## Boundary Value Problems

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# Solvability of a second-order Hamiltonian system with impulsive effects

Boundary Value Problems20132013:151

DOI: 10.1186/1687-2770-2013-151

Accepted: 30 May 2013

Published: 25 June 2013

## Abstract

In this paper, a class of second-order Hamiltonian systems with impulsive effects are considered. By using critical point theory, we obtain some existence theorems of solutions for the nonlinear impulsive problem. We extend and improve some recent results.

MSC:334B18, 34B37, 58E05.

### Keywords

Hamiltonian system impulsive critical point theory

## 1 Introduction and main results

Consider the second-order Hamiltonian systems with impulsive effects
(1.1)
where , , (; ) are continuous and satisfies the following assumption:
1. (A)
is measurable in t for every and continuously differentiable in x for a.e. and there exist , such that

for all and a.e. .

When (; ), (1.1) is the Hamiltonian system
(1.2)

In the past years, the existence of solutions for the second-order Hamiltonian systems (1.2) has been studied extensively via modern variational methods by many authors (see [113]).

When the gradient is bounded, that is, there exists such that
for all and a.e. , Mawhin-Willem in [1] proved the existence of solutions for problem (1.2) under the condition
or
When the gradient is bounded sublinearly, that is, there exist and such that
for all and a.e. , Tang [2] proved the existence of solutions for problem (1.2) under the condition
or

which generalizes Mawhin-Willem’s results.

For , problem (1.1) gives less results (see [1416]). In [14], Zhou and Li extended the results of [2] to impulsive problem (1.1); they proved the following theorems.

Theorem A [14]

Assume that (A) and the following conditions are satisfied:

(h1) There exist and such that

for all and a.e. .

(h2) .

(h3) For any ; ,

Then problem (1.1) has at least one weak solution.

Theorem B [14]

Suppose that (A) and the condition (h1) of Theorem  A hold. Assume that:

(h4) .

(h5) For any ; ,
(h6) There exist and such that

Then problem (1.1) has at least one weak solution.

Let
(1.3)

where is convex in (e.g., ), , satisfying (e.g., ), , and . It is easy to see that satisfies the condition (h2) but does not satisfy the condition (h1). The above example shows that it is valuable to further improve Theorem A.

Let
(1.4)

where satisfies that the gradient is Lipschitz continuous and monotone in (e.g., ), , satisfies (e.g., ), , and . It is easy to see that satisfies the condition (h4) but does not satisfy the condition (h1). The above example shows that it is valuable to further improve Theorem B.

In this paper, we further study the existence of solutions for impulsive problem (1.1). Our main results are the following theorems.

Theorem 1.1 Suppose that satisfies assumption (A) and the following conditions hold:

(H1) There exist and such that
(1.5)

for all and a.e. .

(H2) There exists a positive number such that
(1.6)

for all .

(H3)
(1.7)
(H4) For any ; ,
(1.8)

Then impulsive problem (1.1) has at least one weak solution.

Remark 1.1 Theorem 1.1 generalizes Theorem A, which is a special case of our Theorem 1.1 corresponding to .

Example 1.1 Let , . Consider the following impulsive problem:
where
Take
which is bounded and

, , . Then all the conditions of Theorem 1.1 are satisfied. According to Theorem 1.1, the above problem has at least one weak solution. However, F does not satisfy the condition (h1) in Theorem A. Therefore, our result improves and generalizes the Theorem A.

Theorem 1.2 Suppose that satisfies assumption (A) and the condition (H1) of Theorem  1.1. Furthermore, assume that

(H5) There exist , such that
(1.9)

for all .

(H6)
(1.10)
(H7) For any ; ,
(1.11)
(H8) There exist and such that
(1.12)

for every , ; .

Then impulsive problem (1.1) has at least one weak solution.

Remark 1.2 Theorem 1.2 generalizes Theorem B, which is a special case of our Theorem 1.2 corresponding to .

Example 1.2 Let , . Consider the following impulsive problem:
where
Take
which is bounded from above, and

, , , , (). Then all the conditions of Theorem 1.2 are satisfied. According to Theorem 1.2, the above problem has at least one weak solution. However, F does not satisfy the condition (h4) in Theorem B. Therefore, our result improves and generalizes Theorem B.

Theorem 1.3 Suppose that and satisfy the assumptions (A), (H1), (H2), (H7) and (H8). Furthermore, assume that

(H9)
(1.13)

uniformly for all .

Then impulsive problem (1.1) has at least one weak solution.

Example 1.3 Let , . Consider the following impulsive problem:
where
Take

Then all the conditions of Theorem 1.3 are satisfied. According to Theorem 1.3, the above problem has at least one weak solution. However, is neither superquadratic in X nor subquadratic in X.

## 2 Preliminaries

Let with the inner product
inducing the norm
The corresponding functional ϕ on given by
(2.1)
is continuously differentiable and weakly lower semi-continuous on . For the sake of convenience, we denote
where
and
For any , we have
(2.2)
Definition 2.1 We say that a function is a weak solution of (1.1) if the identity

holds for any .

It is well known that the solutions of impulsive problem (1.1) correspond to the critical point of ϕ.

Definition 2.2 [1]

Let X be a Banach space, and .
1. (1)

ϕ is said to satisfy the -condition on X if the existence of a sequence such that and as implies that c is a critical value of ϕ.

2. (2)

ϕ is said to satisfy the condition on X if any sequence for which is bounded and as possesses a convergent subsequence in X.

Remark 2.1 It is clear that the condition implies the -condition for each .

Lemma 2.1 [1]

If ϕ is weakly lower semi-continuous on a reflexive Banach space X (i.e., if , then ) and has a bounded minimizing sequence, then ϕ has a minimum on X.

Remark 2.2 The existence of a bounded minimizing sequence will be in particular ensured when ϕ is coercive, i.e., such that

Lemma 2.2 [1]

Let X be a Banach space and let . Assume that X splits into a direct sum of closed subspaces with and
where
Let
and

Then if ϕ satisfies the -condition, c is a critical value of ϕ.

Lemma 2.3 [1]

If the sequence converges weakly to u in , then converges uniformly to u on .

Lemma 2.4 [1]

If and , then
and
Lemma 2.5 There exists such that if , then
Moreover, if , then

where and .

## 3 Proof of main results

For , let and .

Proof of Theorem 1.1 It follows from (H1) and Sobolev’s inequality that
(3.1)
for all and some positive constants , and . By (H2) and Wirtinger’s inequality, we have
(3.2)

for all .

From (H4), we obtain
for all . Therefore we have
(3.3)
for all . As if and only if , (3.3) and (H3) imply that

By Lemma 2.1 and Remark 2.1, ϕ has a minimum point on , which is a critical point of ϕ. Therefore, we complete the proof of Theorem 1.1. □

Lemma 3.1 Assume that the conditions of Theorem  1.2 hold. Then ϕ satisfies the condition.

Proof Let be bounded and as . From (H1) and Lemma 2.4, we have
(3.4)
for all large n and some positive constants , and . It follows from (H5) and Lemma 2.4 that
(3.5)
By (3.4), (3.5), (H8) and Young’s inequality, we have
(3.6)
where , . From Wirtinger’s inequality, we obtain
(3.7)
The inequalities (3.6) and (3.7) imply that
(3.8)
for all large n and some positive constants and . It follows from (H5), Cauchy-Schwarz’s inequality and Wirtinger’s inequality that
(3.9)
Like in the proof of Theorem 1.1, we get
(3.10)
From (H7), we have
for all . Since is bounded, from (3.9) and (3.10), there exists a constant C such that
(3.11)
for all large n and some constant . By the above inequality and (H6), we know that is bounded. In fact, if not, without loss of generality, we may assume that as . Then, from (3.8) and the above inequality, we have
which contradicts (H6). Hence is bounded. Furthermore, is bounded from (3.7) and (3.8). Hence, there exists a subsequence of defined by such that
By Lemma 2.3, we have
On the other hand, we get
(3.12)
It follows from the above equality, (A) and the continuity of that

Thus, we conclude that ϕ satisfies the condition. □

Now, we give the proof of our Theorem 1.2.

Proof of Theorem 1.2 Let W be the subspace of given by
Then . Firstly, we show that
(3.13)
In fact, for , then , by the proof of Theorem 1.1, we have
(3.14)
Like in the proof of Lemma 3.1, we obtain
(3.15)
By (H8) and Lemma 2.5, we find
(3.16)
for all . By (3.14), (3.15) and (3.16), we have
(3.17)

for all .

By Lemma 2.4, we have on W. Hence (3.13) follows from (3.17).

On the other hand, by (H7), we get
(3.18)
for all . Therefore, from (3.18) and (H6), we obtain

as in . It follows from Lemma 2.2 and Lemma 3.1 that problem (1.1) has at least one weak solution. □

Proof of Theorem 1.3 First we prove that ϕ satisfies the condition. Suppose that is a sequence for ϕ, that is, as and is bounded. In a way similar to the proof of Theorem 1.1, we have
and
for all . Hence we have
From (3.7) and the above inequalities, we obtain
(3.19)

for some positive constants and .

By (H9) there exists such that
for all and , which implies that
for all and . It follows from assumption (A) that

for all and , where .

Let, then

for all and .

By the boundedness of , (H7) and (3.19), there exists a constant such that

which implies that is bounded. Like in the proof of Lemma 3.1, we know that ϕ satisfies the condition.

Furthermore, we can prove Theorem 1.3 using the same way as in the proof of Theorem 1.2. Here, we omit it. □

## Declarations

### Acknowledgements

We express our gratitude to the referees for their valuable criticism of the manuscript and for helpful suggestions. Supported by the National Natural Science Foundation of China (11271371, 10971229).

## Authors’ Affiliations

(1)
School of Mathematics and Statistics, Central South University

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