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Vol. 1, 2013-14, 66-75

ISSN: 2349-0632 (P), 2349-0640 (online) Published on 22 May 2014

www.researchmathsci.org

66

Journal of

Periodic solutions for a class of nonautomous second

order systems

Er Ye and Yabin Shao

College of Mathematics and Computer Science

Northwest University for Nationalities, Lanzhou – 730030, Lanzhou, China E-mail: yb-shao@163.com

Received 14 April 2014; accepted 20 may 2014

Abstract. In this paper, some multiplicity theorem is obtained for periodic solutions of nonautomous second order systems with partially periodic potential by the minimax methods.

Keywords: Nonautomous second order systems; periodic solutions; minimax methods AMS Mathematics Subject Classification (2010): 34A34, 34A12

1. Introduction and main results

Consider the second-order systems

{

() (, ()) 0 a.e.t [0, ], )

( ) ( ) 0 ( ) ( ) 0 (

T t

u t F t u

t e T u u T u u

∈ =

∇ −

= − = − ɺ ɺ

ɺ

ɺ (1) where T>0, and F [ ]T ×RNR

, 0

: satisfies the following assumption: )

, ( )

(H Ft x is measurable in t for each xRN and continuously differentiable in x for

[ ]T t e

a.. ∈0, and there exist aC(R+,R+), bL1(0,T;R+) such that )

( ) ( ) , ( ), ( ) ( ) ,

(t x a x bt F t x a x bt

F ≤ ∇ ≤

for all x×RN and a.e.t

[ ]

0,T .

The existence of periodic solutions for problem (1) has been studied extensively, a lot of existence and multiplicity results have been obtained, we refer the readers to [1-17] and the reference therein.

Suppose that F( xt, ) is Ti −periodic in

x

i, 1≤ir, that is

F

(

)

(

,

)

1

x

t

F

x

t,

r kTe

i i i i

=

+

= , (2) for alla.e.t

[ ]

0.T , x∈×RN and all integers

i

k

,1≤ir , where

{ }

ei (1≤ir) is the

canonical basis of N R .

With periodic potentials, that is, (2) holding with r=N, the existence and multiplicity theorems are obtained for the nonautonomous second-order system (1) in [1] and [2] respectively. When the nonlinearity is bounded, In [3-4], Chang and Liu have studied the nonautonomous second order system (1) with partially periodic (that is,(2)holding with1≤rN)a- nd partially uniformly coercive potentials (F( xt, )→+∞for every ( x,

r

r R

x

x2⋅⋅⋅, )∈ as(xr+1,xr+2⋅ ⋅⋅xN) tends to infinity inR r N).

In [5], Wu has obtained the following result.

Theorem A. Suppose that F satisfies conditions (H) and (2). Assume that there exist )

; , 0 (

1 +

L T R

(2)

) ( ) ,

(t x g t

F

∇ and

TF t xdt→+∞

0 (, )

,

for every

(

x

1

,

x

2

⋅⋅

,

x

r

)

R

r as

(

x

r+1

,

x

r+2

⋅⋅

,

x

N

)

tends to infinity in

R

Nr .Then problem (1) has at least

r

+

1

geometrically distinct solutions in

H

T

1

.

In 2003, Tang [6] generalized Theorem A and obtained the following results.

Theorem B. Assume that there exist , ∈ 1(0, ; +)

R T L g

f and 0≤α<1 such that ∇F(t,x)≤f(t)

x

+g(t)

α

(3)

for all N

R

x∈ and a.e.t

[ ]

0,T . Suppose that F satisfies conditions (H) and (2). and x

TF t xdt+∞

0 2

) , (

α , (4)

for every

(

x

1

,

x

2

...,

x

r

)

R

r as

(

x

r+1

,

x

r+2

...,

x

N

)

tends to infinity in RN+r .Then

problem (1) has at least r+1 geometrically distinct solutions in HT 1.

In 2011, Zhang and Tang [7] investigated the existence of periodic solutions for problem (1) and obtained the following results.

Theorem C. Assume that there exist f, 1(0, ; )

R T L

g∈ + and 0≤α<1 such that

) g( ) ( ) ,

(t x f t x t

F ≤ +

∇ α ,

for all N R

x∈ and a.e.t

[ ]

0,T . Suppose that F satisfies conditions (H) and (2).and

2

0 0

2

)

)

(

(

8

)

,

(

inf

lim

x

T

F

t

x

dt

T

T

f

t

dt

x

>

α − ∞

→ ,

for every (x1,x2⋅⋅⋅,xr)∈Rras

(

x

r+1

,

x

r+2

⋅⋅

,

x

N

)

tends to infinity in r N

R

−.Then problem (1) has at least r+1 geometrically distinct solutions in H 1T .

In this paper, we consider the following second-order systems:

{

ɺu(uɺ(t0))+Au(T(t))=+uɺ(0F)(tuɺ,u((Tt)))==00, a.e.t

[ ]

0,T , (5)

where T>0, A is antisymmetry constant matric with A

<

2Tπ . The Hilbert space 1

T

H is defined by HT1 =

{

u:

[ ]

0,T ֏RN| u is a absolutely continuous u(0)=u(T) and uɺ∈L2(0,T;RN)

}

and 1

T

H is endowed with the norm

(

)

2

1

0 0

2 2

+

= T T

dt u dt u

u ɺ ,

By sobolev embeded theorems, there exist C>0, for all u HT

1

∈ , such that u Cu

∞ . (6)

where u ∞=tmax[ ]0,T u(t) .Motivated by [1-7], The following main results are obtained by the minimax methods.

Theorem 1.1. Suppose that F satisfies conditions (H), and there exist

,

1

(

0

,

;

)

R

T

L

g

f

+

such that

F(t,x)f(t)x +g(t), (7)

(3)

68

   

 

  

  

=

=

T A

T dt

t f M

T

π π δ

2 1 2 , 2 1 min , )

( 2

2 0

1 .

δ

<

2 1 2M C , Assume that (2) holds and

=

T +∞ dt x t F x

0 2

) ,

( , (8) as x tends to infinity in ∈0×RNr.Then problem (5) has at least r+1geometrically distinct

solutions inHT

1.

Remark 1.1. System (5) generalizes system (1) obviously, for if A= 0, then system (5) becomes system (1). We consider the system (5) with partially periodic potentials and linear nonlinearity.

Theorem 1.1 is a new result, which completes the theorem B and theorem C with α=1 in (3). Theorem 1.1 generalized Theorem A , which is the special case of Theorem 1.1 corresponding to f(t)≡0.

Example 1.1. There are functions F satisfying our Theorem 1.1 and not satisfying the results in [1]-[7]. For example, let

2

1 2

1 sin sin

sin 1 ) ,

(

+ =

+ + ⋅⋅ ⋅ + + + +

= N

r j

j

r x

x x

x r

x t

F ,

where

{

}

N

N R

x x x

x= 1, 2,⋅ ⋅⋅, ∈ .

Theorem 1.2. Suppose that F satisfies conditions (H), (2),(7) and

T −∞ dt x t F

x2 0 ( , ) , (9) as x tends to infinity in ∈0×RNr.Then problem (5) has at least

r

+

1

geometrically

distinct solutions inHT1.

Example 1.2. There are functions F satisfying our Theorem 1.2 and not satisfying the results in [1]-[7]. For example, let

) sin

sin sin 1 ( ) , (

2

1 2

1

+ =

+ + ⋅⋅ ⋅ + + + + −

= N

r j

j

r x

x x

x r

x t

F ,

where

{

}

N

N R

x x x

x= 1, 2,⋅ ⋅⋅, ∈ . 2. Preliminaries

ForuHT1,let 1 0 ( )

T T

u = ∫ u t dt ,and ~u(t)=u(t)−u.Then we have Sobolev’s inequality u T Tut 2dt

0 2

) ( 12

~

ɺ , (10)

and Wirtinger’s inequality

Tu t dt T

Tut dt

0 2 2 2 2

0 () 4 ()

~ ɺ

π (11)

for alluHT1 (see Proposition 1.3 in [8]).

(4)

[

]

i r

i

i i i i

N

r i

i e Qu u e lT e

e u u

P

= +

=

− =

=

1 1

) , ( ,

) , (

and li(1≤i≤r) is the unique integer such that 0≤(u,ei)−liTi <Ti

Define the functional

ϕ

on HT1 by

dt t u t F dt t u t Au dt

t u

u =

T +

T

T

ϕ 0 0

2

0 ( ( ), ( )) ( , ( ))

2 1 ) ( 2 1 )

( ɺ ɺ

Then

ϕ

is continuously differentiable by Lemma 1 in [9] and the solutions of systems (5) correspond to the critical points of ϕ. Moreover, one has

dt t v t u t F dt

t v t Au dt

t v t u v

u =

T +

T

T

ϕ 0 0( (), ( )) 0( (, ( )), ( ))

2 1 )) ( ), ( ( ),

(

Let

1

| ,1

r

i i i i i

G k T e k Z i r

=

 

= ∈ ≤ ≤ 

be a discrete subgroup of H and letT1

π

:

1 1

T T

HH G be the canonical surjection. It is obvious that 1

T

H G = X×V ,

whereX =YZ, = ~1=

{

∈ 1| =0

}

u H u

Y HT T

Z

=

span

{

er+1,⋅⋅⋅,eN

}

,andV span e=

{

1, ,⋅⋅⋅er

}

G is

isomorphic to the torus r

T .Define ψ ;X×YR by

) ( )) (

u ϕ u

ψ = .

It follows from (2) that ψ is well-defined. Moreover,

ψ

is continuously differentiable.

3. Proof of theorem

Now we begin to prove our main result, for the sake of convenience, c denote some constant.

Proof of Theorem 1.1. the proof relies on the generalized saddle point theorem due to Liu [4]. Assume that π((un))→0 is a (PS) sequence for

ψ

,that is, ψ(π(un)) is bounded and ψ′(π(un))→0.Then

ϕ

(un) is bounded and

φ

′(un)→0 .

Let

{ }

1

T

n H

u ∈ be such that

ϕ

(

u

n

)

c

,

ϕ′

(

u

n

)

0

,

(

n

)

(12) First,we shall prove that

{ }

un is bounded in HT1.By contradiction, we assume

∞ →

n

u as n→∞. For

{ }

1

T

n H

u ∈ ,let utdt T u=

T

0 ()

1 ,and u~(t)=u(t)u.By (12), for 1

T H

∈ φ

∀ ,we have

) ( 0 ) ), , ( ( )

, ( )

,

( 0 0

0 φ +

φ −

∇ φ = φ

u dt Au dt T F t un dt

T n T

n

ɺ ɺ

ɺ .

Taking φ=u~n,we have

0Tuɺn2dt+

0T(Aun,uɺn)dt

0T(∇F(t,un),u~n)dtu~n . (13) By the Wirtinger inequality (11) and 2

T

A≤ π, we have

dt

u

Au

dt

u

T n n

T

n

+

0

2

0

(

,

)

(5)

70 dt u A

T T 2

0 ) 2 1 (

π − ≥ ɺ dt u A T T dt u A T T n T n 2 0 2 2 2 0 ~ ) 2 1 ( 4 2 1 ) 2 1 ( 2 1

π − π + π − ≥ ɺ 2 2 2 2 ~ ~ ) 2 1 ( 2 , 2 1

min A un un

T

T  − π =δ

 

 π

≥ (14)

where ) 2 1 ( 2 , 2 1 min 2 2 A T T  − π

    π = δ .

By (7), (11) and the Hoɺɺlder inequality, one has

dt u u t F T n n

0 ∇ )

~ ), , ( (

dt

u

t

g

dt

u

u

t

f

n T n

T n

~

)

(

~

)

(

0 0

+

∞ ∞ ∞

+ ≤ n T n T u dt t g u u dt t

f( ) ~ 0 ( ) ~

0

+

T n n n T

u

dt

t

g

C

u

u

dt

t

f

C

0 0 2

~

)

(

~

)

(

n T n

n

u

C

g

t

dt

u

u

C

M

~

(

)

~

0 2

1

+

=

(15) where =∫T

dt t f

M1 0 () .

By (13),(14) and (15), we have

dt u u t F dt u Au dt u

un

T n +

T n n

Tn n

0 0 2 0 ) ~ ), , ( ( ) , ( ɺ ɺ n T o n n

n M C u u C g t dtu

u~ 2 ~ ( ) ~

1 2

− − δ ≥ , Therefore

u MC u C

n

n ≤ δ +

2 1

~ (16)

It follows from (15) and (16) that

T

n

n u dt

u t F 0 ) ~ ), , ( ( n T n

n u C g t dt u

u C

M ~ () ~

0 2

1 +

C u C u C M n

n + +

δ

12 4 2 , (17)

By (16), 2<δ 1

2MC and the boundedness of Qun , we have

2 1 ~ 2 1 < δ ≤ →m MC u

u

n

(6)

δ − ≥ − → − −

≥ 1 2

2 1 1 ~

1 m MC

u u Q u u u u P n n n n n

n , (n→∞) (18)

By (18), for ∀ε>0,such that 0 2

1 2

1 −ε>

δ

MC ,we have

n

n

P

u

u

C M ε − δ −

2 1 2 1

1

. (19)

It follows from (13),(16), (17) and (19) that

dt

u

Au

dt

u

T n n

T

n

+

0

2

0

(

,

)

ɺ

ɺ

n T

n

n

u

dt

u

u

t

F

(

,

),

~

)

~

(

0

+

M C un +cun +c

δ ≤ 2 4 2 1 c u P c u P C M n n C

M + +

δ ≤ ε − δ − 2 4 2 1 2 1 2 1

1 (20)

By (16) and (19), we have

n n

n Pu

C M c u C M u C M ε − δ − δ ≤ + δ ≤ 2 1 2 1 1

~ 1 2

2

1 . (21)

By (6),(7),(21), the Hoɺɺlder inequality and the boundedness of Qun ,we have

T

T

n

n dt F t Pu dt

u t F

0 (,ˆ ) 0 ( , )

∫ ∫

+

+

T

n n

n

s

Q

u

u

u

P

t

F

0 1 0

))

~

(

,

(

dt u u Q u u Q u P t

f n n

T

n n

n ~ ) ~

)( (

0 + + +

g t Qun undt

T ~ ) ( 0 + +

∞ ∞ ∞+ + +

n n n n n

T u u Q u u Q u P dt t

f( ) ( ~ ) ~

0 +

+ ∞

T n n u u Q dt t g 0 ~ ) ( n n n n n T u u Q u u Q u P dt t f

C () ( ~ ) ~

0

2 + + +

n n T u u Q dt t g

C ( ) ~

0 +

2 2

1 2

1C Pun Qun ~un MC Qun u~n

M + + +

= n n

T u u Q dt t g

C () ~

0 + +

c u P c u P C M C M n

n + +

− −

12 4 2

2 1 2 1 1 ε δ δ 2 2 1 2 2 1 2 4 2 1 ) ( 1 2

1C n

M M C Pu

C M

ε

δ

δ − + 2 4 2 1 2 2 1 2 1 2 1 ) 2 1 ( 2 1 1 n u P C M C M C M C M           − − − − − = ε δ δ ε δ

(7)

72

c u P c u P C

M

n n

C

M + +

δ =

ε − δ −

ε

2

4 2 1

2 2 1

) 2

1 (

2 1

(22)

By (20) and (22), we have

ϕ

(

u

n

)

=

ϕ

(

u

ˆ

n

)

Pu cPu c

C M C

M

n

n + +

− −

12 4 2

2 2 1

) 2

1 (

1

ε δ δ

[

]

− −

T

n T

n

n F t Pu dt F t Pu dt

u t F

0

0 (,ˆ ) ( , ) (, )

t u P C

M C

M

n 2 4

2 1

2 2 1 )

2 1 (

1 ε δ

δ

2

2 2

1 2 1 4 2 1

)

( M1C2 n

u P C

M

ε

ε

δ

δ

− +

dt u P t F c u P

c n + −

T n

+

0 (, )

  

 

  

 

− − −

=

T

n n

dt u P t F u P C

M C

M

0 2 4

2

1 1 (, )

) 2

1 (

1

2 2

1 ε

δ ε

δ

Pun 2+c Pun +c (23)

Since u n → ∞ as n→∞, by (19), we have

n

u

P as n→∞.

Let

ε

small enough, by (8), we get that φ(un)→−∞ as n→∞,this contradicts (12) and hence

{ }

un is bounded in H1

T .Arguing then as in Proposition 4.1 in [9], we conclude that the (PS) condition is satisfied.

Now we check the link condition that (a) inf

{

ψ(π(u))|π(u)∈Y×X

}

>−∞.

(b)

ψ

(

π

(

x

))

−∞

uniformly for π(Qx) as Px→∞,where xRN.

For π(u)∈Y×V,

u

=

u

~

+

Q

u

. In a similar way to (22), we can get that

T

+

T

dt

t

F

dt

u

Q

t

u

t

F

0

(

)

)

0

(

,

0

)

~

,

(

c

u

C

T

M

u

C

M

+

+

~

2

~

1

2 2 2

1 , (24)

It follows from (24), A <2Tπ

and the boundedness of Qun that

)

~

(

))

~

(

(

))

(

(

π

u

=

ψ

π

u

+

Q

u

=

ϕ

u

+

Q

u

ψ

Tu dt T(Au(t),u(t))dt

2 1 2

1

0 2 0

ɺ

ɺ

+

=

−



+

T T T

dt t F dt

t F dt u Q t u t F

0

0 ( ) ) 0 ( ,0) ( ,0)

(8)

2 2 1 2 2

2

~ ~

) 2 1 ( 2 , 2 1 min 2 1

u C M u A T

T  − −

   ≥

π

π +

T

dt t F c u C T M

0

2 ~ (,0)

2 1

By 2<δ 1

2MC , we have ψ(π(x))→+∞, as u →∞, for all π(u)∈Y×V,which

implies (a).

It follows from (7) and the boundedness of Q u n that

)

ˆ

(

)

(

))

(

(

π

x

=

ϕ

x

=

ϕ

x

ψ

=−

0TF(t,Px+Qx)dt

∫ ∫

+

=

T T

dt

Px

t

F

dsdt

Qx

sQx

Px

t

F

0

1

0

(

(

,

),

)

0

(

,

)

dt

Qx

M

dt

Qx

Qx

Px

M

dt

Px

t

F

T T

T

+

+

+

0

(

,

)

0 1 0 2

c Px c dt Px t F

T

+ + −

0 ( , )

   

   

+ + −

=

T

Px c Px

c dt Px t F Px

Px 2 0 2

2

) , (

1 ,

for all N

R

x∈ .Hence we have, by (8),(b) is satisfied. It follows from the generalized saddle point theorem (Theorem 1.7 in [4]) that has at least

r

+

1

critical points .Hence ϕ has at least r+1 geometrically distinct critical points. Therefore, problem (5) has at least r+1 geometrically distinct solutions in 1

T

H .

Proof of Theorem 1.2 the proof relies on Theorem 4.12 in [8]. In a similar way to (22), we can get that

T

T

dt u P t F dt u t F

0 (,ˆ) 0 (, )

M C Pu u~ +c Pu +M C2 u~ 2 +cu~ +c

1 2

1 (25)

It follows from (14) and (25) that ϕ(u)=ϕ(uˆ)

δ

[

T

T

]

T

dt u P t F dt u P t F dt u t F

u~2 0 ( ,ˆ) 0 ( , ) 0 ( , ) 2

MC )u~ MC Pu u~ 2

( 1 2

2 2

1 −

≥ δ +cPu +cu +c

TF t Pu

0 ( , )

~

2 2

1 2

1

) 2 ( 2 ~ ) 2 (

2

1 

 

 

  

 

− − −

= Pu

C M C M u C

M δ

δ +c Pu +cu~ +c

2

0 2 2

1 2

4 2

1 1 (, )

) (

4 MC Pu F t Pu dt Pu

C

M T

   

   

+ −

δ

(9)

74

c u c u P c u P C M u

C M

C M

+ + +

 

 

 

 

− −

δ =

δ ~

) (

2 ~ ) 2

(

2 2

1 2

1

2 1

2

2

0 2 4

2

1 1 ( , )

) (

4 2

1

2

u P dt u P t F u P C

M T

C

M

 

 

 

+

δ (26)

By 2<δ 1

2MC and (9), we get that

ϕ

is bounded from below. Moreover, the functional

ϕ

satisfies the

G

PS)

( condition; that is, for every sequence

(

u

n

)

in HT1 such that ϕ(un)→0 is bounded and ϕ(un)→0,the sequence

π

(

u

n

)

has a convergent subsequence (see Definition 4.2 in [8]). In fact, the boundedness of

ϕ

(

u

n

)

, (9) and (26) imply that (u~n) and (Pu are bounded. Hence n) (uˆn) bounded. As in the proof of

Proposition 4.1 in [8], (ˆ )

n

u has a convergent subsequence, so we have π(un)=π(uˆn). Now the Theorem in this case follows from Theorem 4.12 in [8].

Acknowledgments

This work is supported by the National Natural Scientific Fund of China(11161041).

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