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Existence of fast homoclinic orbits for a class of second-order non-autonomous problems

Qiongfen Zhang1*, Qi-Ming Zhang2 and Xianhua Tang3

Author Affiliations

1 College of Science, Guilin University of Technology, Guilin, Guangxi, 541004, P.R. China

2 College of Science, Hunan University of Technology, Zhuzhou, Hunan, 412000, P.R. China

3 School of Mathematical Sciences and Computing Technology, Central South University, Changsha, Hunan, 410083, P.R. China

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Boundary Value Problems 2014, 2014:89  doi:10.1186/1687-2770-2014-89


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2014/1/89


Received:3 January 2014
Accepted:23 April 2014
Published:6 May 2014

© 2014 Zhang et al.; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Abstract

By applying the mountain pass theorem and the symmetric mountain pass theorem in critical point theory, the existence and multiplicity of fast homoclinic solutions are obtained for the following second-order non-autonomous problem: u ¨ ( t ) + q ( t ) u ˙ ( t ) a ( t ) | u ( t ) | p 2 u ( t ) + W ( t , u ( t ) ) = 0 , where p 2 , t R , u R N , a C ( R , R ) , W C 1 ( R × R N , R ) are not periodic in t and q : R R is a continuous function and Q ( t ) = 0 s q ( s ) d s with lim | t | + Q ( t ) = + .

MSC: 34C37, 35A15, 37J45, 47J30.

Keywords:
fast homoclinic solutions; variational methods; critical point

1 Introduction

Consider fast homoclinic solutions of the following problem:

u ¨ ( t ) + q ( t ) u ˙ ( t ) a ( t ) | u ( t ) | p 2 u ( t ) + W ( t , u ( t ) ) = 0 , t R , (1.1)

where p 2 , t R , u R N , a C ( R , R ) , W C 1 ( R × R N , R ) are not periodic in t, and q : R R is a continuous function and Q ( t ) = 0 s q ( s ) d s with

lim | t | + Q ( t ) = + . (1.2)

When q ( t ) 0 , problem (1.1) reduces to the following special second-order Hamiltonian system:

u ¨ ( t ) a ( t ) | u ( t ) | p 2 u ( t ) + W ( t , u ( t ) ) = 0 , t R . (1.3)

When p = 0 , problem (1.1) reduces to the following second-order damped vibration problem:

u ¨ ( t ) + q ( t ) u ˙ ( t ) a ( t ) u ( t ) + W ( t , u ( t ) ) = 0 , t R . (1.4)

If we take p = 2 and q ( t ) 0 , then problem (1.1) reduces to the following second-order Hamiltonian system:

u ¨ ( t ) a ( t ) u ( t ) + W ( t , u ( t ) ) = 0 , t R . (1.5)

The existence of homoclinic orbits plays an important role in the study of the behavior of dynamical systems. If a system has transversely intersected homoclinic orbits, then it must be chaotic. If it has smoothly connected homoclinic orbits, then it cannot stand the perturbation, and its perturbed system probably produces chaotic phenomena. The first work about homoclinic orbits was done by Poincaré [1].

Recently, the existence and multiplicity of homoclinic solutions and periodic solutions for Hamiltonian systems have been extensively studied by critical point theory. For example, see [2-30] and references therein. In [6,16,17], the authors considered homoclinic solutions for the special Hamiltonian system (1.3) in weighted Sobolev space. Later, Shi et al.[31] obtained some results for system (1.3) with a p-Laplacian, which improved and generalized the results in [6,16,17]. However, there is little research as regards the existence of homoclinic solutions for damped vibration problems (1.4) when q ( t ) 0 . In 2008, Wu and Zhou [32] obtained some results for damped vibration problems (1.4) with some boundary value conditions by variational methods. Zhang and Yuan [33,34] studied the existence of homoclinic solutions for (1.4) when q ( t ) c is a constant. Later, Chen et al.[35] investigated fast homoclinic solutions for (1.4) and obtained some new results under more relaxed assumptions on W ( t , x ) , which resolved some open problems in [33]. Zhang [36] obtained infinitely many solutions for a class of general second-order damped vibration systems by using the variational methods. Zhang [37] investigated subharmonic solutions for a class of second-order impulsive systems with damped term by using the mountain pass theorem.

Motivated by [21,23,32-34,38-42], we will establish some new results for (1.1) in weighted Sobolev space. In order to introduce the concept of fast homoclinic solutions for problem (1.1), we first state some properties of the weighted Sobolev space E on which the certain variational functional associated with (1.1) is defined and the fast homoclinic solutions are the critical points of the certain functional.

Let

X = { u H 1 , 2 ( R , R N ) | R e Q ( t ) [ | u ˙ ( t ) | 2 + | u ( t ) | 2 ] d t < + } ,

where Q ( t ) is defined in (1.2) and for u , v X , let

u , v = R e Q ( t ) [ ( u ˙ ( t ) , v ˙ ( t ) ) + ( u ( t ) , v ( t ) ) ] d t .

Then X is a Hilbert space with the norm given by

u = ( R e Q ( t ) [ | u ˙ ( t ) | 2 + | u ( t ) | 2 ] d t ) 1 / 2 .

It is obvious that

X L 2 ( e Q ( t ) )

with the embedding being continuous. Here L p ( e Q ( t ) ) ( 2 p < + ) denotes the Banach spaces of functions on ℝ with values in R N under the norm

u p = { R e Q ( t ) | u ( t ) | p d t } 1 / p .

If σ is a positive, continuous function on ℝ and 1 < s < + , let

L σ s ( e Q ( t ) ) = { u L loc 1 ( e Q ( t ) ) | R σ ( t ) e Q ( t ) | u ( t ) | s d t < + } .

L σ s equipped with the norm

u s , σ = ( R σ ( t ) e Q ( t ) | u ( t ) | s d t ) 1 / s

is a reflexive Banach space.

Set E = X L a p ( e Q ( t ) ) , where a is the function given in condition (A). Then E with its standard norm is a reflexive Banach space. Similar to [33,35], we have the following definition of fast homoclinic solutions.

Definition 1.1 If (1.2) holds, a solution of (1.1) is called a fast homoclinic solution if u E .

The functional φ corresponding to (1.1) on E is given by

φ ( u ) = R e Q ( t ) [ 1 2 | u ˙ ( t ) | 2 + a ( t ) p | u ( t ) | p W ( t , u ( t ) ) ] d t , u E . (1.6)

Clearly, it follows from (W1) or (W1)′ that φ : E R . By Theorem 2.1 of [43], we can deduce that the map

u a ( t ) e Q ( t ) | u ( t ) | p 2 u ( t )

is continuous from L a p ( e Q ( t ) ) in the dual space L a 1 / ( p 1 ) p 1 ( e Q ( t ) ) , where p 1 = p p 1 . As the embeddings E X L γ ( e Q ( t ) ) for all γ 2 are continuous, if (A) and (W1) or (W1)′ hold, then φ C 1 ( E , R ) and one can easily check that

φ ( u ) , v = R e Q ( t ) [ ( u ˙ ( t ) , v ˙ ( t ) ) + a ( t ) | u ( t ) | p 2 ( u ( t ) , v ( t ) ) ( W ( t , u ( t ) ) , v ( t ) ) ] d t , u E . (1.7)

Furthermore, the critical points of φ in E are classical solutions of (1.1) with u ( ± ) = 0 .

Now, we state our main results.

Theorem 1.1Suppose thata, q, andWsatisfy (1.2) and the following conditions:

(A) Let p > 2 , a ( t ) is a continuous, positive function onsuch that for all t R

a ( t ) α | t | β , α > 0 , β > ( p 2 ) / 2 .

(W1) W ( t , x ) = W 1 ( t , x ) W 2 ( t , x ) , W 1 , W 2 C 1 ( R × R N , R ) , and there exists a constant R > 0 such that

1 a ( t ) | W ( t , x ) | = o ( | x | p 1 ) as  x 0

uniformly in t ( , R ] [ R , + ) .

(W2) There is a constant μ > p such that

0 < μ W 1 ( t , x ) ( W 1 ( t , x ) , x ) , ( t , x ) R × R N { 0 } .

(W3) W 2 ( t , 0 ) = 0 and there exists a constant ϱ ( p , μ ) such that

W 2 ( t , x ) 0 , ( W 2 ( t , x ) , x ) ϱ W 2 ( t , x ) , ( t , x ) R × R N .

Then problem (1.1) has at least one nontrivial fast homoclinic solution.

Theorem 1.2Suppose thata, q, andWsatisfy (1.2), (A), (W2), and the following conditions:

(W1)′ W ( t , x ) = W 1 ( t , x ) W 2 ( t , x ) , W 1 , W 2 C 1 ( R × R N , R ) , and

1 a ( t ) | W ( t , x ) | = o ( | x | p 1 ) as  x 0

uniformly in t R .

(W3)′ W 2 ( t , 0 ) = 0 and there exists a constant ϱ ( p , μ ) such that

( W 2 ( t , x ) , x ) ϱ W 2 ( t , x ) , ( t , x ) R × R N .

Then problem (1.1) has at least one nontrivial fast homoclinic solution.

Theorem 1.3Suppose thata, q, andWsatisfy (1.2), (A), (W1)-(W3), and the following assumption:

(W4) W ( t , x ) = W ( t , x ) , ( t , x ) R × R N .

Then problem (1.1) has an unbounded sequence of fast homoclinic solutions.

Theorem 1.4Suppose thata, q, andWsatisfy (1.2), (A), (W1)′, (W2), (W3)′, and (W4). Then problem (1.1) has an unbounded sequence of fast homoclinic solutions.

Remark 1.1 It is easy to see that our results hold true even if p = 2 . To the best of our knowledge, similar results for problem (1.1) cannot be seen in the literature, from this point, our results are new. As pointed out in [17], condition (A) can be replaced by more general assumption: a ( t ) + as | t | + .

The rest of this paper is organized as follows: in Section 2, some preliminaries are presented. In Section 3, we give the proofs of our results. In Section 4, some examples are given to illustrate our results.

2 Preliminaries

Let E and be given in Section 1, by a similar argument in [41], we have the following important lemma.

Lemma 2.1For any u E ,

u 1 2 e 0 u = 1 2 e 0 { R e Q ( s ) [ | u ˙ ( s ) | 2 + | u ( s ) | 2 ] d s } 1 / 2 , (2.1)

| u ( t ) | { t + e Q ( s ) e Q ( s ) [ | u ˙ ( s ) | 2 + | u ( s ) | 2 ] d s } 1 / 2 | u ( t ) | 1 e 0 4 { t + e Q ( s ) [ | u ˙ ( s ) | 2 + | u ( s ) | 2 ] d s } 1 / 2 (2.2)

and

| u ( t ) | { t e Q ( s ) e Q ( s ) [ | u ˙ ( s ) | 2 + | u ( s ) | 2 ] d s } 1 / 2 1 e 0 4 { t e Q ( s ) [ | u ˙ ( s ) | 2 + | u ( s ) | 2 ] d s } 1 / 2 , (2.3)

where u = ess sup t R | u ( t ) | , e 0 = e min { Q ( t ) : t R } .

The following lemma is an improvement result of [16].

Lemma 2.2Ifasatisfies assumption (A), then

the embedding  L a p ( e Q ( t ) ) L 2 ( e Q ( t ) )  is continuous . (2.4)

Moreover, there exists a Hilbert spaceZsuch that

the embeddings  L a p ( e Q ( t ) ) Z L 2 ( e Q ( t ) )  are continuous ; (2.5)

the embedding  X Z L 2 ( e Q ( t ) )  is compact . (2.6)

Proof Let θ = p / ( p 2 ) , θ = p / 2 , we have

u 2 2 = R e Q ( t ) | u ( t ) | 2 d t = R a 1 / θ a 1 / θ e Q ( t ) / θ e Q ( t ) / θ | u ( t ) | 2 d t ( R a θ / θ e Q ( t ) d t ) 1 / θ ( R a e Q ( t ) | u ( t ) | 2 θ d t ) 1 / θ = a 1 ( R a e Q ( t ) | u ( t ) | p d t ) 2 / p = a 1 u p , a 2 ,

where from (A) and (1.2), a 1 = ( R a 2 / ( p 2 ) e Q ( t ) d t ) ( p 2 ) / p < + . Then (2.4) holds.

By (A), there exists a continuous positive function ρ such that ρ ( t ) + as | t | + and

a 2 = ( R ρ θ a θ / θ e Q ( t ) d t ) 1 / θ < + .

Since

u 2 , ρ 2 = R ρ e Q ( t ) | u ( t ) | 2 d t = R ρ a 1 / θ a 1 / θ e Q ( t ) / θ e Q ( t ) / θ | u ( t ) | 2 d t ( R ρ θ a θ / θ e Q ( t ) d t ) 1 / θ ( R a e Q ( t ) | u ( t ) | p d t ) 1 / θ = a 2 u p , a 2 ,

(2.5) holds by taking Z = L ρ 2 ( e Q ( t ) ) .

Finally, as X Z is the weighted Sobolev space Γ 1 , 2 ( R , ρ , 1 ) , it follows from [43] that (2.6) holds. □

The following two lemmas are the mountain pass theorem and the symmetric mountain pass theorem, which are useful in the proofs of our theorems.

Lemma 2.3[44]

LetEbe a real Banach space and I C 1 ( E , R ) satisfying (PS)-condition. Suppose I ( 0 ) = 0 and:

(i) There exist constants ρ , α > 0 such that I B ρ ( 0 ) α .

(ii) There exists an e E B ¯ ρ ( 0 ) such that I ( e ) 0 .

ThenIpossesses a critical value c α which can be characterized as

c = inf h Φ max s [ 0 , 1 ] I ( h ( s ) ) ,

where Φ = { h C ( [ 0 , 1 ] , E ) | h ( 0 ) = 0 , h ( 1 ) = e } and B ρ ( 0 ) is an open ball inEof radiusρcentered at 0.

Lemma 2.4[44]

LetEbe a real Banach space and I C 1 ( E , R ) withIeven. Assume that I ( 0 ) = 0 andIsatisfies (PS)-condition, assumption (i) of Lemma 2.3 and the following condition:

(iii) For each finite dimensional subspace E E , there is r = r ( E ) > 0 such that I ( u ) 0 for u E B r ( 0 ) , B r ( 0 ) is an open ball inEof radiusrcentered at 0.

ThenIpossesses an unbounded sequence of critical values.

Lemma 2.5Assume that (W2) and (W3) or (W3)′ hold. Then for every ( t , x ) R × R N ,

(i) s μ W 1 ( t , s x ) is nondecreasing on ( 0 , + ) ;

(ii) s ϱ W 2 ( t , s x ) is nonincreasing on ( 0 , + ) .

The proof of Lemma 2.5 is routine and we omit it.

3 Proofs of theorems

Proof of Theorem 1.1 Step 1. The functional φ satisfies the (PS)-condition. Let { u n } E satisfying φ ( u n ) is bounded and φ ( u n ) 0 as n . Then there exists a constant C 1 > 0 such that

| φ ( u n ) | C 1 , φ ( u n ) E μ C 1 . (3.1)

From (1.6), (1.7), (3.1), (W2), and (W3), we have

2 C 1 + 2 C 1 u n 2 φ ( u n ) 2 μ φ ( u n ) , u n = μ 2 μ u ˙ n 2 2 + 2 R e Q ( t ) [ W 2 ( t , u n ( t ) ) 1 μ ( W 2 ( t , u n ( t ) ) , u n ( t ) ) ] d t 2 R e Q ( t ) [ W 1 ( t , u n ( t ) ) 1 μ ( W 1 ( t , u n ( t ) ) , u n ( t ) ) ] d t + ( 2 p 2 μ ) R a ( t ) e Q ( t ) | u n ( t ) | p d t μ 2 μ u ˙ n 2 2 + ( 2 p 2 μ ) u n p , a p .

It follows from Lemma 2.2, μ > p > 2 , and the above inequalities that there exists a constant C 2 > 0 such that

u n C 2 , n N . (3.2)

Now we prove that u n u 0 in E. Passing to a subsequence if necessary, it can be assumed that u n u 0 in E. Since Q ( t ) as | t | , we can choose T > R such that

Q ( t ) ln ( C 2 2 ξ 2 ) for  | t | T . (3.3)

It follows from (2.2), (3.2), and (3.3) that

| u n ( t ) | 2 t + e Q ( s ) e Q ( s ) [ | u ˙ n ( s ) | 2 + | u n ( s ) | 2 ] d s ξ 2 C 2 2 u n 2 ξ 2 for  t T  and  n N . (3.4)

Similarly, by (2.3), (3.2), and (3.3), we have

| u n ( t ) | 2 ξ 2 for  t T  and  n N . (3.5)

Since u n u 0 in E, it is easy to verify that u n ( t ) converges to u 0 ( t ) pointwise for all t R . Hence, it follows from (3.4) and (3.5) that

| u 0 ( t ) | ξ for  t ( , T ] [ T , + ) . (3.6)

Since e Q ( t ) e 0 > 0 on [ T , T ] = J , the operator defined by S : E X ( J ) : u u | J is a linear continuous map. So u n u 0 in X ( J ) . The Sobolev theorem implies that u n u 0 uniformly on J, so there is n 0 N such that

T T e Q ( t ) | W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) | | u n ( t ) u 0 ( t ) | d t < ε for  n n 0 . (3.7)

For any given number ε > 0 , by (W1), we can choose ξ > 0 such that

| W ( t , x ) | ε a ( t ) | x | p 1 for  | t | R  and  | x | ξ . (3.8)

From (3.8), we have

e Q ( t ) | W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) | 2 e Q ( t ) [ ε a ( t ) ( | u n ( t ) | p 1 + | u 0 ( t ) | p 1 ) ] 2 e Q ( t ) [ ε 2 p 1 a ( t ) | u n ( t ) u 0 ( t ) | p 1 + ε ( 1 + 2 p 1 ) a ( t ) | u 0 ( t ) | p 1 ] 2 2 2 p ε 2 a 2 ( t ) e Q ( t ) | u n ( t ) u 0 ( t ) | 2 ( p 1 ) + ( 2 ε ) 2 ( 1 + 2 p 1 ) 2 a 2 ( t ) e Q ( t ) | u 0 ( t ) | 2 ( p 1 ) : = g n ( t ) . (3.9)

Moreover, since a ( t ) is a positive continuous function on ℝ and u n ( t ) converges to u 0 ( t ) pointwise for all t R , it follows from (3.9) that

lim n g n ( t ) = ( 2 ε ) 2 ( 1 + 2 p 1 ) 2 a 2 ( t ) e Q ( t ) | u 0 ( t ) | 2 ( p 1 ) : = g ( t ) for a.e.  t R

and

lim n R ( T , T ) g n ( t ) d t = lim n R ( T , T ) e Q ( t ) [ 2 2 p ε 2 a 2 ( t ) | u n ( t ) u 0 ( t ) | 2 ( p 1 ) + ( 2 ε ) 2 ( 1 + 2 p 1 ) 2 a 2 ( t ) | u 0 ( t ) | 2 ( p 1 ) ] d t = 2 2 p ( ε ) 2 lim n R ( T , T ) a 2 ( t ) e Q ( t ) | u n ( t ) u 0 ( t ) | 2 ( p 1 ) d t + ( 2 ε ) 2 ( 1 + 2 p 1 ) 2 R ( T , T ) a 2 ( t ) e Q ( t ) | u 0 ( t ) | 2 ( p 1 ) d t = ( 2 ε ) 2 ( 1 + 2 p 1 ) 2 R ( T , T ) a 2 ( t ) e Q ( t ) | u 0 ( t ) | 2 ( p 1 ) d t = R g ( t ) d t < + .

From Lebesgue’s dominated convergence theorem, (3.4), (3.5), (3.6), (3.9), and the above inequalities, we have

lim n R ( T , T ) e Q ( t ) | W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) | 2 d t = 0 . (3.10)

From Lemma 2.2, we have u n u 0 in L 2 ( e Q ( t ) ) . Hence, by (3.10),

R ( T , T ) e Q ( t ) | W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) | | u n ( t ) u 0 ( t ) | d t ( R ( T , T ) e Q ( t ) | W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) | 2 d t ) 1 / 2 × ( R ( T , T ) e Q ( t ) | u n ( t ) u 0 ( t ) | 2 d t ) 1 / 2

tends to 0 as n + , which together with (3.7) shows that

R e Q ( t ) | W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) | | u n ( t ) u 0 ( t ) | d t 0 as  n . (3.11)

From (1.7), we have

0 φ ( u n ) φ ( u 0 ) , u n u 0 = u ˙ n u ˙ 0 2 2 + R a ( t ) e Q ( t ) ( | u n ( t ) | p 2 u n ( t ) | u 0 ( t ) | p 2 u 0 ( t ) ) ( u n ( t ) u 0 ( t ) ) d t R e Q ( t ) ( W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) , u n ( t ) u 0 ( t ) ) d t u ˙ n u ˙ 0 2 2 + C 3 R a ( t ) e Q ( t ) ( | u n ( t ) u 0 ( t ) | p ) d t R e Q ( t ) ( W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) , u n ( t ) u 0 ( t ) ) d t , n , (3.12)

where C 3 is a positive constant. It follows from (3.11) and (3.12) that

u ˙ n 2 u ˙ 0 2 as  n (3.13)

and

R a ( t ) e Q ( t ) | u n ( t ) | p d t R a ( t ) e Q ( t ) | u 0 ( t ) | p d t as  n . (3.14)

Hence, u n u 0 in E by (3.13) and (3.14). This shows that φ satisfies (PS)-condition.

Step 2. From (W1), there exists δ ( 0 , 1 ) such that

| W ( t , x ) | 1 2 a ( t ) | x | p 1 for  | t | R , | x | δ . (3.15)

By (3.15) and W ( t , 0 ) = 0 , we have

| W ( t , x ) | 1 2 p a ( t ) | x | p for  | t | R , | x | δ . (3.16)

Let

C 4 = sup { W 1 ( t , x ) a ( t ) | t [ R , R ] , x R , | x | = 1 } . (3.17)

Set σ = min { 1 / ( 2 p C 4 + 1 ) 1 / ( μ p ) , δ } and u = 2 e 0 σ : = ρ , it follows from Lemma 2.1 that | u ( t ) | σ δ < 1 for t R . From Lemma 2.5(i) and (3.17), we have

R R e Q ( t ) W 1 ( t , u ( t ) ) d t { t [ R , R ] : u ( t ) 0 } e Q ( t ) W 1 ( t , u ( t ) | u ( t ) | ) | u ( t ) | μ d t C 4 R R a ( t ) e Q ( t ) | u ( t ) | μ d t C 4 σ μ p R R a ( t ) e Q ( t ) | u ( t ) | p d t 1 2 p R R a ( t ) e Q ( t ) | u ( t ) | p d t . (3.18)

By (W3), (3.16), and (3.18), we have

φ ( u ) = 1 2 R e Q ( t ) | u ˙ ( t ) | 2 d t + R a ( t ) p e Q ( t ) | u ( t ) | p d t R e Q ( t ) W ( t , u ( t ) ) d t = 1 2 u ˙ 2 2 + 1 p u p , a p R ( R , R ) e Q ( t ) W ( t , u ( t ) ) d t R R e Q ( t ) W ( t , u ( t ) ) d t 1 2 u ˙ 2 2 + 1 p u p , a p R ( R , R ) e Q ( t ) W ( t , u ( t ) ) d t R R e Q ( t ) W 1 ( t , u ( t ) ) d t 1 2 u ˙ 2 2 + 1 p u p , a p 1 2 p R ( R , R ) a ( t ) e Q ( t ) | u ( t ) | p d t 1 2 p R R a ( t ) e Q ( t ) | u ( t ) | p d t = 1 2 u ˙ 2 2 + 1 2 p u p , a p .

Therefore, we can choose a constant α > 0 depending on ρ such that φ ( u ) α for any u E with u = ρ .

Step 3. From Lemma 2.5(ii) and (2.1), we have for any u E

3 3 e Q ( t ) W 2 ( t , u ( t ) ) d t = { t [ 3 , 3 ] : | u ( t ) | > 1 } e Q ( t ) W 2 ( t , u ( t ) ) d t + { t [ 3 , 3 ] : | u ( t ) | 1 } e Q ( t ) W 2 ( t , u ( t ) ) d t { t [ 3 , 3 ] : | u ( t ) | > 1 } e Q ( t ) W 2 ( t , u ( t ) | u ( t ) | ) | u ( t ) | ϱ d t + 3 3 e Q ( t ) max | x | 1 W 2 ( t , x ) d t u ϱ 3 3 e Q ( t ) max | x | = 1 W 2 ( t , x ) d t + 3 3 e Q ( t ) max | x | 1 W 2 ( t , x ) d t ( 1 2 e 0 ) ϱ u ϱ 3 3 e Q ( t ) max | x | = 1 W 2 ( t , x ) d t + 3 3 e Q ( t ) max | x | 1 W 2 ( t , x ) d t = C 5 u ϱ + C 6 , (3.19)

where C 5 = ( 1 2 e 0 ) ϱ 3 3 e Q ( t ) max | x | = 1 W 2 ( t , x ) d t , C 6 = 3 3 e Q ( t ) max | x | 1 W 2 ( t , x ) d t . Take ω E such that

| ω ( t ) | = { 1 for  | t | 1 , 0 for  | t | 3 , (3.20)

and | ω ( t ) | 1 for | t | ( 1 , 3 ] . For s > 1 , from Lemma 2.5(i) and (3.20), we get

1 1 e Q ( t ) W 1 ( t , s ω ( t ) ) d t s μ 1 1 e Q ( t ) W 1 ( t , ω ( t ) ) d t = C 7 s μ , (3.21)

where C 7 = 1 1 e Q ( t ) W 1 ( t , ω ( t ) ) d t > 0 . From (W3), (1.6), (3.19), (3.20), and (3.21), we get for s > 1

φ ( s ω ) = s 2 2 ω ˙ 2 2 + s p p ω p , a p + R e Q ( t ) [ W 2 ( t , s ω ( t ) ) W 1 ( t , s ω ( t ) ) ] d t s 2 2 ω ˙ 2 2 + s p p ω p , a p + 3 3 e Q ( t ) W 2 ( t , s ω ( t ) ) d t 1 1 e Q ( t ) W 1 ( t , s ω ( t ) ) d t s 2 2 ω ˙ 2 2 + s p p ω p , a p + C 5 s ϱ ω ϱ + C 6 C 7 s μ . (3.22)

Since μ > ϱ > p and C 7 > 0 , it follows from (3.22) that there exists s 1 > 1 such that s 1 ω > ρ and φ ( s 1 ω ) < 0 . Set e = s 1 ω ( t ) , then e E , e = s 1 ω > ρ , and φ ( e ) = φ ( s 1 ω ) < 0 . It is easy to see that φ ( 0 ) = 0 . By Lemma 2.3, φ has a critical value c > α given by

c = inf g Φ max s [ 0 , 1 ] φ ( g ( s ) ) , (3.23)

where

Φ = { g C ( [ 0 , 1 ] , E ) : g ( 0 ) = 0 , g ( 1 ) = e } .

Hence, there exists u E such that

φ ( u ) = c , φ ( u ) = 0 .

The function u is the desired solution of problem (1.1). Since c > 0 , u is a nontrivial fast homoclinic solution. The proof is complete. □

Proof of Theorem 1.2 In the proof of Theorem 1.1, the condition W 2 ( t , x ) 0 in (W3) is only used in the proofs of (3.2) and Step 2. Therefore, we only need to prove that (3.2) and Step 2 still hold if we use (W1)′ and (W3)′ instead of (W1) and (W3). We first prove that (3.2) holds. From (W2), (W3)′, (1.6), (1.7), and (3.1), we have

2 C 1 + 2 C 1 μ ϱ u n 2 φ ( u n ) 2 ϱ φ ( u n ) , u n = ϱ 2 ϱ u ˙ n 2 2 + 2 R e Q ( t ) [ W 2 ( t , u n ( t ) ) 1 ϱ ( W 2 ( t , u n ( t ) ) , u n ( t ) ) ] d t 2 R e Q ( t ) [ W 1 ( t , u n ( t ) ) 1 ϱ ( W 1 ( t , u n ( t ) ) , u n ( t ) ) ] d t + 2 ( 1 p 1 ϱ ) R a ( t ) e Q ( t ) | u n ( t ) | p d t ϱ 2 ϱ u ˙ n 2 2 + 2 ( 1 p 1 ϱ ) u n p , a p ,

which implies that there exists a constant C 3 > 0 such that (3.2) holds. Next, we prove that Step 2 still holds. From (W1)′, there exists δ ( 0 , 1 ) such that

| W ( t , x ) | 1 2 a ( t ) | x | p 1 for  t R , | x | δ . (3.24)

By (3.24) and W ( t , 0 ) = 0 , we have

| W ( t , x ) | 1 2 p a ( t ) | x | p for  t R , | x | δ . (3.25)

Let u = 2 e 0 δ : = ρ , it follows from Lemma 2.1 that | u ( t ) | δ . It follows from (1.6) and (3.25) that

φ ( u ) = 1 2 R e Q ( t ) | u ˙ ( t ) | 2 d t + R a ( t ) e Q ( t ) p | u ( t ) | p d t R e Q ( t ) W ( t , u ( t ) ) d t 1 2 u ˙ 2 2 + 1 p u p , a p R 1 2 p a ( t ) e Q ( t ) | u ( t ) | p d t = 1 2 u ˙ 2 2 + 1 2 p u p , a p .

Therefore, we can choose a constant α > 0 depending on ρ such that φ ( u ) α for any u E with u = ρ . The proof of Theorem 1.2 is complete. □

Proof of Theorem 1.3 Condition (W4) shows that φ is even. In view of the proof of Theorem 1.1, we know that φ C 1 ( E , R ) and satisfies (PS)-condition and assumption (i) of Lemma 2.3. Now, we prove that (iii) of Lemma 2.4. Let E be a finite dimensional subspace of E. Since all norms of a finite dimensional space are equivalent, there exists d > 0 such that

u d u . (3.26)

Assume that dim E = m and { u 1 , u 2 , , u m } is a basis of E such that

u i = d , i = 1 , 2 , , m . (3.27)

For any u E , there exists λ i R , i = 1 , 2 , , m such that

u ( t ) = i = 1 m λ i u i ( t ) for  t R . (3.28)

Let

u = i = 1 m | λ i | u i . (3.29)

It is easy to see that is a norm of E . Hence, there exists a constant d > 0 such that d u u . Since u i E , by Lemma 2.1, we can choose R 1 > R such that

| u i ( t ) | < d δ 1 + m , | t | > R 1 , i = 1 , 2 , , m , (3.30)

where δ is given in (3.25). Let

Θ = { i = 1 m λ i u i ( t ) : λ i R , i = 1 , 2 , , m ; i = 1 m | λ i | = 1 } = { u E : u = d } . (3.31)

Hence, for u Θ , let t 0 = t 0 ( u ) R such that

| u ( t 0 ) | = u . (3.32)

Then by (3.26)-(3.29), (3.31), and (3.32), we have

d = u u d d d u = d d | u ( t 0 ) | = d d | i = 1 m λ i u i ( t 0 ) | d i = 1 m | λ i | | u i ( t 0 ) | , u Θ . (3.33)

This shows that | u ( t 0 ) | d and there exists i 0 { 1 , 2 , , m } such that | u i 0 ( t 0 ) | d / m , which together with (3.30), implies that | t 0 | R 1 . Let R 2 = R 1 + 1 and

γ = min { e Q ( t ) W 1 ( t , x ) : R 2 t R 2 , d 2 | x | d 2 e 0 } . (3.34)

Since W 1 ( t , x ) > 0 for all t R and x R N { 0 } , and W 1 C 1 ( R × R N , R ) , it follows that γ > 0 . For any u E , from Lemma 2.1 and Lemma 2.5(i), we have

R 2 R 2 e Q ( t ) W 2 ( t , u ( t ) ) d t = { t [ R 2 , R 2 ] : | u ( t ) | > 1 } e Q ( t ) W 2 ( t , u ( t ) ) d t + { t [ R 2 , R 2 ] : | u ( t ) | 1 } e Q ( t ) W 2 ( t , u ( t ) ) d t { t [ R 2 , R 2 ] : | u ( t ) | > 1 } e Q ( t ) W 2 ( t , u ( t ) | u ( t ) | ) | u ( t ) | ϱ d t + R 2 R 2 e Q ( t ) max | x | 1 W 2 ( t , x ) d t u ϱ R 2 R 2 e Q ( t ) max | x | = 1 W 2 ( t , x ) d t + R 2 R 2 e Q ( t ) max | x | 1 W 2 ( t , x ) d t ( 1 2 e 0 ) ϱ u ϱ R 2 R 2 e Q ( t ) max | x | = 1 W 2 ( t , x ) d t + R 2 R 2 e Q ( t ) max | x | 1 W 2 ( t , x ) d t = C 8 u ϱ + C 9 , (3.35)

where C 8 = ( 1 2 e 0 ) ϱ R 2 R 2 e Q ( t ) max | x | = 1 W 2 ( t , x ) d t , C 9 = R 2 R 2 e Q ( t ) max | x | 1 W 2 ( t , x ) d t . Since u ˙ i L 2 ( e Q ( t ) ) , i = 1 , 2 , , m , it follows that there exists ε ( 0 , ( ( d ) 2 e 0 ) / ( 32 m 2 d 2 ) ) such that

t + ε t ε | u ˙ i ( s ) | d s = t + ε t ε e Q ( s ) 2 e Q ( s ) 2 | u ˙ i ( s ) | d s 1 e 0 t + ε t ε e Q ( s ) 2 | u ˙ i ( s ) | d s 1 e 0 ( 2 ε ) 1 / 2 ( t + ε t ε e Q ( s ) | u ˙ i ( s ) | 2 d s ) 1 / 2 ( 2 ε e 0 ) 1 / 2 u ˙ i 2 d 4 m for  t R , i = 1 , 2 , , m . (3.36)

Then for u Θ with | u ( t 0 ) | = u and t [ t 0 ε , t 0 + ε ] , it follows from (3.28), (3.31), (3.32), (3.33), and (3.36) that

| u ( t ) | 2 = | u ( t 0 ) | 2 + 2 t 0 t ( u ˙ ( s ) , u ( s ) ) d s | u ( t 0 ) | 2 2 t 0 ε t 0 + ε | u ( s ) | | u ˙ ( s ) | d s | u ( t 0 ) | 2 2 | u ( t 0 ) | t 0 ε t 0 + ε | u ˙ ( s ) | d s | u ( t 0 ) | 2 2 | u ( t 0 ) | i = 1 m | λ i | t 0 ε t 0 + ε | u ˙ i ( s ) | d s ( d ) 2 2 . (3.37)

On the other hand, since u d for u Θ , then

| u ( t ) | u d 2 e 0 , t R , u Θ . (3.38)

Therefore, from (3.34), (3.37), and (3.38), we have

R 2 R 2 e Q ( t ) W 1 ( t , u ( t ) ) d t t 0 ε t 0 + ε e Q ( t ) W 1 ( t , u ( t ) ) d t 2 ε γ for  u Θ . (3.39)

By (3.30) and (3.31), we have

| u ( t ) | i = 1 m | λ i | | u i ( t ) | δ for  | t | R 1 , u Θ . (3.40)

By (1.6), (3.16), (3.35), (3.39), (3.40), and Lemma 2.5, we have for u Θ and r > 1

φ ( r u ) = r 2 2 u ˙ 2 2 + r p p u p , a p + R e Q ( t ) [ W 2 ( t , r u ( t ) ) W 1 ( t , r u ( t ) ) ] d t r 2 2 u ˙ 2 2 + r p p u p , a p + r ϱ R e Q ( t ) W 2 ( t , u ( t ) ) d t r μ R e Q ( t ) W 1 ( t , u ( t ) ) d t = r 2 2 u ˙ 2 2 + r p p u p , a p + r ϱ R ( R 2 , R 2 ) e Q ( t ) W 2 ( t , u ( t ) ) d t r μ R ( R 2 , R 2 ) e Q ( t ) W 1 ( t , u ( t ) ) d t + r ϱ R 2 R 2 e Q ( t ) W 2 ( t , u ( t ) ) d t r μ R 2 R 2 e Q ( t ) W 1 ( t , u ( t ) ) d t r 2 2 u ˙ 2 2 + r p p u p , a p r ϱ R ( R 2 , R 2 ) e Q ( t ) W ( t , u ( t ) ) d t r μ R 2 R 2 e Q ( t ) W 1 ( t , u ( t ) ) d t + r ϱ R 2 R 2 e Q ( t ) W 2 ( t , u ( t ) ) d t r 2 2 u ˙ 2 2 + r p p u p , a p + r ϱ 2 p R ( R 2 , R 2 ) a ( t ) e Q ( t ) | u ( t ) | p d t + r ϱ ( C 8 u ϱ + C 9 ) 2 ε γ r μ r 2 2 u ˙ 2 2 + r p p u p , a p + r ϱ 2 p u p , a p + r ϱ ( C 8 u ϱ + C 9 ) 2 ε γ r μ r 2 2 d 2 + r p p d p + r ϱ 2 p d p + C 8 ( r d ) ϱ + C 9 r ϱ 2 ε γ r μ . (3.41)

Since μ > ϱ > p > 2 , we deduce that there exists r 0 = r 0 ( d , d , C 8 , C 9 , R 1 , R 2 , ε , γ ) = r 0 ( E ) > 1 such that

φ ( r u ) < 0 for  u Θ  and  r r 0 .

It follows that

φ ( u ) < 0 for  u E  and  u d r 0 ,

which shows that (iii) of Lemma 2.4 holds. By Lemma 2.4, φ possesses an unbounded sequence { c n } n = 1 of critical values with c n = φ ( u n ) , where u n is such that φ ( u n ) = 0 for n = 1 , 2 ,  . If { u n } is bounded, then there exists C 10 > 0 such that

u n C 10 for  n N . (3.42)

In a similar fashion to the proof of (3.4) and (3.5), for the given δ in (3.16), there exists R 3 > R such that

| u n ( t ) | δ for  | t | R 3 , n N . (3.43)

Hence, by (1.6), (2.1), (3.16), (3.42), and (3.43), we have

1 2 u ˙ n 2 2 + 1 p u n p , a p = c n + R e Q ( t ) W ( t , u n ( t ) ) d t = c n + R [ R 3 , R 3 ] e Q ( t ) W ( t , u n ( t ) ) d t + R 3 R 3 e Q ( t ) W ( t , u n ( t ) ) d t c n 1 2 p R [ R 3 , R 3 ] a ( t ) e Q ( t ) | u n ( t ) | p d t R 3 R 3 e Q ( t ) | W ( t , u n ( t ) ) | d t c n 1 2 p u n p , a p R 3 R 3 e Q ( t ) max | x | 2 e 0 C 10 | W ( t , x ) | d t ,

which, together with (3.42), implies that

c n 1 2 u ˙ n 2 2 + 3 2 p u n p , a p + R 3 R 3 max | x | 2 e 0 C 10 e Q ( t ) | W ( t , x ) | d t < + .

This contradicts the fact that { c n } n = 1 is unbounded, and so { u n } is unbounded. The proof is complete. □

Proof of Theorem 1.4 In view of the proofs of Theorem 1.2 and Theorem 1.3, the conclusion of Theorem 1.4 holds. The proof is complete. □

4 Examples

Example 4.1 Consider the following system:

u ¨ ( t ) + t u ˙ ( t ) a ( t ) | u ( t ) | 1 / 2 u ( t ) + W ( t , u ( t ) ) = 0 , a.e.  t R , (4.1)

where q ( t ) = t , p = 5 / 2 , t R , u R N , a C ( R , ( 0 , ) ) , and a satisfies (A). Let

W ( t , x ) = a ( t ) ( i = 1 m a i | x | μ i j = 1 n b j | x | ϱ j ) ,

where μ 1 > μ 2 > > μ m > ϱ 1 > ϱ 2 > > ϱ n > 5 / 2 , a i , b j > 0 , i = 1 , , m , j = 1 , , n . Let

W 1 ( t , x ) = a ( t ) i = 1 m a i | x | μ i , W 2 ( t , x ) = a ( t ) j = 1 n b j | x | ϱ j .

Then it is easy to check that all the conditions of Theorem 1.3 are satisfied with μ = μ m and ϱ = ϱ 1 . Hence, problem (4.1) has an unbounded sequence of fast homoclinic solutions.

Example 4.2 Consider the following system:

u ¨ ( t ) + ( t + t 3 ) u ˙ ( t ) a ( t ) | u ( t ) | 4 u ( t ) + W ( t , u ( t ) ) = 0 , a.e.  t R , (4.2)

where q ( t ) = t + t 3 , p = 6 , t R , u R N , a C ( R , ( 0 , ) ) , and a satisfies (A). Let

W ( t , x ) = a ( t ) [ a 1 | x | μ 1 + a 2 | x | μ 2 b 1 ( cos t ) | x | ϱ 1 b 2 | x | ϱ 2 ] ,

where μ 1 > μ 2 > ϱ 1 > ϱ 2 > 6 , a 1 , a 2 > 0 , b 1 , b 2 > 0 . Let

W 1 ( t , x ) = a ( t ) ( a 1 | x | μ 1 + a 2 | x | μ 2 ) , W 2 ( t , x ) = a ( t ) [ b 1 ( cos t ) | x | ϱ 1 + b 2 | x | ϱ 2 ] .

Then it is easy to check that all the conditions of Theorem 1.4 are satisfied with μ = μ 2 and ϱ = ϱ 1 . Hence, by Theorem 1.4, problem (4.2) has an unbounded sequence of fast homoclinic solutions.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors read and approved the final manuscript.

Acknowledgements

Qiongfen Zhang was supported by the NNSF of China (No. 11301108), Guangxi Natural Science Foundation (No. 2013GXNSFBA019004) and the Scientific Research Foundation of Guangxi Education Office (No. 201203YB093). Qi-Ming Zhang was supported by the NNSF of China (No. 11201138). Xianhua Tang was supported by the NNSF of China (No. 11171351).

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