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New potential condition on homoclinic orbits for a class of discrete Hamiltonian systems

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Abstract

In the present paper, we establish an existence criterion to guarantee that the second-order self-adjoint discrete Hamiltonian system [p(n)u(n1)]L(n)u(n)+W(n,u(n))=0 has a nontrivial homoclinic solution, which does not need periodicity and coercivity conditions on L(n).

MSC:39A11, 58E05, 70H05.

1 Introduction

Consider the second-order self-adjoint discrete Hamiltonian system

[ p ( n ) u ( n 1 ) ] L(n)u(n)+W ( n , u ( n ) ) =0,
(1.1)

where nZ, u R N , u(n)=u(n+1)u(n) is the forward difference, p,L:Z R N × N and W:Z× R N R.

Discrete Hamiltonian systems can be applied in many areas, such as physics, chemistry, and so on. For more discussions on discrete Hamiltonian systems, we refer the reader to [1, 2].

As usual, we say that a solution u(n) of system (1.1) is homoclinic (to 0) if u(n)0 as n±. In addition, if u(n)0 then u(n) is called a nontrivial homoclinic solution.

The existence and multiplicity of homoclinic solutions of system (1.1) or its special forms have been investigated by many authors. Papers [38] deal with the periodic case where p, L and W are periodic in n or independent of n. In contrast, if the periodicity is lost, because of lack of compactness of the Sobolev embedding, up to our knowledge, all existence results require a coercivity condition on L:

lim | n | [ inf x R N , | x | = 1 ( L ( n ) x , x ) ] =.
(1.2)

For example, see [914]. In the above mentioned papers, except [14], L was always required to be positive definite.

In this paper, we derive an existence result which does not need periodicity and coercivity conditions on L(n). To state our results precisely, we make the following assumptions.

  1. (P)

    p(n) is N×N real symmetric positive definite matrix for all nZ.

  2. (L)

    L(n) is N×N real symmetric nonnegative definite matrix for all nZ, and there exist a positive integer N 0 Z and β>0 such that

    min x R N , | x | = 1 ( L ( n ) x , x ) β,|n| N 0 ,

where here and in the sequel, (,) denotes the standard inner product in R N and || is the induced norm.

  • (W1) W(n,x) is continuously differentiable in x for every nZ, W(n,0)=0, W(n,x)0 for all (n,x)Z× R N .

  • (W2) lim | x | 0 W ( n , x ) | x | =0 uniformly for all nZ.

  • (W3) lim | x | | W ( n , x ) | | x | 2 = uniformly for all nZ.

  • (W4) W ˜ (n,x):= 1 2 (W(n,x),x)W(n,x)0, (n,x)Z× R N , and there exist ε(0,1), c 0 >0, and R 0 >0 such that

    ( W ( n , x ) , x ) β ( 1 ε ) 2 | x | 2 ,(n,x)Z× R N ,|x| R 0

    and

    ( W ( n , x ) , x ) c 0 | x | 2 W ˜ (n,x),(n,x)Z× R N ,|x| R 0 .

Now, we are ready to state the main result of this paper.

Theorem 1.1 Assume that p, L and W satisfy (P), (L), (W1), (W2), (W3), and (W4). If there exist n 0 Z and x 0 R N such that

β2 c 0 sup s 0 [ s 2 2 ( ( p ( n 0 ) + p ( n 0 + 1 ) + L ( n 0 ) ) x 0 , x 0 ) W ( n 0 , s x 0 ) ] ,
(1.3)

then system (1.1) possesses a nontrivial homoclinic solution.

In Theorem 1.1, we replace (L) and (W4) by the following assumptions:

(L′) L(n) is N×N real symmetric nonnegative definite matrix for all nZ, and it satisfies (1.2).

(W4′) W ˜ (n,x):= 1 2 (W(n,x),x)W(n,x)0, (n,x)Z× R N , and there exist c 0 >0 and R 0 >0 such that

( W ( n , x ) , x ) c 0 | x | 2 W ˜ (n,x),(n,x)Z× R N ,|x| R 0 .

Then we have the following corollary immediately.

Corollary 1.2 Assume that p, L and W satisfy (P), (L′), (W1), (W2), (W3) and (W4′). Then system (1.1) possesses a nontrivial homoclinic solution.

Remark 1.3 If W(n,x) satisfies the well-known global Ambrosetti-Rabinowitz superquadratic condition:

(AR) there exists μ>2 such that

0<μW(n,x) ( W ( n , x ) , x ) ,(n,x)Z× R N {0},

then there exists a constant C 0 >0 such that

W(n,x) C 0 | x | μ ,(n,x)Z× R N ,|x|1;

moreover W ˜ (n,x)>0 for all (n,x)Z×( R N {0}), and

( W ( n , x ) , x ) 2 μ μ 2 | x | 2 W ˜ (n,x),(n,x)Z× R N ,|x|1.

In addition, by virtue of (W2), there exists β 1 >0 such that

( W ( n , x ) , x ) β 1 2 | x | 2 ,(n,x)Z× R N ,|x|1.

These show that (W3) and (W4) hold with R 0 =1, c 0 =2μ/(μ2) and β> β 1 . Let p(n)= I N and L(n)=λ n 2 /(1+ n 2 ) I N and choose n 0 =0 and x 0 =(1,0,,0) R N . In view of Theorem 1.1, if

λ>max { 4 μ μ 2 sup s 0 [ s 2 W ( 0 , s x 0 ) ] , β 1 } ,

then system (1.1) possesses a nontrivial homoclinic solution.

Example 1.4 Let p(n)= I N , L(n)=[1+λ n 2 /(1+ n 2 )] I N and

W(n,x)= | x | 2 ln ( 1 + | x | 2 ) .
(1.4)

Then

( W ( n , x ) , x ) =2 | x | 2 ln ( 1 + | x | 2 ) + 2 | x | 4 1 + | x | 2 .

It is easy to see that W ˜ (n,x)0 for all (n,x)Z× R N , and

( W ( n , x ) , x ) ( 2 ln 2 + 1 ) | x | 2 , ( n , x ) Z × R N , | x | 1 , ( W ( n , x ) , x ) 6 | x | 2 W ˜ ( n , x ) , ( n , x ) Z × R N , | x | 1 .

These show that (W3) and (W4) hold with R 0 =1, c 0 =6 and β>2(2ln2+1). We choose n 0 =0 and x 0 =(1,0,,0) R N . Then

sup s 0 [ s 2 2 ( ( p ( n 0 ) + p ( n 0 + 1 ) + L ( n 0 ) ) x 0 , x 0 ) W ( n 0 , s x 0 ) ] = sup s 0 [ 3 s 2 2 s 2 ln ( 1 + s 2 ) ] < 6 ln 2 .

In view of Theorem 1.1, if λ12(6ln2), then system (1.1) possesses a nontrivial homoclinic solution.

2 Preliminaries

Throughout this section, we always assume that p and L satisfy (P) and (L). Let

S = { { u ( n ) } n Z : u ( n ) R N , n Z } , E = { u S : n Z [ ( p ( n + 1 ) u ( n ) , u ( n ) ) + ( L ( n ) u ( n ) , u ( n ) ) ] < + } ,

and for u,vE, let

u,v= n Z [ ( p ( n + 1 ) u ( n ) , v ( n ) ) + ( L ( n ) u ( n ) , v ( n ) ) ] .

Then E is a Hilbert space with the above inner product, and the corresponding norm is

u= { n Z [ ( p ( n + 1 ) u ( n ) , u ( n ) ) + ( L ( n ) u ( n ) , u ( n ) ) ] } 1 / 2 ,uE.

As usual, for 1s<+, set

l s ( Z , R N ) = { u S : n Z | u ( n ) | s < + }

and

l ( Z , R N ) = { u S : sup n Z | u ( n ) | < + } ,

and their norms are defined by

u s = ( n Z | u ( n ) | s ) 1 / s , u l s ( Z , R N ) ; u = sup n Z | u ( n ) | , u l ( Z , R N ) ,

respectively.

Lemma 2.1 Suppose that (L) is satisfied. Then

u 1 β u+ | s | N 0 1 | u ( s ) | ,uE,
(2.1)

and

u max { 2 β , 2 N 0 α } u,uE,
(2.2)

where α= min | n | N 0 , | x | = 1 (p(n)x,x).

Proof Since uE, it follows that lim | n | |u(n)|=0. Hence, there exists n Z such that u =|u( n )|. There are two possible cases.

Case (i). | n | N 0 . According to (L), one has

u 2 = | u ( n ) | 2 1 β | s | N 0 ( L ( s ) u ( s ) , u ( s ) ) 1 β u.

Case (ii). | n |< N 0 . Without loss of generality, we can assume that n 0, then

u | u ( N 0 ) | + s = n N 0 1 | u ( s ) | [ 1 β | s | N 0 ( L ( s ) u ( s ) , u ( s ) ) ] 1 / 2 + N 0 α ( s = n N 0 1 α | u ( s ) | 2 ) 1 / 2 2 [ 1 β | s | N 0 ( L ( s ) u ( s ) , u ( s ) ) + N 0 α s = n N 0 1 ( p ( s + 1 ) u ( s ) , u ( s ) ) ] 1 / 2 max { 2 β , 2 N 0 α } u .
(2.3)

Cases (i) and (ii) imply that (2.1) and (2.2) hold. □

Now we define a functional Φ on E by

Φ(u)= 1 2 n Z [ ( p ( n + 1 ) u ( n ) , u ( n ) ) + ( L ( n ) u ( n ) , u ( n ) ) ] n Z W ( n , u ( n ) ) .
(2.4)

For any uE, there exists an n 1 N such that |u(n)|1 for |n| n 1 . Hence, under assumptions (P), (L), (W1), and (W2), the functional Φ is of class C 1 (E,R). Moreover,

Φ(u)= 1 2 u 2 n Z W(n,u),uE
(2.5)

and

Φ ( u ) , v =u,v n Z ( W ( n , u ) , v ) ,u,vE.
(2.6)

Furthermore, the critical points of Φ in E are solutions of system (1.1) with u(±)=0, see [5, 6].

Let e= { e ( n ) } n Z E with e( n 0 )= x 0 and e(n)=0 R N for n n 0 .

Lemma 2.2 Suppose that (L), (W1) and (W2) are satisfied. Then

sup { Φ ( s e ) : s 0 } sup s 0 [ s 2 2 ( ( p ( n 0 ) + p ( n 0 + 1 ) + L ( n 0 ) ) x 0 , x 0 ) W ( n 0 , s x 0 ) ] .
(2.7)

Proof From (2.4) and the definition of e, we get

Φ ( s e ) = s 2 2 n Z [ ( p ( n + 1 ) e ( n ) , e ( n ) ) + ( L ( n ) e ( n ) , e ( n ) ) ] n Z W ( n , s e ( n ) ) = s 2 2 [ ( p ( n 0 + 1 ) e ( n 0 ) , e ( n 0 ) ) + ( p ( n 0 ) e ( n 0 1 ) , e ( n 0 1 ) ) + ( L ( n 0 ) e ( n 0 ) , e ( n 0 ) ) ] W ( n 0 , s e ( n 0 ) ) = s 2 2 ( ( p ( n 0 ) + p ( n 0 + 1 ) + L ( n 0 ) ) x 0 , x 0 ) W ( n 0 , s x 0 ) .
(2.8)

Now the conclusion of Lemma 2.1 follows by (2.8). □

Applying the mountain-pass lemma without the (PS) condition, by standard arguments, we can prove the following lemma.

Lemma 2.3 Let W(n,x)0, (n,x)Z× R N . Suppose that (P), (L), (W1), (W2) and (W3) are satisfied. Then there exist a constant c(0, sup s 0 Φ(se)] and a sequence { u k }E satisfying

Φ( u k )c, Φ ( u k ) ( 1 + u k ) 0.
(2.9)

Lemma 2.4 Suppose that (P), (L), (W1), (W2), (W3), and (W4) are satisfied. Then any sequence { u k }E satisfying

Φ( u k )c>0, Φ ( u k ) , u k 0
(2.10)

is bounded in E.

Proof To prove the boundedness of { u k }, arguing by contradiction, suppose that u k . Let v k = u k / u k . Then v k =1. By virtue of (2.5), (2.6), and (2.10), we have

Φ( u k ) 1 2 Φ ( u k ) , u k = n Z W ˜ (n, u k )=c+o(1).
(2.11)

If δ:= lim sup k v k =0, then it follows from (L), (W4) and (2.11) that

| u k | < R 0 | ( W ( n , u k ) , u k ) | β 2 | u k | < R 0 | u k | 2 β 2 | s | N 0 | u k ( s ) | 2 + β 2 | s | < N 0 | u k ( s ) | 2 1 2 u k 2 + N 0 β u k 2 v k 2 [ 1 2 + o ( 1 ) ] u k 2
(2.12)

and

| u k | R 0 | ( W ( n , u k ) , u k ) | u k 2 c 0 | u k | R 0 | v k | 2 W ˜ ( n , u k ) c 0 v k 2 | u k | R 0 W ˜ ( n , u k ) c 0 ( c + 1 ) v k 2 0 , k .
(2.13)

Combining (2.12) with (2.13) and using (2.5) and (2.10), we have

1 + o ( 1 ) u k 2 Φ ( u k ) , u k u k 2 n Z | ( W ( n , u k ) , u k ) | u k 2 = | u k | < R 0 | ( W ( n , u k ) , u k ) | u k 2 + | u k | R 0 | ( W ( n , u k ) , u k ) | u k 2 1 2 + o ( 1 ) .
(2.14)

This contradiction shows that δ>0.

Going if necessary to a subsequence, we may assume the existence of n k Z such that

| v k ( n k ) | = v k > δ 2 .

Let w k (n)= v k (n+ n k ), then

| w k ( 0 ) | > δ 2 ,kN.
(2.15)

Now we define u ˜ k (n)= u k (n+ n k ). Then u ˜ k (n)/ u k = w k (n) and w k 2 = v k 2 . Passing to a subsequence, we have w k w in l 2 (Z, R N ), then w k (n)w(n) for all nZ. Clearly, (2.15) implies that w(0)0.

It is obvious that w(n)0 implies lim k | u ˜ k (n)|=. Hence, it follows from (2.5), (2.10), and (W3) that

0 = lim k c + o ( 1 ) u k 2 = lim k Φ ( u k ) u k 2 = lim k [ 1 2 n Z W ( n , u k ) | u k | 2 | v k | 2 ] = lim k [ 1 2 n Z W ( n + k n , u ˜ k ) | u ˜ k | 2 | w k | 2 ] 1 2 lim inf k n Z W ( n + k n , u ˜ k ) | u ˜ k | 2 | w k | 2 = ,

which is a contradiction. Thus { u k } is bounded in E. □

3 Proof of theorem

Proof of Theorem 1.1 Applying Lemmas 2.3 and 2.4, we deduce that there exists a bounded sequence { u k }E satisfying (2.9). By Lemma 2.2 and (1.3), one has

c β 2 c 0 .
(3.1)

Going if necessary to a subsequence, we can assume that u k u ¯ in E and Φ ( u k )0. Next, we prove that u ¯ 0.

Arguing by contradiction, suppose that u ¯ =0, i.e. u k 0 in E, and so u k (n)0 for every nZ. Hence,

u k 2 2 = | n | N 0 | u k ( n ) | 2 + | n | < N 0 | u k ( n ) | 2 1 β u k 2 +o(1).
(3.2)

According to (W4) and (3.2), one gets

| u k | < R 0 ( W ( n , u k ) , u k ) β ( 1 ε ) 2 | u k | < R 0 | u k | 2 1 ε 2 u k 2 +o(1).
(3.3)

By virtue of (2.5), (2.6), and (2.9), we have

Φ( u k ) 1 2 Φ ( u k ) , u k = n Z W ˜ (n, u k )=c+o(1).
(3.4)

Using (W4), (2.1), (3.1), (3.2), and (3.4), we obtain

| u k | R 0 ( W ( n , u k ) , u k ) c 0 | u k | R 0 | u k | 2 W ˜ ( n , u k ) c 0 u k 2 | u k | R 0 W ˜ ( n , u k ) c 0 c u k 2 + o ( 1 ) c 0 c ( 1 β u k + | s | N 0 1 | u k ( s ) | ) 2 + o ( 1 ) = c 0 c β u k 2 + o ( 1 ) 1 2 u k 2 + o ( 1 ) ,
(3.5)

which, together with (2.6), (2.9), and (3.3), yields

o ( 1 ) = Φ ( u k ) , u k = u k 2 n Z ( W ( n , u k ) , u k ) ε 2 u k 2 + o ( 1 ) ,
(3.6)

resulting in the fact that u k 0. Consequently, it follows from (W1), (2.5), and (2.9) that

0<c= lim k Φ( u k )=Φ(0)=0.

This contradiction shows u ¯ 0. By standard arguments, we easily prove that u ¯ is a nontrivial solution of (1.1). □

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Acknowledgements

This work is partially supported by Scientific Research Fund of Hunan Provincial Education Department (13A093, 07A066) and supported by Hunan Provincial Natural Science Foundation of China (No. 14JJ2133).

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Correspondence to Xiaoping Wang.

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Keywords

  • homoclinic solution
  • discrete Hamiltonian system
  • critical point