Research

# Existence of nontrivial weak homoclinic orbits for second-order impulsive differential equations

Hui Fang* and Hongbo Duan

Author Affiliations

Department of Mathematics, Kunming University of Science and Technology, Kunming, Yunnan, 650500, China

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Boundary Value Problems 2012, 2012:138  doi:10.1186/1687-2770-2012-138

 Received: 25 July 2012 Accepted: 7 November 2012 Published: 26 November 2012

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 cited.

### Abstract

A sufficient condition is obtained for the existence of nontrivial weak homoclinic orbits of second-order impulsive differential equations by employing the mountain pass theorem, a weak convergence argument and a weak version of Lieb’s lemma.

### 1 Introduction

Fečkan [1], Battelli and Fečkan [2] studied the existence of homoclinic solutions for impulsive differential equations by using perturbation methods. Tang et al.[3-6] studied the existence of homoclinic solutions for Hamiltonian systems via variational methods. In recent years, many researchers have paid much attention to multiplicity and existence of solutions of impulsive differential equations via variational methods (for example, see [7-12]). However, few papers have been published on the existence of homoclinic solutions for second-order impulsive differential equations via variational methods.

In this paper, we consider the following impulsive differential equations:

(1.1)

(1.2)

where V : R × R R is of class C 1 , V ( t , 0 ) = V ( t , 0 ) = 0 with V ( t , x ) = ( V / x ) ( t , x ) , and I C ( R , R ) with I ( 0 ) = 0 . ℤ denotes the set of all integers, and t j ( j Z ) are impulsive points. Moreover, there exist a positive integer p and a positive constant T such that 0 < t 0 < t 1 < < t p 1 < T , t l + k p = t l + k T , k Z , l = 0 , 1 , , p 1 . q ( t j + ) = lim h 0 + q ( t j + h ) and q ( t j ) = lim h 0 + q ( t j h ) represent the right and left limits of q ( t ) at t = t j respectively.

We say that a function q ( t ) is a weak homoclinic orbit of Eqs. (1.1) and (1.2) if q satisfies (1.1) and

q { q C ( R , R ) : j = + | q ( t j ) | 2 < + , q L 2 ( R ) , q ( ± ) = 0 , q ( k T ) = 0 , k Z } .

Motivated by the works of Nieto and Regan [7], Smets and Willem [13], in this paper we study the existence of nontrivial weak homoclinic orbits of (1.1)-(1.2) by using the mountain pass theorem, a weak version of Lieb’s lemma and a weak convergence argument. Our method is different from those of [8,9].

The main result is the following.

Theorem 1.1Assume that Eqs. (1.1) and (1.2) satisfy the following conditions:

(H1) There exists a positive numberTsuch that

V ( t + T , x ) = V ( t , x ) , V ( t + T , x ) = V ( t , x ) , ( t , x ) R 2 ;

(H2) lim x 0 V ( t , x ) x = 0 uniformly for t R ;

(H3) There exists a constant μ > 2 such that

x V ( t , x ) μ V ( t , x ) > 0 , ( t , x ) R × R { 0 } ;

(H4) There exist constants a 0 > 0 and a 1 > 0 such that

(H5) There exists a constantb, with 0 < b < μ 2 ( μ + 2 ) T p , such that

| I ( x ) | b | x | ,

and

2 0 x I ( t ) d t I ( x ) x 0 .

Then there exists a nontrivial weak homoclinic orbit of Eqs. (1.1) and (1.2).

Remark 1.1 (H2) implies that q ( t ) 0 is an equilibrium of (1.1)-(1.2).

Remark 1.2 Set V ( t , x ) = ( 2 + sin t ) x 4 , I ( x ) = x 10 π p . It is easy to see that V ( t , x ) , I ( x ) satisfy (H1)-(H5).

### 2 Proof of main results

Lemma 2.1 (Mountain pass lemma [14])

LetEbe a Banach space and φ C 1 ( E , R ) , e E , r > 0 be such that e > r and

b : = inf y = r φ ( y ) > φ ( 0 ) φ ( e ) .

Let

Then, for each ε > 0 , δ > 0 , there exists y E such that

(V1) d 2 ε φ ( y ) d + 2 ε ;

(V2) dist ( y , E ) 2 δ ;

(V3) φ ( y ) 8 ε δ .

In what follows, l 2 denotes the space of sequences whose second powers are summable on ℤ (the set of all integers), that is,

j Z | a j | 2 < + , a = { a j } j = + l 2 .

The space l 2 is equipped with the following norm:

a l 2 = ( j Z | a j | 2 ) 1 2 .

We now prove some technical lemmas.

Lemma 2.2The space

H : = { q C ( R , R ) : { q ( t j ) } j = + l 2 , q L 2 ( R ) , q ( ± ) = 0 , q ( k T ) = 0 , k Z } (2.1)

is a Hilbert space with the inner product

( q 1 , q 2 ) H = R q 1 ( t ) q 2 ( t ) d t , (2.2)

and the corresponding norm

q H = ( R | q ( t ) | 2 d t ) 1 2 . (2.3)

Proof Let { q n } be a Cauchy sequence in H, then { q n } is a Cauchy sequence in L 2 ( R ) and there exists y L 2 ( R ) such that { q n } converges to y in L 2 ( R ) . Define the function q ( t ) as follows:

q ( t ) = k T t y ( s ) d s , k T t < ( k + 1 ) T , k Z .

It is easy to see that

lim h 0 + q ( k T h ) = ( k 1 ) T k T y ( s ) d s .

Since q n ( k T ) = 0 , k Z , we have

| ( k 1 ) T k T y ( s ) d s | = | ( k 1 ) T k T y ( s ) d s [ q n ( k T ) q n ( ( k 1 ) T ) ] | = | ( k 1 ) T k T [ y ( s ) q n ( s ) ] d s | ( k 1 ) T k T | y ( s ) q n ( s ) | d s T 1 2 [ ( k 1 ) T k T | y ( s ) q n ( s ) | 2 d s ] 1 2 T 1 2 [ R | y ( s ) q n ( s ) | 2 d s ] 1 2 ,

which implies that ( k 1 ) T k T y ( s ) d s = 0 , that is, q ( k T ) = 0 , k Z . Therefore, q is continuous. Thus, q C ( R , R ) and q = y .

Noticing that, for k T t < ( k + 1 ) T , we have

| q ( t ) | 2 = | k T t y ( s ) d s | 2 [ k T ( k + 1 ) T | y ( s ) | d s ] 2 T k T ( k + 1 ) T | y ( s ) | 2 d s = T k T + | y ( s ) | 2 d s T ( k + 1 ) T + | y ( s ) | 2 d s ,

which implies q ( ± ) = 0 . On the other hand, since

j = + | q ( t j ) | 2 = l = 0 p 1 k = + | q ( t l + k p ) | 2 ,

and k T < t l + k p = t l + k T < ( k + 1 ) T ( l = 0 , 1 , , p 1 ), we have

| q ( t l + k p ) | 2 = | k T t l + k p y ( s ) d s | 2 T k T ( k + 1 ) T | y ( s ) | 2 d s .

Therefore,

j = + | q ( t j ) | 2 l = 0 p 1 k = + T k T ( k + 1 ) T | y ( s ) | 2 d s = p T R | y ( s ) | 2 d s < + .

Consequently, q H and { q n } converges to q in H. The proof is complete. □

Lemma 2.3For any q H , the following inequalities hold:

| q | : = sup t R | q ( t ) | T 1 2 q H , | q | 2 : = [ R | q ( t ) | 2 d t ] 1 2 T q H .

Furthermore, q H 1 ( R ) and

Proof For any t R , there exists an integer k such that ( k 1 ) T t < k T . Then it follows from Cauchy-Schwarz inequality that

| q ( t ) | = | q ( k T ) q ( t ) | t k T | q ( s ) | d s ( k 1 ) T k T | q ( s ) | d s T 1 2 ( ( k 1 ) T k T | q ( s ) | 2 d s ) 1 2 T 1 2 q H ,

which implies | q | T 1 2 q H .

Furthermore, from the above argument, we have

k = + ( k 1 ) T k T | q ( t ) | 2 d t T 2 k = + ( k 1 ) T k T | q ( t ) | 2 d t = T 2 q H 2 ,

that is, | q | 2 T q H .

Since

Finally, we obtain that

j = + | q ( t j ) | 2 = l = 0 p 1 k = + | q ( t l + k T ) | 2 = l = 0 p 1 k = + | t l + k T ( k + 1 ) T q ( s ) d s | 2 l = 0 p 1 k = + [ k T ( k + 1 ) T | q ( s ) | d s ] 2 p T k = + k T ( k + 1 ) T | q ( s ) | 2 d s = p T q H 2 .

The proof is complete. □

Define the functional φ : H R as follows:

φ ( q ) = 1 2 R | q ( t ) | 2 d t R V ( t , q ( t ) ) d t + j = + 0 q ( t j ) I ( s ) d s , q H . (2.4)

Lemma 2.4If (H1)-(H5) hold, then φ C 1 ( H , R ) and

φ ( q ) , h = R q ( t ) h ( t ) d t R V ( t , q ( t ) ) h ( t ) d t + j = + I ( q ( t j ) ) h ( t j ) , h H . (2.5)

Proof From the continuity of V, V and (H2)-(H3), we see that, for each γ > 0 , there exists C γ > 0 , such that

| V ( t , x ) | C γ | x | , | V ( t , x ) | 1 2 C γ | x | 2 , t R , | x | γ .

Since q ( t ) 0 as t , there exists ρ γ > 0 such that

| q ( t ) | γ , whenever  | t | ρ γ .

Therefore, we have

| V ( t , q ( t ) ) | C γ | q ( t ) | , | V ( t , q ( t ) ) | 1 2 C γ | q ( t ) | 2 , for all  | t | ρ γ .

It follows from (H5) that, q , h H ,

| j = + I ( q ( t j ) ) h ( t j ) | j = + | I ( q ( t j ) ) | | h ( t j ) | j = + b | q ( t j ) | | h ( t j ) | b ( j = + | q ( t j ) | 2 ) 1 2 ( j = + | h ( t j ) | 2 ) 1 2 < + ,

and

j = + | 0 q ( t j ) I ( s ) d s | j = + min { 0 , q ( t j ) } max { 0 , q ( t j ) } | I ( s ) | d s b 2 j = + | q ( t j ) | 2 < + . (2.6)

Thus, φ and the right hand of (2.5) is well defined on H. By the definition of Fréchet derivative, it is easy to see that φ C 1 ( H , R ) and (2.5) holds. □

Lemma 2.5If q H is a critical point of the functionalφ, thenqsatisfies (1.1).

Proof If q H is a critical point of the functional φ, then for any h C 0 ( R ) , we have

0 = φ ( q ) , h = R q ( t ) h ( t ) d t R V ( t , q ( t ) ) h ( t ) d t + j = + I ( q ( t j ) ) h ( t j ) .

j Z , take h C 0 ( R ) such that h ( t ) = 0 for any t ( , t j ] [ t j + 1 , + ) , and h C 0 ( [ t j , t j + 1 ] ) . Therefore, we have

0 = t j t j + 1 q ( t ) h ( t ) d t t j t j + 1 V ( t , q ( t ) ) h ( t ) d t ,

by the definition of the weak derivative, which implies

q ( t ) + V ( t , q ( t ) ) = 0 a.e. on  ( t j , t j + 1 ) . (2.7)

Hence, the critical point q H of the functional φ satisfies (1.1). The proof is complete. □

Lemma 2.6Under the assumptions (H1)-(H5), there exists e H and r > 0 such that e H > r and

b : = inf y H = r φ ( y ) > φ ( 0 ) φ ( e ) .

Proof If q H and q H 1 T 1 2 , then, by Lemma 2.3, | q | 1 . Hence, by (H5) and Lemma 2.3, we have

j = + 0 q ( t j ) I ( s ) d s j = + min { 0 , q ( t j ) } max { 0 , q ( t j ) } | I ( s ) | d s 1 2 j = + b | q ( t j ) | 2 1 2 b T p q H 2 , (2.8)

and

j = + I ( q ( t j ) ) q ( t j ) j = + | I ( q ( t j ) ) | | q ( t j ) | j = + b | q ( t j ) | 2 b T p q H 2 . (2.9)

It follows from (2.8), (H4) and Lemma 2.3 that

φ ( q ) = 1 2 q H 2 R V ( t , q ( t ) ) d t + j = + 0 q ( t j ) I ( s ) d s 1 2 q H 2 a 1 R | q ( t ) | μ d t 1 2 b T p q H 2 1 2 q H 2 a 1 | q | μ 2 R | q ( t ) | 2 d t 1 2 b T p q H 2 1 2 ( 1 b T p ) q H 2 a 1 T μ + 2 2 q H μ .

Therefore, as μ > 2 and b < μ 2 ( μ + 2 ) T p < 1 T p , there exists r > 0 such that inf q H = r φ ( q ) > 0 .

Now, let v H { 0 } and λ > 1 . Then there exists a subset ( a , b ) of ℝ and λ large enough such that

λ | v ( t ) | > 1 , for all  t ( a , b ) .

Since V ( t , λ v ( t ) ) 0 , by (2.4), (H4) and Lemma 2.3, we have

φ ( λ v ) λ 2 2 R | v ( t ) | 2 d t a b V ( t , λ v ( t ) ) d t + j = + 0 λ v ( t j ) I ( s ) d s λ 2 2 v H 2 a 0 λ μ a b | v ( t ) | μ d t + λ 2 2 b T p v H 2 = λ 2 2 ( 1 + b T p ) v H 2 a 0 λ μ a b | v ( t ) | μ d t .

Since μ > 2 , the right-hand member is negative of λ sufficiently large, and there exists e : = λ v H such that e H > r , φ ( e ) 0 . The proof is complete. □

Lemma 2.7Under the assumptions (H1)-(H5), there exists a bounded sequence { q n } inHsuch that

φ ( q n ) d , φ ( q n ) 0 , dist ( q n , H ) 0 ,

where d : = inf γ Γ sup t [ 0 , 1 ] φ ( γ ( t ) ) , Γ = { γ C ( [ 0 , 1 ] , H ) : γ ( 0 ) = 0 , γ ( 1 ) = e } . Furthermore, q n does not converge to 0 in measure.

Proof All we have to prove is that any sequence { q n } obtained by taking ε = 1 / n 2 and δ = 1 / n in Lemma 2.1 is bounded and q n does not converge to 0 in measure. For n sufficiently large, it follows from (H3), (H5), (2.4), (2.5), (2.8) and (2.9) that

d + 1 + q n H φ ( q n ) 1 μ φ ( q n ) , q n = ( 1 2 1 μ ) R | q n ( t ) | 2 d t R [ V ( t , q n ( t ) ) 1 μ V ( t , q n ( t ) ) q n ( t ) ] d t + j = + 0 q n ( t j ) I ( s ) d s 1 μ j = + I ( q n ( t j ) ) q n ( t j ) = ( 1 2 1 μ ) q n H 2 1 μ R [ μ V ( t , q n ( t ) ) V ( t , q n ( t ) ) q n ( t ) ] d t + j = + 0 q n ( t j ) I ( s ) d s 1 μ j = + I ( q n ( t j ) ) q n ( t j ) ( 1 2 1 μ ) q n H 2 b T p 2 q n H 2 b T p μ q n H 2 = ( 1 2 1 μ b T p 2 b T p μ ) q n H 2 .

Since b < μ 2 ( μ + 2 ) T p , { q n } is bounded in H.

Let a 2 : = sup n N { q n H } . By (H2) and (H3), we have

1 2 V ( t , u ) u V ( t , u ) = o ( u 2 ) , as  u 0 ,

which implies

a 3 : = sup | u | T 1 2 a 2 1 2 V ( t , u ) u V ( t , u ) u 2 < .

For any ε > 0 , there exists δ > 0 such that, for | u | δ , we have

| 1 2 V ( t , u ) u V ( t , u ) | ε u 2 .

Therefore, by Lemma 2.3, we have

(2.10)

If q n converges to 0 in measure on R, then it follows from (H5) and (2.10) that

0 < d = φ ( q n ) 1 2 φ ( q n ) , q n + o ( 1 ) = R [ 1 2 V ( t , q n ) q n V ( t , q n ) ] d t + j = + 0 q n ( t j ) I ( s ) d s 1 2 j = + I ( q n ( t j ) ) q n ( t j ) + o ( 1 ) meas { | q n ( t ) | > δ } T a 2 2 a 3 + ε T 2 a 2 2 + 1 2 j = + [ 2 0 q n ( t j ) I ( s ) d s I ( q n ( t j ) ) q n ( t j ) ] + o ( 1 ) meas { | q n ( t ) | > δ } T a 2 2 a 3 + ε T 2 a 2 2 + o ( 1 ) = o ( 1 ) ,

a contradiction. The proof is complete. □

The following lemma is similar to a weak version of Lieb’s lemma [15], which will play an important role in the proof of Theorem 1.1.

Lemma 2.8If { u n } is bounded inHand u n does not converge to 0 in measure, then there exist a sequence { x n k } Z and a subsequence { u n k } of { u n } such that

u n k ( + x n k T ) u 0 in   H 1 ( R ) .

Proof If

lim n sup q Z sup t [ q T T , q T + T ] | u n ( t ) | = 0 ,

then, for any ε > 0 , there exists n 0 > 0 such that, for n n 0 , we have

sup q Z sup t [ q T T , q T + T ] | u n ( t ) | ε .

Therefore, for all t R and n n 0 , we have

| u n ( t ) | ε ,

which implies

lim n meas { t R : | u n ( t ) | > ε } = 0 ,

a contradiction. Therefore, there exist a constant ρ > 0 and a subsequence { n k } of { n } such that

sup x Z sup t [ x T T , x T + T ] | u n k ( t ) | > ρ , k N ,

where ℕ denotes the set of all positive integers. So, for k N , there exists x n k Z such that

sup t [ x n k T T , x n k T + T ] | u n k ( t ) | > ρ .

Let v n k ( t ) = u n k ( t + x n k T ) , t R . Since { u n } is bounded in H, by Lemma 2.3, it is easy to see that { v n k } is bounded in H 1 ( R ) . Therefore, { v n k } has a subsequence which weakly converges to u in H 1 ( R ) . Without loss of generality, we assume that v n k u in H 1 ( R ) . Thus, v n k u in H 1 ( [ T , T ] ) . Therefore, v n k uniformly converges to u in [ T , T ] . Noticing that

sup t [ T , T ] | v n k ( t ) | = sup t [ T , T ] | u n k ( t + x n k T ) | = sup t [ x n k T T , x n k T + T ] | u n k ( t ) | > ρ ,

we have

sup t [ T , T ] | u ( t ) | ρ ,

that is, u 0 . □

Proof of Theorem 1.1 By Lemma 2.7, there exists a bounded { q n } in H such that

φ ( q n ) d , φ ( q n ) 0 , dist ( q n , H ) 0 ,

and { q n } does not converge to 0 in measure on ℝ, where d is the mountain pass value. By Lemma 2.8, there exists a sequence { x n k } in ℤ such that

ω k : = q n k ( + x n k T ) ω 0 in  H 1 ( R ) .

For any fixed k N , set s = t + x n k T and h k ( s ) : = h ( s x n k T ) . Then s j : = t j + x n k T ( j Z ) are impulsive points and

h k H = ( R | h k ( s ) | 2 d s ) 1 2 = ( R | h ( s ) | 2 d s ) 1 2 = h H .

For any h C 0 ( R ) with h ( k T ) = 0 , we have

φ ( ω k ) , h = R ω k ( t ) h ( t ) d t R V ( t , ω k ( t ) ) h ( t ) d t + j = + I ( ω k ( t j ) ) h ( t j ) = R [ q n k ( t + x n k T ) h ( t ) V ( t , q n k ( t + x n k T ) ) h ( t ) ] d t + j = + I ( q n k ( t j + x n k T ) ) h ( t j ) = R [ q n k ( s ) h ( s x n k T ) V ( s x n k T , q n k ( s ) ) h ( s x n k T ) ] d s + j = + I ( q n k ( s j ) ) h ( s j x n k T ) = R [ q n k ( s ) h ( s x n k T ) V ( s , q n k ( s ) ) h ( s x n k T ) ] d s + j = + I ( q n k ( s j ) ) h ( s j x n k T ) = R [ q n k ( s ) h k ( s ) V ( s , q n k ( s ) ) h k ( s ) ] d s + j = + I ( q n k ( s j ) ) h k ( s j ) = φ ( q n k ) , h k .

Hence, we have

| φ ( ω k ) , h | = | φ ( q n k ) , h k | φ ( q n k ) h k H = φ ( q n k ) h H ,

which implies

φ ( ω k ) , h 0 as  k . (2.11)

Since H H 1 ( R ) , ω k ω in H, therefore

R ω k h R ω h . (2.12)

As ω k ω in H 1 ( R ) , { ω k } is bounded in H 1 ( R ) and hence | ω k | c for some c > 0 and all k N . Also, { ω k } uniformly converges to ω on supp ( h ) and, V being uniformly continuous on supp ( h ) × [ c , c ] , V ( t , ω k ) h uniformly converges to V ( t , ω ) h on supp ( h ) × [ c , c ] . By the Lebesgue dominated convergence theorem, this implies that

R V ( t , ω k ) h R V ( t , ω ) h . (2.13)

For any h H and ε > 0 , take J 0 sufficiently large such that

( j = J 0 + 1 + | h ( t j ) | 2 ) 1 2 ε , ( j = J 0 1 | h ( t j ) | 2 ) 1 2 ε .

Since ω k ω in H 1 ( R ) , ω k ω in H 1 ( [ t J 0 , t J 0 ] ) , therefore ω k uniformly converges to ω in [ t J 0 , t J 0 ] . By the continuity of I, there exists K > 0 such that, when k > K , we have

| j = J 0 J 0 [ I ( ω k ( t j ) ) I ( ω ( t j ) ) ] h ( t j ) | ε .

Since

| I ( ω k ( t j ) ) | b | ω k ( t j ) | , | I ( ω ( t j ) ) | b | ω ( t j ) | ,

it follows from Lemma 2.3 that

Similarly, we have

( j = J 0 1 [ I ( ω k ( t j ) ) I ( ω ( t j ) ) ] 2 ) 1 2 2 b T p max { sup k ω k H , ω H } .

By the Cauchy-Schwarz inequality, we have

Therefore,

lim k j = + I ( ω k ( t j ) ) h ( t j ) = j = + I ( ω ( t j ) ) h ( t j ) . (2.14)

From (2.11)-(2.14), we have

φ ( ω ) , h = lim k φ ( ω k ) , h = 0 .

Thus, φ ( ω ) = 0 and ω is a nontrivial weak homoclinic orbit of (1.1)-(1.2). □

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and read and approved the final manuscript.

### Acknowledgements

This research is supported by the National Natural Science Foundation of China (Grant No. 10971085). The authors would like to thank the anonymous reviewers for their valuable comments and suggestions to improve the manuscript.

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