Lazer-Leach type condition for second order differential equations at resonance with impulsive effects via variational method

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Abstract

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

1 Introduction

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

(1.1)

where $m∈N$, $g:R→R$ is a continuous function, $e∈ L 1 (0,2π)$, $0< t 1 < t 2 <⋯< t p <2π$, and $I j :[0,2π]×R→R$ is continuous for every j.

When $Δ x ′ ( t j )≡0$, problem (1.1) becomes the well-known periodic boundary value problem at resonance

(1.2)

Assume that

$lim x → ± ∞ g(x)=g(±∞)$
(g)

exist and are finite. Lazer and Leach  proved that (1.2) has at least one 2π-periodic solution provided that the following condition holds:

$2 [ g ( + ∞ ) − g ( − ∞ ) ] ≠ ∫ 0 2 π e(t)sin(mt+θ)dt,∀θ∈R.$
(1.3)

From then on, a series of relevant resonant problems were studied (see – and the references cited therein) by some classical tools such as topological degree method, variational method, etc. Recently, the periodic problem of the second order differential equation with impulses has been widely studied because of its background in applied sciences (see – and the references cited therein). In this paper, we investigate problem (1.1) under a more general Lazer-Leach type condition. Define

$G(x)= ∫ 0 x g(s)ds$

and for $j=1,2,…,p$,

$J j (t,x)= ∫ 0 x I j (t,s)ds.$

Throughout this paper, we give the following fundamental assumptions.

(H1): The limits

$lim x → ± ∞ G ( x ) x =G(±∞)$

exist and are finite.

(H2): There exist continuous, 2π-periodic functions $K 1 (t), K 2 (t),…, K p (t)$ such that for $j=1,2,…,p$,

(H3): For all $θ∈R$,

$2 [ G ( + ∞ ) − G ( − ∞ ) ] ≠ ∫ 0 2 π e(t)sin(mt+θ)dt+ ∑ j = 1 p K j ( t j )sin(m t j +θ).$

For the sake of convenience, we decompose (H3) into the following two conditions.

($H 3 +$): For all $θ∈R$,

$2 [ G ( + ∞ ) − G ( − ∞ ) ] > ∫ 0 2 π e(t)sin(mt+θ)dt+ ∑ j = 1 p K j ( t j )sin(m t j +θ).$

($H 3 −$): For all $θ∈R$,

$2 [ G ( + ∞ ) − G ( − ∞ ) ] < ∫ 0 2 π e(t)sin(mt+θ)dt+ ∑ j = 1 p K j ( t j )sin(m t j +θ).$

We now can state the main theorems of this paper.

Theorem 1.1

Assume that conditions (H1), (H2) and ($H 3 +$) hold. Then problem (1.1) has at least one 2π-periodic solution.

Theorem 1.2

Assume that conditions (H1), (H2) and ($H 3 −$) hold. Then problem (1.1) has at least one 2π-periodic solution.

From Theorem 1.1 and Theorem 1.2, we obtain the following theorem.

Theorem 1.3

Assume that conditions (H1), (H2) and (H3) hold. Then problem (1.1) has at least one 2π-periodic solution.

Moreover, we have the following corollary.

Corollary 1.4

Assume that conditions (H1) and

($H 3 ′$): for all$θ∈R$,

$2 [ G ( + ∞ ) − G ( − ∞ ) ] ≠ ∫ 0 2 π e(t)sin(mt+θ)dt$
(1.4)

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

Remark 1.5

It is easy to find a function $g(x)$ such that (g) is not satisfied and (H1) holds. For example, we can take $g(x)=cosx$. Hence, Corollary 1.4 improves the related results in the literature mentioned above. Moreover, since we consider the problem with impulses, Theorem 1.3 is also a complement of the pioneering works.

Remark 1.6

When condition ($H 3 ′$) is not satisfied, i.e., there exists $θ 0 ∈R$ such that

$2 [ G ( + ∞ ) − G ( − ∞ ) ] = ∫ 0 2 π e(t)sin(mt+ θ 0 )dt,$

problem (1.2) may have no solution. For example, we consider the resonant differential equation

$x ″ + m 2 x+arctanx=4cosmt.$
(1.5)

Obviously, $g(x)=arctanx$, $e(t)=4cosmt$ and $G(+∞)= π 2$, $G(−∞)=− π 2$. We have

$2 [ G ( + ∞ ) − G ( − ∞ ) ] − ∫ 0 2 π e ( t ) sin ( m t + θ ) d t = 2 π − 4 ∫ 0 2 π cos m t sin ( m t + θ ) d t = 2 π − 4 π sin θ .$

We take $θ 0 ∈R$ such that $sin θ 0 = 1 2$. Then ($H 3 ′$) is not satisfied. From now on, we prove that (1.5) has no 2π-periodic solution by contradiction. Assume that (1.5) has 2π-periodic solution. Multiplying both sides of (1.5) by $cosmt$ and integrating over $[0,2π]$, we get

$4π= ∫ 0 2 π arctanxcosmtdt⩽ ∫ 0 2 π |arctanxcosmt|dt⩽ π 2 ∫ 0 2 π dt= π 2 ,$

which is impossible. Hence, problem (1.2) may have no solution if condition ($H 3 ′$) is not satisfied. Now, we give the following boundary value condition:

$x(0)−x(2π)= x ′ (0)− x ′ (2π)=0$
(1.6)

and the impulsive condition

$Δ x ′ ( π m ) =3π.$
(1.7)

Clearly, $p=1$ and $K 1 ( π m )=3π$. Then

Hence, (H1), (H2) and (H3) hold. Equivalently, Eq. (1.5) with conditions (1.6) and (1.7) has at least one 2π-periodic solution. Therefore, the impulses in problem (1.1) 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 and Theorem 1.2.

2 Preliminaries

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

Define

$H= { x ∈ H 1 ( 0 , 2 π ) : x ( 0 ) = x ( 2 π ) } ,$

with the norm

$∥x∥= ( ∫ 0 2 π ( x ′ 2 ( t ) + x 2 ( t ) ) d t ) 1 2 .$

Consider the functional $φ(x)$ defined on H by

$φ ( x ) = 1 2 ∫ 0 2 π x ′ 2 ( t ) d t − m 2 2 ∫ 0 2 π x 2 ( t ) d t − ∫ 0 2 π G ( x ( t ) ) d t + ∫ 0 2 π e ( t ) x ( t ) d t + ∑ j = 1 p J j ( t j , x ( t j ) ) .$
(2.1)

Similarly as in , $φ(x)$ is continuously differentiable on H, and

(2.2)

Now, we have the following lemma.

Lemma 2.1

If$x∈H$is a critical point of φ, then x is a 2π-periodic solution of Eq. (1.1).

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

We say that φ satisfies (PS) if every sequence $( x n )$ for which $φ( x n )$ is bounded in and $φ ′ ( x n )→0$ (as $n→∞$) possesses a convergent subsequence.

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

Theorem 2.2

Let$φ∈ C 1 (H,R)$and$H= H − ⊕ H +$, $dim( H − )<∞$, $dim( H + )=∞$. We suppose that:

1. (a)

there exist a bounded neighborhood D of 0 in $H −$ and a constant α such that $φ | ∂ D ⩽α$;

2. (b)

there exists a constant $β>α$ such that $φ | H + ⩾β$;

3. (c)

φ satisfies (PS).

Then the functional φ has a critical point in H.

3 The proof of the main results

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

Lemma 3.1

Assume that conditions (H1), (H2) and (H3) hold. Then φ defined by (2.1) satisfies (PS).

Proof

Let $M>0$ be a constant and ${ x n }⊂H$ be a sequence satisfying

$| φ ( x n ) | = | 1 2 ∫ 0 2 π x n ′ 2 d t − m 2 2 ∫ 0 2 π x n 2 d t − ∫ 0 2 π G ( x n ) d t + ∫ 0 2 π e ( t ) x n ( t ) d t + ∑ j = 1 p J j ( t j , x n ( t j ) ) | ⩽ M$
(3.1)

and

$lim n → ∞ ∥ φ ′ ( x n ) ∥ =0.$
(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 $k∈N$,

$lim n → ∞ | φ ′ ( x n ) z k |⩽ lim n → ∞ ∥ φ ′ ( x n ) ∥ ∥ z k ∥=0.$

Thus

Hence

$lim n → ∞ ( ∫ 0 2 π ( x n ′ z k ′ − m 2 x n z k ) d t − ∫ 0 2 π ( g ( x n ) z k − e ( t ) z k ) d t + ∑ j = 1 p I j ( t j , x n ( t j ) ) z k ( t j ) ) = 0 .$
(3.3)

By (H1) and (H2), we have

$lim n → ∞ ( ∫ 0 2 π g ( x n ) z k − e ( t ) z k ∥ x n ∥ d t − ∑ j = 1 p I j ( t j , x n ( t j ) ) z k ( t j ) ∥ x n ∥ ) =0.$
(3.4)

From (3.3) and (3.4), we obtain

$lim n → ∞ ∫ 0 2 π ( x n ′ ∥ x n ∥ z k ′ − m 2 x n ∥ x n ∥ z k ) dt=0.$
(3.5)

Set

$y n = x n ∥ x n ∥ .$

Then we have

$lim n → ∞ ∫ 0 2 π ( y n ′ z k ′ − m 2 y n z k ) dt=0,$

and furthermore,

$lim n → ∞ i → ∞ ∫ 0 2 π [ ( y n − y i ) ′ z k ′ − m 2 ( y n − y i ) z k ] dt=0.$
(3.6)

Replacing $z k$ in (3.6) by $( y n − y i )$, we get

$lim n → ∞ i → ∞ ( ∥ y n − y i ∥ 2 − ( m 2 + 1 ) ∥ y n − y i ∥ 2 2 ) =0.$

Due to the compact imbedding $H↪ L 2 (0,2π)$, going to a subsequence,

Therefore,

$lim n → ∞ i → ∞ ∥ y n − y i ∥ 2 2 =0.$

Furthermore, we have

$lim n → ∞ i → ∞ ∥ y n − y i ∥ 2 =0,$

which implies that ${ y n }$ is a Cauchy sequence in H. Thus, $y n → y 0$ in H. It follows from (3.5) and the usual regularity argument for ordinary differential equations (see ) that

$y 0 = k 1 sinmt+ k 2 cosmt,$
(3.7)

where $k 1 2 + k 2 2 = 1 ( m 2 + 1 ) π$ ($∥ y 0 ∥=1$). (Different subsequences of ${ y n }$ correspond to different $k 1$ and $k 2$.)

Write (3.7) as

$y 0 = 1 ( m 2 + 1 ) π sin(mt+θ),$

where θ satisfies $sinθ= k 2 k 1 2 + k 2 2$ and $cosθ= k 1 k 1 2 + k 2 2$.

Taking $z k = 1 ( m 2 + 1 ) π sin(mt+θ)$, we get, for any $n∈N$,

$∫ 0 2 π ( x n ′ z k ′ − m 2 x n z k ) dt=0.$
(3.8)

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

$lim n → ∞ [ ∫ 0 2 π ( g ( x n ) − e ( t ) ) 1 ( m 2 + 1 ) π sin ( m t + θ ) d t − ∑ j = 1 p I j ( t j , x n ( t j ) ) 1 ( m 2 + 1 ) π sin ( m t j + θ ) ] = 0 .$
(3.9)

By (H1) and (H2), we obtain

$lim n → ∞ [ ∫ 0 2 π ( g ( x n ) − e ( t ) ) ( 1 ( m 2 + 1 ) π sin ( m t + θ ) − y n ) d t − ∑ j = 1 p I j ( t j , x n ( t j ) ) ( 1 ( m 2 + 1 ) π sin ( m t j + θ ) − y n ( t j ) ) ] = 0 .$
(3.10)

It follows from (3.9) and (3.10) that

$lim n → ∞ [ ∫ 0 2 π ( g ( x n ) − e ( t ) ) y n d t − ∑ j = 1 p I j ( t j , x n ( t j ) ) y n ( t j ) ] =0.$

Hence, replacing $z k$ in (3.3) by $y n$, we have

$lim n → ∞ ∫ 0 2 π ( x n ′ x n ′ ∥ x n ∥ − m 2 x n x n ∥ x n ∥ ) dt=0.$
(3.11)

Now, dividing (3.1) by $∥ x n ∥$, we get

$| 1 2 ∫ 0 2 π ( x n ′ 2 ∥ x n ∥ − m 2 x n 2 ∥ x n ∥ ) dt− ∫ 0 2 π G ( x n ) − e ( t ) x n ∥ x n ∥ dt+ ∑ j = 1 p J j ( t j , x n ( t j ) ) ∥ x n ∥ |⩽ M ∥ x n ∥ .$

Passing to the limits, we have

$0 = lim n → ∞ ∫ 0 2 π ( G ( x n ) − e ( t ) x n ) d t ∥ x n ∥ − lim n → ∞ ∑ j = 1 p J j ( t j , x n ( t j ) ) ∥ x n ∥ = lim n → ∞ ∫ 0 2 π G ( x n ) x n ⋅ x n ∥ x n ∥ d t − lim n → ∞ ∫ 0 2 π e ( t ) ⋅ x n ∥ x n ∥ d t − lim n → ∞ ∑ j = 1 p J j ( t j , x n ( t j ) ) x n ( t j ) ⋅ x n ( t j ) ∥ x n ∥ .$

Noting that $x n ∥ x n ∥ → 1 ( m 2 + 1 ) π sin(mt+θ)$ in H as $n→∞$ and

$lim n → ∞ x n (t)= { + ∞ , ∀ t ∈ I + , − ∞ , ∀ t ∈ I − ,$

where $I + :={t∈[0,2π]∣sin(mt+θ)>0}$, $I − :={t∈[0,2π]∣sin(mt+θ)<0}$, we get from the Lebesgue domain convergence theorem that

$0 = ∫ I + G ( + ∞ ) 1 ( m 2 + 1 ) π sin + ( m t + θ ) d t − ∫ I + G ( − ∞ ) 1 ( m 2 + 1 ) π sin − ( m t + θ ) d t − ∫ 0 2 π e ( t ) 1 ( m 2 + 1 ) π sin ( m t + θ ) d t − ∑ j = 1 p K j ( t j ) 1 ( m 2 + 1 ) π sin ( m t j + θ ) ,$

i.e.,

$0=2 [ G ( + ∞ ) − G ( − ∞ ) ] − ∫ 0 2 π e(t)sin(mt+θ)dt− ∑ j = 1 p K j ( t j )sin(m t j +θ),$

which contradicts (H3). This implies that the sequence ${ x n }$ is bounded. Thus, there exists $x 0 ∈H$ such that $x n ⇀ x 0$ weakly in H. Due to the compact imbedding $H↪ L 2 (0,2π)$ and $H↪C(0,2π)$, going to a subsequence,

From (3.3), we obtain

$lim n → ∞ i → ∞ ( ∫ 0 2 π ( ( x n ′ − x i ′ ) z k ′ − m 2 ( x n − x i ) z k ) d t − ∫ 0 2 π ( g ( x n ) − g ( x i ) ) z k d t + ∑ j = 1 p ( I j ( t j , x n ( t j ) ) − I j ( t j , x i ( t j ) ) ) z k ( t j ) ) = 0 .$

Replacing $z k$ by $x n − x i$ in the above equality, we get

$lim n → ∞ i → ∞ ( ∫ 0 2 π ( ( x n ′ − x i ′ ) 2 − m 2 ( x n − x i ) 2 ) d t − ∫ 0 2 π ( g ( x n ) − g ( x i ) ) ( x n − x i ) d t + ∑ j = 1 p ( I j ( t j , x n ( t j ) ) − I j ( t j , x i ( t j ) ) ) ( x n ( t j ) − x i ( t j ) ) ) = 0 .$
(3.12)

By (H1) and (H2), we have

$lim n → ∞ i → ∞ ∫ 0 2 π ( g ( x n ) − g ( x i ) ) ( x n − x i )dt=0$
(3.13)

and

$lim n → ∞ i → ∞ ∑ j = 1 p ( I j ( t j , x n ( t j ) ) − I j ( t j , x i ( t j ) ) ) ( x n ( t j ) − x i ( t j ) ) =0.$
(3.14)

Thus, it follows from (3.12), (3.13) and (3.14) that

$lim n → ∞ i → ∞ ∫ 0 2 π [ ( x n ′ − x i ′ ) 2 − m 2 ( x n − x i ) 2 ] dt=0.$

Therefore,

$lim n → ∞ i → ∞ ∥ x n − x i ∥ 2 =0,$

which implies $x n → x 0$ in H. It shows that φ satisfies (PS). □

Remark 3.2

If conditions (H1), (H2) and ($H 3 +$) (or ($H 3 −$)), φ defined by (2.1) still satisfies (PS).

Now, we can give the proof of Theorem 1.1.

Proof of Theorem 1.1

Denote

$H − =R⊕span{sint,cost,sin2t,cos2t,…,sinmt,cosmt}$

and

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

We first prove that

(3.15)

by contradiction. Assume that there exists a sequence $( x n )⊂ H −$ such that $∥ x n ∥→∞$ (as $n→∞$) and there exists a constant $c −$ satisfying

$lim inf n → ∞ φ( x n )⩾ c − .$
(3.16)

By (H1), we have

$lim n → ∞ ∫ 0 2 π G ( x n ) − e ( t ) x n ∥ x n ∥ 2 dt=0.$
(3.17)

By (H2), we get

$lim n → ∞ ∑ j = 1 p J j ( t j , x n ( t j ) ) ∥ x n ∥ 2 =0.$
(3.18)

From (3.16) and the definition of φ, we obtain

$lim inf n → ∞ [ 1 2 ∫ 0 2 π x n ′ 2 − m 2 x n 2 ∥ x n ∥ 2 d t − ∫ 0 2 π G ( x n ) − e ( t ) x n ∥ x n ∥ 2 d t + ∑ j = 1 p J j ( t j , x n ( t j ) ) ∥ x n ∥ 2 ] ⩾0.$
(3.19)

For $x∈ H −$, we get that there exist constants $a 0 , a 1 ,…, a m$, $b 1 , b 2 ,…, b m$ such that

$x(t)= ∑ j = 0 m a j cosjt+ ∑ j = 1 m b j sinjt.$

Since $∫ 0 2 π sin 2 jtdt= ∫ 0 2 π cos 2 jtdt$ for $j=1,2,…$ , we have, for $x∈ H −$,

$∫ 0 2 π x ′ 2 d t = ∫ 0 2 π ∑ j = 1 m j 2 a j 2 sin 2 j t d t + ∫ 0 2 π ∑ j = 1 m j 2 b j 2 cos 2 j t d t ⩽ m 2 [ ∫ 0 2 π ∑ j = 1 m a j 2 sin 2 j t d t + ∫ 0 2 π ∑ j = 1 m b j 2 cos 2 j t d t ] = m 2 [ ∫ 0 2 π ∑ j = 1 m a j 2 cos 2 j t d t + ∫ 0 2 π ∑ j = 1 m b j 2 sin 2 j t d t ] ⩽ m 2 ∫ 0 2 π x 2 d t .$

Hence, for $x∈ H −$,

$∫ 0 2 π ( x ′ 2 − m 2 x 2 ) dt⩽0.$
(3.20)

The equality in (3.20) holds only for

$x= 1 ( m 2 + 1 ) π sin(mt+θ),θ∈R.$

Set $y n = x n ∥ x n ∥$. Since $dim H − <∞$, going to a subsequence, there exists $y 0 ∈ H −$ such that $y n → y 0$ in H and $y n → y 0$ in $L 2 (0,2π)$. Then (3.17), (3.18), (3.19) and (3.20) imply that

$y 0 = 1 ( m 2 + 1 ) π sin(mt+θ),θ∈R.$

By (3.16), we have, for n large enough,

$1 2 ∫ 0 2 π x n ′ 2 − m 2 x n 2 ∥ x n ∥ dt− ∫ 0 2 π G ( x n ) − e ( t ) x n ∥ x n ∥ dt+ ∑ j = 1 p J j ( t j , x n ( t j ) ) ∥ x n ∥ ⩾ c − ∥ x n ∥ .$
(3.21)

It follows from $x n ∈ H −$ that

$∫ 0 2 π x n ′ 2 − m 2 x n 2 ∥ x n ∥ ⩽0.$
(3.22)

From (3.21) and (3.22), we get, for n large enough,

$c − ∥ x n ∥ ⩽− ∫ 0 2 π G ( x n ) − e ( t ) x n ∥ x n ∥ dt+ ∑ j = 1 p J j ( t j , x n ( t j ) ) ∥ x n ∥ .$

Passing to the limits and using an argument similarly as in the proof of Lemma 3.1, we get

$2 [ G ( + ∞ ) − G ( − ∞ ) ] ⩽ ∫ 0 2 π e(t)sin(mt+θ)dt+ ∑ j = 1 p K j ( t j )sin(m t j +θ),$

which is a contradiction to ($H 3 +$).

Then (3.15) holds.

Next, we prove that

and φ is bounded on bounded sets.

Because of the compact imbedding of $H↪C(0,2π)$ and $H↪ L 2 (0,2π)$, there exist constants $m 1$, $m 2$ such that

$∥ x ∥ ∞ ⩽ m 1 ∥x∥, ∥ x ∥ 2 ⩽ m 2 ∥x∥.$

Then by (H1) and (H2), one has that there exist positive constants $c g , c 1 , c 2 ,…, c p$ such that

$| φ ( x ) | = | 1 2 ∫ 0 2 π x ′ 2 d t − m 2 2 ∫ 0 2 π x 2 d t − ∫ 0 2 π [ G ( x ) − e ( t ) x ] d t + ∑ j = 1 p J j ( t j , x ( t j ) ) | ⩽ 1 2 ∥ x ∥ 2 + m 2 2 m 2 2 ∥ x ∥ 2 + ∫ 0 2 π ( c g | x | + | e ( t ) | | x | ) d t + ∑ j = 1 p c j | x ( t j ) | ⩽ 1 + m 2 m 2 2 2 ∥ x ∥ 2 + m 1 ( c g + ∥ e ∥ 1 ) ∥ x ∥ + ∑ j = 1 p c j m 1 ∥ x ∥ .$
(3.23)

Hence, φ is bounded on the bounded sets of H.

For $x∈ H +$, using an argument similar to the case $x∈ H −$, we have

$∥ x ∥ 2 ⩾ ( ( m + 1 ) 2 + 1 ) ∥ x ∥ 2 2 .$
(3.24)

Thus, from (3.23) and (3.24), we obtain

$φ ( x ) = 1 2 ∫ 0 2 π x ′ 2 d t − m 2 2 ∫ 0 2 π x 2 d t − ∫ 0 2 π [ G ( x ) − e ( t ) x ] d t + ∑ j = 1 p J j ( t j , x ( t j ) ) ⩾ 2 m + 1 2 ( ( m + 1 ) 2 + 1 ) ∥ x ∥ 2 − m 1 ( c g + ∥ e ∥ 1 + ∑ j = 1 p c j ) ∥ x ∥ ,$

which implies

Up to now, the conditions (a) and (b) of Theorem 2.2 are satisfied. According to Remark 3.2, (c) is also satisfied. Hence, by Theorem 2.2, problem (1.1) has at least one solution. This completes the proof. □

Next, we prove Theorem 1.2 slightly differently from Theorem 1.1.

Proof of Theorem 1.2

Denote

$H − =R⊕span { sin t , cos t , sin 2 t , cos 2 t , … , sin ( m − 1 ) t , cos ( m − 1 ) t }$

and

$H + =span{sinmt,cosmt,…}.$

We first prove that

(3.25)

For $x∈ H −$, we get that there exist constants $a 0 , a 1 ,…, a m − 1$, $b 1 , b 2 ,…, b m − 1$ such that

$x(t)= ∑ j = 0 m − 1 a j cosjt+ ∑ j = 1 m − 1 b j sinjt.$

Since $∫ 0 2 π sin 2 jtdt= ∫ 0 2 π cos 2 jtdt$ for $j=1,2,…$ , we have, for $x∈ H −$,

$∫ 0 2 π x ′ 2 d t = ∫ 0 2 π ∑ j = 1 m − 1 j 2 a j 2 sin 2 j t d t + ∫ 0 2 π ∑ j = 1 m − 1 j 2 b j 2 cos 2 j t d t ⩽ ( m − 1 ) 2 [ ∫ 0 2 π ∑ j = 1 m − 1 a j 2 sin 2 j t d t + ∫ 0 2 π ∑ j = 1 m − 1 b j 2 cos 2 j t d t ] = ( m − 1 ) 2 [ ∫ 0 2 π ∑ j = 1 m − 1 a j 2 cos 2 j t d t + ∫ 0 2 π ∑ j = 1 m − 1 b j 2 sin 2 j t d t ] ⩽ ( m − 1 ) 2 ∫ 0 2 π x 2 d t .$

Hence, for $x∈ H −$,

$∥ x ∥ 2 = ∫ 0 2 π ( x ′ 2 + x 2 ) dt⩽ [ ( m − 1 ) 2 + 1 ] ∫ 0 2 π x 2 dt= [ ( m − 1 ) 2 + 1 ] ∥ x ∥ 2 2 .$

The equality holds only for

$x= 1 ( ( m − 1 ) 2 + 1 ) π sin ( ( m − 1 ) t + θ ) ,θ∈R.$

If $x∈ H −$ and $∥x∥→∞$, then

$∥ x ∥ 2 →∞.$

For $x∈ H −$, we have

$∫ 0 2 π ( x ′ 2 − m 2 x 2 ) dt⩽ ( m − 1 ) 2 ∫ 0 2 π x 2 dt− m 2 ∫ 0 2 π x 2 dt=−(2m−1) ∥ x ∥ 2 2 .$

By (H1) and (H2), we get that there exists a constant $c 0 >0$ such that

$| − ∫ 0 2 π [ G ( x ) − e ( t ) x ] d t + ∑ j = 1 p J j ( t j , x ( t j ) ) | ⩽ c 0 ∥ x ∥ 2 .$

Hence, for $x∈ H −$, we obtain

Therefore, (3.25) holds.

Next, we prove that

and φ is bounded on bounded sets.

Because of the compact imbedding of $H↪C(0,2π)$ and $H↪ L 2 (0,2π)$, there exist constants $m 1$, $m 2$ such that

$∥ x ∥ ∞ ⩽ m 1 ∥x∥, ∥ x ∥ 2 ⩽ m 2 ∥x∥.$

Then, by (H1) and (H2), one has that there exist positive constants $c g , c 1 , c 2 ,…, c p$ such that

$| φ ( x ) | = | 1 2 ∫ 0 2 π x ′ 2 d t − m 2 2 ∫ 0 2 π x 2 d t − ∫ 0 2 π [ G ( x ) − e ( t ) x ] d t + ∑ j = 1 p J j ( t j , x ( t j ) ) | ⩽ 1 2 ∥ x ∥ 2 + m 2 2 m 2 2 ∥ x ∥ 2 + ∫ 0 2 π ( c g | x | + | e ( t ) | | x | ) d t + ∑ j = 1 p c j | x ( t j ) | ⩽ 1 + m 2 m 2 2 2 ∥ x ∥ 2 + m 1 ( c g + ∥ e ∥ 1 ) ∥ x ∥ + ∑ j = 1 p c j m 1 ∥ x ∥ .$

Hence, φ is bounded on the bounded sets of H.

In what follows, we prove that

by contradiction. Assume that there exists a sequence $( x n )⊂ H −$ such that $∥ x n ∥→∞$ (as $n→∞$), and there exists a constant $c +$ satisfying

$lim sup n → ∞ φ( x n )⩽ c + .$
(3.26)

By (H1), we have

$lim n → ∞ ∫ 0 2 π G ( x n ) − e ( t ) x n ∥ x n ∥ 2 dt=0.$
(3.27)

By (H2), we get

$lim n → ∞ ∑ j = 1 p J j ( t j , x n ( t j ) ) ∥ x n ∥ 2 =0.$
(3.28)

From (3.26) and the definition of φ, we obtain

$lim sup n → ∞ [ 1 2 ∫ 0 2 π x n ′ 2 − m 2 x n 2 ∥ x n ∥ 2 d t − ∫ 0 2 π G ( x n ) − e ( t ) x n ∥ x n ∥ 2 d t + ∑ j = 1 p J j ( t j , x n ( t j ) ) ∥ x n ∥ 2 ] ⩽0.$
(3.29)

For $x∈ H +$, we get

$∫ 0 2 π x ′ 2 dt⩾ m 2 ∫ 0 2 π x 2 dt.$

Hence, for $x∈ H +$, we have

$∫ 0 2 π ( x ′ 2 − m 2 x 2 ) dt⩾0.$
(3.30)

The equality in (3.30) holds only for

$x= 1 ( m 2 + 1 ) π sin(mt+θ),θ∈R.$

Set $y n = x n ∥ x n ∥$. There exists $y 0 ∈ H +$ such that $y n ⇀ y 0$ weakly in H. Due to the compact imbedding $H↪ L 2 (0,2π)$, going to a subsequence, $y n → y 0$ in $L 2 (0,2π)$. Then (3.27), (3.28), (3.29) and (3.30) imply that

$y 0 = 1 ( m 2 + 1 ) π sin(mt+θ),θ∈R.$

By (3.26), we have, for n large enough,

$1 2 ∫ 0 2 π x n ′ 2 − m 2 x n 2 ∥ x n ∥ dt− ∫ 0 2 π G ( x n ) − e ( t ) x n ∥ x n ∥ dt+ ∑ j = 1 p J j ( t j , x n ( t j ) ) ∥ x n ∥ ⩽ c + ∥ x n ∥ .$
(3.31)

It follows from $x n ∈ H +$ that

$∫ 0 2 π x n ′ 2 − m 2 x n 2 ∥ x n ∥ dt⩾0.$
(3.32)

From (3.31) and (3.32), we get, for n large enough,

$c + ∥ x n ∥ ⩾− ∫ 0 2 π G ( x n ) − e ( t ) x n ∥ x n ∥ dt+ ∑ j = 1 p J j ( t j , x n ( t j ) ) ∥ x n ∥ .$

Passing to the limits and using an argument similarly as in the proof of Lemma 3.1, we get

$2 [ G ( + ∞ ) − G ( − ∞ ) ] ⩾ ∫ 0 2 π e(t)sin(mt+θ)dt+ ∑ j = 1 p K j ( t j )sin(m t j +θ),$

which is a contradiction to ($H 3 −$). This completes the proof. □

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Acknowledgements

The author would like to express their thanks to the editor of the journal and the referees for their careful reading of the first draft of the manuscript and making many helpful comments and suggestions which improved the presentation of the paper. This research was supported by the National Natural Science Foundation of China (11401274), Science and Technology Landing Project of colleges and universities in Jiangxi Province (KJLD14092) and Natural Science Foundation Project of Science and Technology Department of Jiangxi Province (20132BAB201012, 20142BDH80027).

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Correspondence to Jin Li.

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The author declares that they have no competing interests.

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The author has contributed in obtaining new results and written the whole article.

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