$ directly. Hence Theorem 1.3 implies [24, Theorem2.2]. There are functions F satisfying Theorem 1.3 but not satisfying [24, Theorem 2.2]. For example, let

where $b\in L^\infty (0,T;{\mathbb {R}})$ with $k^2\omega ^2\le b(t)\le (k+1)^2\omega ^2$ $(k\ge 1)$ for a.e. $t\in [0,T]$, and $\mu \in L^\infty (0,T;{\mathbb {R}})$ with $\mu (t)\le \omega ^2$ for a.e. $t\in [0,T]$, $\frac{6}{T}\le \int _0^T\mu ^-(t)dt<\int _0^T\mu ^+(t)dt\le \frac{12}{T},$ and $\mu ^\pm =max\left\{ \pm \mu (t),0\right\}$. Then one has

$$\begin{aligned} F(t,x)\le \frac{1}{2}\mu (t)|x|^2+p(t)|x|+c_1,\ \ \ \ \ \ \ \ \forall x\in {\mathbb {R}}^N \text { and a.e. }t\in [0,T], \end{aligned}$$

where $p(t)=3(b(t)-\mu (t))/4\in L^\infty (0,T;{\mathbb {R}}^+)$ and $c_1=2(k+1)^2\omega ^2$, and hence

$$\begin{aligned} F(t,x)&\le \frac{1}{2}\mu ^+(t)|x|^2+p(t)|x|+c_1\\&\le \frac{1}{2}\mu ^+(t)|x|^2+\frac{1}{2}p(t)\left( \frac{12-T|\mu ^+|_1}{T(|p|_1+1)}|x|^2+\frac{T(|p|_1+1)}{12-T|\mu ^+|_1}\right) +c_1, \end{aligned}$$

which is just (1.2) with

$$\begin{aligned} \alpha (t)=\mu ^+(t)+\frac{12-T|\mu ^+|_1}{T(|p|_1+1)}p(t) \text { and } \gamma (t)=\frac{T(|p|_1+1)}{2(12-T|\mu ^+|_1)}p(t)+c_1. \end{aligned}$$

Thus F satisfies all the conditions of Theorem 1.3. But it does not satisfy the corresponding conditions of [24, Theorem 2.2], since $\int _0^T|\mu (t)|dt>12/T$ and ($A_6'$) does not hold.

Theorem 1.4

The conclusion of Theorem 1.3 holds if we replace $(A_6)$ by $(A_8)$.

Remark 1.4

There are functions F satisfying Theorem 1.4 but not satisfying Theorem 1.3 and [24, Theorem 2.2]. For example, let

where $b\in L^\infty (0,T;{\mathbb {R}})$ with $\omega ^2\le b\le 4\omega ^2$ for a.e. $t\in [0,T]$, and $\mu \in L^\infty (0,T;{\mathbb {R}})$ with $\mu (t)\le \omega ^2$ for a.e. $t\in [0,T]$, $\int _0^T\mu (t)dt>0$, meas$\left\{ t\in [0,T]:\mu ^+(t)<\omega ^2\right\} >0$ and $b(t)-\mu (t)\le 4(\omega ^2-\mu ^+(t))$ for a.e. $t\in [0,T]$. Hence we obtain

$$\begin{aligned} F(t,x)\le \frac{1}{2}\mu (t)|x|^2+p(t)|x|+c_1,\ \ \ \ \ \ \ \ \forall x\in {\mathbb {R}}^N \text { and a.e. }t\in [0,T], \end{aligned}$$

where $p(t)=3(b(t)-\mu (t))/4\in L^\infty (0,T;{\mathbb {R}}^+)$ and $c_1=8\omega ^2$. Then one has

$$\begin{aligned} F(t,x)&\le \frac{1}{2}\mu ^+(t)|x|^2+p(t)|x|+c_1\\&\le \frac{1}{2}(\mu ^+(t)+\frac{1}{6}p(t))|x|^2+3p(t)+c_1 \end{aligned}$$

for all $x\in {\mathbb {R}}^N$ and a.e. $t\in [0,T]$. Consequently, F satisfies all the assumptions of Theorem 1.4. But in the case $\int _0^T\mu ^+(t)dt>12/T$, it does not satisfy the conditions of Theorem 1.3 and [24, Theorem 2.2].

Theorem 1.5

The conclusion of Theorem 1.3 holds if we replace $(A_6)$ by $(A_9)$.

Remark 1.5

Theorem 1.5 is a complement of Theorem 4.2 in [19], where the existence of one solution of (1.1) was proved under conditions (A), $(A_5)$ and $(A_9)$. There are functions F satisfying Theorem 1.5 but not satisfying the results mentioned above. For example, let

where $b\in L^\infty (0,T;{\mathbb {R}}^+)$ with $k^2\omega ^2\le b(t)\le (k+1)^2\omega ^2$ $(k\ge 2)$ for a.e. $t\in [0,T]$, and $\mu \in L^\infty (0,T;{\mathbb {R}})$ with $\mu (t)\le \omega ^2$ for a.e. $t\in [0,T]$, $\int _0^T\mu (t)dt>0$ and

$$\begin{aligned} \text {meas}\left\{ t\in [0,T]:\mu (t)<\omega ^2\right\} >0. \end{aligned}$$

Then F satisfies Theorem 1.5 with $\alpha =\mu ^+(t)$. However, in the case

$$\begin{aligned}\text {meas}\left\{ t\in [0,T]:\mu (t)=\omega ^2\right\} >0 \end{aligned}$$

and $\int _0^T\mu ^+(t)dt> 12/T$, it does not satisfy the conditions of Theorems 1.3, 1.4 and [24, Theorem 2.2].

The paper is organized as follows. Section 2 is devoted to some related preliminaries. In Sect. 3, we show that the reduction functional $\psi$ is coercive and satisfies the (PS) condition. Then we apply the three-critical-point theorem to prove the theorems.

2 Preliminaries

Under assumption (A), the energy functional associated to problem (1.1) given by

$$\begin{aligned} \varphi (u)=-\frac{1}{2}\int _0^T|\dot{u}(t)|^2dt+\int _0^T[F(t,u(t))-F(t,0)]dt \end{aligned}$$

(2.1)

is continuously differentiable and weakly lower semi-continuous on $H_T^1$, where

$$\begin{aligned} H^1_T =\left\{ u:[0,T] \rightarrow {\mathbb {R}}^N \left| \begin{array}{l} u \text { is absolutely continuous,}\\ u(0) = u(T)\text { and }\dot{u} \in L^2(0,T;{\mathbb {R}}^N) \end{array} \right. \right\} \end{aligned}$$

is a Hilbert space with the norm defined by

$$\begin{aligned} \Vert u\Vert =\left( \int _0^T|u(t)|^2dt+\int _0^T|\dot{u}(t)|^2dt\right) ^{1/2}. \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \langle \varphi '(u),v\rangle =-\int _0^T(\dot{u}(t),\dot{v}(t))dt+\int _0^T(\nabla F(t,u(t)),v(t))dt \end{aligned}$$

for all $u,v\in H_T^1$. It is well known that the weak solutions of problem (1.1) correspond to the critical points of $\varphi$ (see [14]).

In view of [14, Proposition 1.3], for $u\in \widetilde{H}_T^1=\left\{ u\in H_T^1:\int _0^Tu(t)dt=0\right\}$, we have

$$\begin{aligned} \int _0^T|u(t)|^2dt\le \frac{T^2}{4\pi ^2}\int _0^T|\dot{u}(t)|^2dt\ \ \ \ \ \ \text {(Wirtinger's inequality)} \end{aligned}$$

and

$$\begin{aligned} \Vert u\Vert _\infty ^2\le \frac{T}{12}\int _0^T|\dot{u}(t)|^2dt\ \ \ \ \ \ \text {(Sobolev's inequality)}, \end{aligned}$$

which implies that

$$\begin{aligned} \Vert u\Vert _\infty \le C\Vert u\Vert \end{aligned}$$

(2.2)

for all $u\in H_T^1$ and some $C>0$, where $\Vert u\Vert _\infty =\max _{t\in [0,T]}|u(t)|$.

Lemma 2.1

If $\alpha \in L^\infty (0,T;{\mathbb {R}}^+)$ such that $\alpha (t)\le \omega ^2$ for a.e. $t\in [0,T]$ and

$$\begin{aligned} \text {meas}\left\{ t\in [0,T]:\alpha (t)<\omega ^2\right\} >0, \end{aligned}$$

then there exists $a<1$ such that

$$\begin{aligned} \int _0^T\alpha (t)|u|^2dt\le a\int _0^T|\dot{u}|^2dt,\ \ \ \ \ \ \forall u\in \widetilde{H}_T^1. \end{aligned}$$

Proof

We prove this assertion by contradiction. In fact, if not, there exists a sequence $(u_n)\subset \widetilde{H}_T^1$ such that

$$\begin{aligned} \int _0^T\alpha (t)|u_n|^2dt>\left( 1-\frac{1}{n}\right) \int _0^T|\dot{u}_n|^2dt,\ \ \ \ \ \ \ \ \ \forall n\in {\mathbb {N}}, \end{aligned}$$

which implies that $u_n\ne 0$ for all n. By the homogeneity of the above inequality, we may assume that $\int _0^T|\dot{u}_n|^2dt=1$ and

$$\begin{aligned} \int _0^T\alpha (t)|u_n|^2dt\ge 1-\frac{1}{n},\ \ \ \ \ \ \ \ \forall n\in {\mathbb {N}}. \end{aligned}$$

(2.3)

It follows from the weak compactness of the unit ball of $\widetilde{H}_T^1$ that there exists a subsequence, still denoted by $(u_n)$, such that

$$\begin{aligned}&u_n\rightharpoonup u\ \ \ \ \ \ \ \text {in } \widetilde{H}_T^1,\\&u_n\rightarrow u\ \ \ \ \ \ \ \text {in } C(0,T;{\mathbb {R}}^N). \end{aligned}$$

Combining this with (2.3), we obtain

$$\begin{aligned} \int _0^T\alpha (t)|u|^2dt\ge 1, \end{aligned}$$

and then,

$$\begin{aligned} 1\ge \int _0^T|\dot{u}|^2dt\ge \omega ^2\int _0^T|u|^2dt\ge \int _0^T\alpha (t)|u|^2dt\ge 1. \end{aligned}$$

Hence we obtain

$$\begin{aligned}1=\int _0^T|\dot{u}|^2dt=\omega ^2\int _0^T|u|^2dt \end{aligned}$$

and

$$\begin{aligned} \int _0^T(\omega ^2-\alpha (t))|u|^2dt=0, \end{aligned}$$

which yields that $u=a\cos \omega t+b\sin \omega t$, a, $b\in {\mathbb {R}}^N$, $u\ne 0$ and $u=0$ on a positive measure subset. This contradicts the fact that $u=a\cos \omega t+b\sin \omega t$ only has finite zeros if $u\ne 0$. $\square$

To end this section, we state the reduction method developed by Tang and Wu [19, Lemma 2.1] and the three-critical-point theorem proposed by Brezis and Nirenberg (see [3, Theorem4]).

Proposition 2.1

(see [19]) Suppose that V is a reflexive Banach space and W is a Banach space, $\varphi \in C^1(V\times W,{\mathbb {R}})$. Assume that there exists $\mu >0$ such that

$$\begin{aligned} \langle D_1\varphi (v_1,w)-D_1\varphi (v_2,w),v_1-v_2\rangle \le -\mu \Vert v_1-v_2\Vert ^2 \end{aligned}$$

(2.4)

for all $v_1$, $v_2\in V$ and $w\in W$. Then there exists a map $\theta \in C(W,V)$ such that $\theta (w)$ is the unique maxmum of $\varphi (\cdot ,w)$ for all $w\in W$ and the functional $\psi$, given by

$$\begin{aligned} \psi (w)=\varphi (\theta (w),w)=\sup _{v\in V}\varphi (v,w), \end{aligned}$$

is continuously differentiable and

$$\begin{aligned} \psi '(w)=D_2\varphi (\theta (w),w),\ \ \ \ \forall w\in W. \end{aligned}$$

Moreover, $(\theta (w),w)$ is a critical point of $\varphi$ if and only if w is a critical point of $\psi$.

Remark 2.1

When V and W are Hilbert spaces, Proposition 2.1 is a corollary of Amann [1, Theorem 2.3].

Proposition 2.2

(Brezis and Nirenberg [3]) Let X be a Banach space with a direct sum decomposition $X=X_1\bigoplus X_2$ with dim$X_2<+\infty$ and let $\psi$ be a $C^1$-functional on X with $\psi (0)=0$, satisfying the (PS) condition. Assume that, for some $\delta _0>0$,

$$\begin{aligned} \psi (u)\ge 0,\ \ \ \ \ \ \ \ \forall u\in X_1 \text { with }\Vert u\Vert \le \delta _0 \end{aligned}$$

and

$$\begin{aligned} \psi (u)\le 0,\ \ \ \ \ \ \ \ \forall u\in X_2 \text { with }\Vert u\Vert \le \delta _0. \end{aligned}$$

Assume also that $\psi$ is bounded from below and $\inf _{u\in X}\psi (u)<0$. Then $\psi$ has at least two nonzero critical points.

3 Proofs of the Theorems

First, we consider the subquadratic k(t)-concave case. Let $\varphi$ be the functional introduced in (2.1) and decompose the Hilbert space $H_T^1$ as $H_T^1=V\bigoplus W$ with

$$\begin{aligned} V=\widetilde{H}_T^1\ \ \ \ \ \ \ \ \text {and}\ \ \ \ \ \ \ W={\mathbb {R}}^N. \end{aligned}$$

Proof of Theorem 1.1

The proof will be divided into several steps.

• (1) By condition ($A_1$), since

  $$\begin{aligned} (\nabla (-F(t,x))-\nabla (-F(t,y)),x-y)\ge -k(t)|x-y|^2,\ \ \ \ \ \ \forall x,y\in {\mathbb {R}}^N, \end{aligned}$$

  we obtain, for $v_1$, $v_2\in \widetilde{H}_T^1$ and $x\in {\mathbb {R}}^N$,

  $$\begin{aligned}&\langle D_1\varphi (v_1,x)-D_1\varphi (v_2,x),v_1-v_2\rangle \\&= -\int _0^T|\dot{v}_1-\dot{v}_2|^2dt+\int _0^T(\nabla F(t,v_1+x)-\nabla F(t,v_2+x),v_1-v_2)dt\\&\le -\int _0^T|\dot{v}_1-\dot{v}_2|^2dt+\int _0^Tk(t)|v_1-v_2|^2dt\\&\le -\int _0^T|\dot{v}_1-\dot{v}_2|^2dt+\int _0^Tk(t)dt\cdot \Vert v_1-v_2\Vert _\infty ^2\\&\le -\left( 1-\frac{T}{12}\int _0^Tk(t)dt\right) |\dot{v}_1-\dot{v}_2|_2^2, \end{aligned}$$

  that is, $\varphi$ satisfies (2.4) with $\mu =1-(T/12)\int _0^Tk(t)dt$. Applying Proposition 2.1, we obtain a continuous map $\theta :{\mathbb {R}}^N\rightarrow \widetilde{H}_T^1$ and a $C^1$-functional $\psi :{\mathbb {R}}^N\rightarrow {\mathbb {R}}$ such that

  $$\begin{aligned} \psi (x)=\varphi (\theta (x)+x)=\sup _{v\in \widetilde{H}_T^1}\varphi (v+x). \end{aligned}$$

  It suffices to find two nonzero critical points of $\psi$.

• (2) It follows from the definition of $\psi$ and condition ($A_2$) that

  $$\begin{aligned} \psi (x)\ge \varphi (x)=\int _0^T[F(t,x)-F(t,0)]dt\rightarrow +\infty \ \ \ \ \ \ \ \ \text {as }|x|\rightarrow \infty . \end{aligned}$$

  Hence $\psi$ is coercive on ${\mathbb {R}}^N$, and then $\psi$ is bounded from below and satisfies the (PS) condition, i.e., $(u_n)\subset W$ possesses a convergent subsequence whenever $\left\{ \psi (u_n)\right\}$ is bounded and $\psi '(u_n)\rightarrow 0$ as $n\rightarrow \infty$.

• (3) Set $W_1=\left\{ 0\right\}$ and $W_2={\mathbb {R}}^N$. The continuity of $\theta$ implies that there exists $\delta _0>0$ small enough such that

  $$\begin{aligned} \Vert \theta (x)+x\Vert _\infty \le C\Vert \theta (x)+x\Vert \le \delta \end{aligned}$$

  for all $x\in W_2$ with $|x|\le \delta _0$. Thus, using ($A_3$), we obtain

  $$\begin{aligned} \psi (x)=\varphi (\theta (x)+x)\le \int _0^T[F(t,\theta (x)+x)-F(t,0)]dt\le 0 \end{aligned}$$

  (3.1)

  for all $x\in W_2$ with $|x|\le \delta _0$. On the other hand, noting

  $$\begin{aligned} \psi (0)\ge \varphi (0)=0, \end{aligned}$$

  we have

  $$\begin{aligned} \psi (w)\ge 0,\ \ \ \ \ \ \ \ \forall w\in W_1 \text { with }\Vert w\Vert \le \delta _0. \end{aligned}$$

  This, jointly with (3.1), shows that $\varphi$ has a local linking at 0 with respect to $(W_1,W_2)$.

• (4) If $\displaystyle {\inf _{x\in {\mathbb {R}}^N}\psi (x)=0}$, then all $x\in W_2$ with $|x|\le \delta _0$ are minimums of $\psi$, which implies that $\psi$ has infinitely many critical points. If $\displaystyle {\inf _{x\in {\mathbb {R}}^N}\psi (x)<0}$, Proposition 2.2 implies that $\psi$ has at least two nontrivial critical points. In any case, we can find two nonzero critical points $x_1$, $x_2$ of $\psi$. Consequently, $u_1=\theta (x_1)+x_1$ and $u_2=\theta (x_2)+x_2$ are two nontrivial solutions of problem (1.1). This completes the proof.

$\square$

Proof of Theorem 1.2

The proof is identical to that of Theorem 1.1 except that now (2.4) follows from condition $(A_4)$. Indeed, using Lemma 2.1, we obtain

$$\begin{aligned} \langle D_1\varphi (v_1,x)-D_1\varphi (v_2,x),v_1-v_2\rangle&\le -\int _0^T|\dot{v}_1-\dot{v}_2|^2dt+\int _0^Tk(t)|v_1-v_2|^2dt\\&\le -(1-a)|\dot{v}_1-\dot{v}_2|_2^2 \end{aligned}$$

for all $v_1$, $v_2\in \widetilde{H}_T^1$ and $x\in {\mathbb {R}}^N$. Therefore, (2.4) holds with $\mu =1-a$. $\square$

Next we consider the subquadratic $\mu (t)$-convex case. Let $\Phi =-\varphi$ and decompose the Hilbert space $H_T^1$ as $H_T^1=V\bigoplus W$ with

$$\begin{aligned} V={\mathbb {R}}^N \ \ \ \ \ \ \ \text {and}\ \ \ \ \ \ W=\widetilde{H}_T^1. \end{aligned}$$

Proof of Theorem 1.3

We divide the proof into several steps.

• (1) It follows from $(A_5)$ that

  $$\begin{aligned} (\nabla F(t,x)-\nabla F(t,y),x-y)\ge \mu (t)|x-y|^2,\ \ \ \ \ \ \forall x,y\in {\mathbb {R}}^N, \end{aligned}$$

  and hence,

  $$\begin{aligned} \langle D_1\Phi (x_1,w)-D_1\Phi (x_2,w),x_1-x_2\rangle&=-\int _0^T(\nabla F(t,x_1+w)-\nabla F(t,x_2+w),x_1-x_2)dt\nonumber \\&\le -\int _0^T\mu (t)|x_1-x_2|^2dt\nonumber \\&\le -{\frac{1}{T}}\int _0^T\mu (t)dt\Vert x_1-x_2\Vert ^2, \end{aligned}$$

  (3.2)

  for all $x_1$, $x_2\in {\mathbb {R}}^N$ and $w\in \widetilde{H}_T^1$. Therefore, (2.4) holds with $\mu =T^{-1}\int _0^T\mu (t)dt$. In view of Proposition 2.1, we obtain a continuous map $\theta :\widetilde{H}_T^1\rightarrow {\mathbb {R}}^N$ and a $C^1$-functional $\psi :\widetilde{H}_T^1\rightarrow {\mathbb {R}}$ such that

  $$\begin{aligned} \psi (w)=\Phi (\theta (w)+w)=\sup _{x\in {\mathbb {R}}^N}\Phi (x+w). \end{aligned}$$

  It suffices to find two nonzero critical points of $\psi$.

• (2) Since the reduced functional $\psi$ is defined on an infinite dimensional subspace, it is difficult to verify the coerciveness and (PS) condition for $\psi$. To overcome this difficulty, we need to study the properties of $\theta$. We claim that the functional $\theta$ obtained above is bounded, i.e., it maps bounded sets into bounded sets. Indeed, note that $D_1\Phi (\theta (w),w)=0$. Setting $x_1=\theta (w)$ and $x_2=0$ in (3.2), we obtain

  $$\begin{aligned} {\frac{1}{T}}\int _0^T\mu (t)dt\Vert \theta (w)\Vert ^2&\le -\langle D_1\Phi (\theta (w),w)-D_1\Phi (0,w),\theta (w)\rangle \\&=\langle D_1\Phi (0,w),\theta (w)\rangle \\&=-\int _0^T(\nabla F(t,w),\theta (w))dt\\&\le \max _{s\in [0,C\Vert w\Vert ]}a(s)\int _0^Tb(t)dt\cdot C\Vert \theta (w)\Vert \end{aligned}$$

  by assumption (A) and (2.2). Hence the desired result follows from the last inequality. Now letting $(w_n)\subset \widetilde{H}_T^1$ be a bounded sequence such that $\psi '(w_n)\rightarrow 0$ as $n\rightarrow \infty$, we show that $(w_n)$ has a convergent subsequence. Without loss of generality, we can assume that

  $$\begin{aligned}&w_n\rightharpoonup w\ \ \ \ \ \ \ \text {in }\widetilde{H}_T^1,\\&w_n\rightarrow w\ \ \ \ \ \ \ \text {in } C(0,T;{\mathbb {R}}^N). \end{aligned}$$

  Since $(w_n)$ is bounded, then $\left\{ \theta (w_n)\right\}$ and hence $\left\{ \theta (w_n)+w_n\right\}$ are also bounded in $H_T^1$. Hence, taking

  $$\begin{aligned} a_1=\max _{s\in [0,C(\Vert \theta (w_n)+w_n\Vert )]}a(s),\ \ \ \ \ a_2=\max _{s\in [0,C(\Vert \theta (w)+w\Vert )]}a(s), \end{aligned}$$

  we have

  $$\begin{aligned}&|\dot{w}_n-\dot{w}|_2^2\\&\quad =\langle \psi '(w_n)-\psi '(w),w_n-w\rangle +\int _0^T(\nabla F(t,\theta (w_n)+w_n)-\nabla F(t,\theta (w)+w),w_n-w)dt\\&\quad \le \Vert \psi '(w_n)\Vert \Vert w_n-w\Vert -\langle \psi '(w),w_n-w\rangle +\left( a_1\int _0^Tb(t)dt+ a_2\int _0^Tb(t)dt\right) \Vert w_n-w\Vert _\infty \\&\quad \rightarrow 0 \end{aligned}$$

  because of assumption (A) and (2.2). The equivalence of the $L^2$-norm for $\dot{w}$ and the $H_T^1$-norm on $\widetilde{H}_T^1$ implies that

  $$\begin{aligned} w_n\rightarrow w\ \ \ \ \ \ \text {in }\widetilde{H}_T^1. \end{aligned}$$

  Hence the (PS) condition holds. Moreover, it follows from $(A_6)$ and Sobolev’s inequality that

  $$\begin{aligned} \Phi (w)&=\frac{1}{2}\int _0^T|\dot{w}|^2dt-\int _0^T[F(t,w)-F(t,0)]dt\\&\ge \frac{1}{2}\int _0^T|\dot{w}|^2dt-\frac{1}{2}\int _0^T\alpha (t)|w|^2dt-\int _0^T\gamma (t)dt+\int _0^TF(t,0)dt\\&\ge \frac{1}{2}\int _0^T|\dot{w}|^2dt-\frac{1}{2}\int _0^T\alpha (t)dt\cdot \Vert w\Vert _\infty ^2-\int _0^T\gamma (t)dt+\int _0^TF(t,0)dt\\&\ge \frac{1}{2}\left( 1-\frac{T}{12}\int _0^T\alpha (t)dt\right) |\dot{w}|_2^2-\int _0^T\gamma (t)dt+\int _0^TF(t,0)dt \end{aligned}$$

  for all $w\in \widetilde{H}_T^1$, which implies that

  $$\begin{aligned} \Phi (w)\rightarrow +\infty \ \ \ \ \ \ \ \text {as }\Vert w\Vert \rightarrow \infty ,\ \ w\in \widetilde{H}_T^1. \end{aligned}$$

  (3.3)

  This, jointly with the definition of $\psi$, shows that

  $$\begin{aligned} \psi (w)\rightarrow +\infty \ \ \ \ \ \ \ \ \ \text {as}\ \ \Vert w\Vert \rightarrow \infty ,\ \ w\in \widetilde{H}_T^1. \end{aligned}$$

  So $\psi$ is bounded from below.

• (3) Letting

  $$\begin{aligned} W_2=\left\{ \sum _{j=1}^k(a_j\cos j\omega t+b_j \sin j\omega t)\left| a_j,b_j\in {\mathbb {R}}^N, j=1,2,\cdots ,k\right. \right\} \end{aligned}$$

  and $W_1=({\mathbb {R}}^N\bigoplus W_2)^\bot$. Then $\widetilde{H}_T^1=W_1\bigoplus W_2$, and the reduction functional $\psi$ has a local linking at 0 with respect to $(W_1,W_2)$. In fact, condition $(A_7)$ and (2.2) imply that

  $$\begin{aligned} \psi (w)&=\sup _{x\in {\mathbb {R}}^N}\Phi (x+w)\\&\ge \Phi (w)\\&=\frac{1}{2}\int _0^T|\dot{w}|^2dt-\int _0^T[F(t,w)-F(t,0)]dt\\&\ge \frac{1}{2}\int _0^T|\dot{w}|^2dt-\frac{(k+1)^2\omega ^2}{2}\int _0^T|w|^2dt\\&\ge 0 \end{aligned}$$

  for all $w\in W_1$ with $\Vert w\Vert \le \delta /C$. Furthermore, by the continuity of $\theta$, the map $w\longmapsto \theta (w)+w$ from $\widetilde{H}_T^1$ to $H_T^1$ is continuous. Hence, using (2.2), there exists a constant $\delta _0\in (0,\delta /C)$ such that

  $$\begin{aligned} \Vert \theta (w)+w\Vert _\infty \le C\Vert \theta (w)+w\Vert \le \delta \end{aligned}$$

  for $w\in \widetilde{H}_T^1$ with $\Vert w\Vert \le \delta _0$, so that, using $(A_7)$, we obtain

  $$\begin{aligned} \psi (w)&=\Phi (\theta (w)+w)\nonumber \\&= \frac{1}{2}\int _0^T|\dot{w}|^2dt-\int _0^T[F(t,\theta (w)+w)-F(t,0)]dt\nonumber \\&\le \frac{1}{2}\int _0^T|\dot{w}|^2dt-\frac{1}{2}k^2\omega ^2\int _0^T|\theta (w)+w|^2dt\nonumber \\&\le 0 \end{aligned}$$

  (3.4)

  for $w\in W_2$ with $\Vert w\Vert \le \delta _0$. Thus, $\psi$ has a local linking at 0 with respect to $(W_1,W_2)$.

• (4) It follows from (3.4) that $\inf _{w\in \widetilde{H}_T^1}\psi (w)\le 0$. If $\inf _{w\in \widetilde{H}_T^1}\psi (w)=0$, all $w\in W_2$ with $\Vert w\Vert \le \delta _0$ are minimums of $\psi$, which implies that $\psi$ has infinitely many critical points. If $\inf _{w\in \widetilde{H}_T^1}\psi (w)<0$, Proposition 2.2 implies that $\psi$ has at least two nontrivial critical points. In any case, the functional $\psi$ possesses two nonzero critical points, denoted by $w_1$, $w_2$. Consequently, $u_1=\theta (w_1)+w_1$ and $u_2=\theta (w_2)+w_2$ are two nonzero solutions of problem (1.1). This completes the proof.

$\square$

Proof of Theorem 1.4

The proof is similar to that of Theorem 1.3 with the exception that now (3.3) follows from condition $(A_8)$. We verify this assertion by contradiction. If not, there exist $c_2\in {\mathbb {R}}$ and a sequence $(u_n)\subset \widetilde{H}_T^1$ such that $\Vert u_n\Vert \rightarrow \infty$ as $n\rightarrow \infty$ and

$$\begin{aligned} \Phi (u_n)\le c_2,\ \ \ \ \ \ \ \forall n\in {\mathbb {N}}. \end{aligned}$$

(3.5)

Set

$$\begin{aligned} H_1={\mathbb {R}}^N\bigoplus (\sin \omega t{\mathbb {R}}^N)\bigoplus (\cos \omega t{\mathbb {R}}^N). \end{aligned}$$

Then $H_T^1=H_1\bigoplus H_1^\bot$, and for $u_n\in \widetilde{H}_T^1$, we can rewrite

$$\begin{aligned} u_n=(a_n\Vert u_n\Vert \cos \omega t+b_n\Vert u_n\Vert \sin \omega t)+w_n, \end{aligned}$$

where $a_n$, $b_n\in {\mathbb {R}}^N$ and $w_n\in H_1^\bot$. It is obvious that

$$\begin{aligned} \int _0^T |\dot{w}_n|^2dt\ge 4\omega ^2\int _0^T |w_n|^2dt. \end{aligned}$$

Hence, using $(A_8)$, we have

$$\begin{aligned} c_2&\ge \Phi (u_n)\\&\ge \frac{1}{2}\int _0^T|\dot{u}_n|^2dt-\frac{\omega ^2}{2}\int _0^T|u_n|^2dt-\int _0^T\gamma (t)dt+\int _0^TF(t,0)dt\\&= \frac{1}{2}\int _0^T|\dot{w}_n|^2dt-\frac{\omega ^2}{2}\int _0^T|w_n|^2dt-\int _0^T\gamma (t)dt+\int _0^TF(t,0)dt\\&\ge \frac{3}{8}\int _0^T|\dot{w}_n|^2dt-\int _0^T\gamma (t)dt+\int _0^TF(t,0)dt, \end{aligned}$$

which implies that $(w_n)$ is bounded. Taking $v_n=u_n/\Vert u_n\Vert$, then $\Vert v_n\Vert =1$, and hence $(a_n)$ and $(b_n)$ are bounded. Up to a subsequence, we can assume that

$$\begin{aligned} a_n\rightarrow a\ \ \ \ \ \ \text {and }\ \ \ \ \ \ \ b_n\rightarrow b\ \ \ \text {as}\ \ n\rightarrow \infty , \end{aligned}$$

for some a, $b\in {\mathbb {R}}$. By the boundedness of $(w_n)$, one has $w_n/\Vert u_n\Vert \rightarrow 0$. Hence

$$\begin{aligned} v_n\rightarrow a\cos \omega t+b\sin \omega t\text { in } H_T^1, \end{aligned}$$

and $|a|+|b|\ne 0$, which yields that

$$\begin{aligned} v_n(t)\rightarrow a\cos \omega t+b\sin \omega t \text { uniformly for a.e. }t\in [0,T] \end{aligned}$$

by Sobolev’s inequality. Therefore,

$$\begin{aligned} |u_n(t)|\rightarrow \infty \text { as}\ \ n\rightarrow \infty \end{aligned}$$

for a.e. $t\in [0,T]$, since $a\cos \omega t+b\sin \omega t$ only has finite zeros. Now set

$$\begin{aligned} E=\left\{ t\in [0,T]\left| F(t,x)-\frac{1}{2}\omega ^2|x|^2\rightarrow -\infty \text { as}\ \ |x|\rightarrow \infty \right. \right\} . \end{aligned}$$

It follows from Fatou’s lemma (see [25]) that

$$\begin{aligned} \liminf _{n\rightarrow \infty }\Phi (u_n)&\ge \liminf _{n\rightarrow \infty }\int _0^T\left( \frac{\omega ^2}{2}|u_n|^2-F(t,u_n)\right) dt+\int _0^TF(t,0)dt\\&\ge \liminf _{n\rightarrow \infty }\int _E\left( \frac{\omega ^2}{2}|u_n|^2-F(t,u_n)\right) dt-\int _0^T\gamma (t)dt+\int _0^TF(t,0)dt\\&=+\,\infty , \end{aligned}$$

a contradiction with (3.5). Thus

$$\begin{aligned} \Phi (u)\rightarrow +\infty \text { as }\Vert u\Vert \rightarrow \infty ,\ \ u\in \widetilde{H}_T^1.\ \ \ \ \ \ \ \ \ \end{aligned}$$

$\square$

Proof of Theorem 1.5

It follows from (A) and $(A_9)$ that, for $\varepsilon \in (0,1-a)$, there exists $M_\varepsilon >0$ such that

$$\begin{aligned} F(t,x)\le \frac{1}{2}(\alpha (t)+\varepsilon \omega ^2)|x|^2+\max _{s\in [0,M_\varepsilon ]}a(s)b(t) \end{aligned}$$

for all $x\in {\mathbb {R}}^N$ and a.e. $t\in [0,T]$. Hence, we obtain, by Lemma 2.1 and Wirtinger’s inequality,

$$\begin{aligned} \Phi (w)&=\frac{1}{2}\int _0^T|\dot{w}|^2dt-\int _0^T[F(t,w)-F(t,0)]dt\\&\ge \frac{1}{2}\int _0^T|\dot{w}|^2dt-\frac{1}{2}\int _0^T(\alpha (t)+\varepsilon \omega ^2)|w|^2dt-\max _{s\in [0,M_\varepsilon ]}a(s)\int _0^Tb(t)dt+\int _0^TF(t,0)dt\\&\ge \frac{1}{2}(1-a-\varepsilon )|\dot{w}|_2^2-\max _{s\in [0,M_\varepsilon ]}a(s)\int _0^Tb(t)dt+\int _0^TF(t,0)dt \end{aligned}$$

for all $w\in \widetilde{H}_T^1$. The equivalence of the norm $|\dot{w}|_2$ and the $H_T^1$-norm on $\widetilde{H}_T^1$ shows that

$$\begin{aligned} \Phi (w)\rightarrow +\infty \text { as}\ \ \Vert w\Vert \rightarrow \infty ,\ \ w\in \widetilde{H}_T^1. \end{aligned}$$

The remainder of the proof is the same as that of Theorem 1.3. Consequently, Theorem 1.5 holds. $\square$