Periodic solutions for a kind of second-order neutral differential systems with deviating arguments

Periodic solutions for a kind of second-order neutral differential systems with deviating arguments

Applied Mathematics and Computation 156 (2004) 719–732 www.elsevier.com/locate/amc Periodic solutions for a kind of second-order neutral differential ...

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Applied Mathematics and Computation 156 (2004) 719–732 www.elsevier.com/locate/amc

Periodic solutions for a kind of second-order neutral differential systems with deviating arguments Lu Shiping

a,*

, Ge Weigao

b

a

b

Department of Mathematics, Anhui Normal University, Wuhu 241000, Anhui, PR China Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081, PR China

Abstract By means of the generalized continuation theorem of coincidence degree theory, a kind of second-order neutral differential systems with deviating arguments as follows d2 d ðxðtÞ þ Cxðt  rÞÞ þ grad F ðxðtÞÞ þ grad Gðxðt  sðtÞÞÞ ¼ pðtÞ dt dt2 is studied. A new result on the existence of periodic solutions is obtained, which relates to the deviating arguments sðtÞ and r. Ó 2003 Elsevier Inc. All rights reserved. Keywords: Periodic solution; Generalized continuation theorem; Deviating argument; Neutral differential system

1. Introduction We discuss the existence of periodic solutions to a kind of second-order neutral differential systems with deviating arguments in the following form: d2 d ðxðtÞ þ Cxðt  rÞÞ þ grad F ðxðtÞÞ þ grad Gðxðt  sðtÞÞÞ ¼ pðtÞ; dt dt2

*

Corresponding author. E-mail addresses: [email protected], [email protected] (L. Shiping).

0096-3003/$ - see front matter Ó 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.06.018

ð1Þ

720

L. Shiping, G. Weigao / Appl. Math. Comput. 156 (2004) 719–732

where F 2 C 2 ðRn ; RÞ, G 2 C 1 ðRn ; RÞ, p 2 C 0 ðR; Rn Þ, s 2 C 0 ðR; RÞ and pðt þ T Þ  2 pðtÞ, sðt þ T Þ  sðtÞ for a constant T > 0, r 2 R is a constant, C 2 Rn is a symmetric matrix. In recent years, by using the continuation theorem of coincidence degree theory, the existence of periodic solutions to ordinary system(or equation)were extensively studied, see [1–8] and their reference lists. But the case of differential system(or equation) with delay were studied far less often, see [9–13]. One of significant reasons is that some methods to estimate a priori bound of periodic solutions in the case of ordinary differential equation cannot be adapted directly to the case of differential equation with delay. For example,R if xðtÞ is a differentiable continuous periodic function with period of T T , then 0 x0T ðtÞgrad GðxðtÞÞ dt ¼ 0, which is very important to estimate a priori bound of periodic R T solutions. But in the case of differential equation with delay, the formula 0 x0T ðtÞgrad Gðxðt  sðtÞÞÞ dt ¼ 0 no longer holds generally. In paper [9–13], sever types of second-order scalar differential equation with delay such as x00 ðtÞ þ ax0 ðtÞ þ bxðtÞ þ gðxðt  1ÞÞ ¼ pðtÞ; x00 ðtÞ þ gðxðt  sÞÞ ¼ pðtÞ; x00 ðtÞ þ m2 xðtÞ þ gðxðt  sÞÞ ¼ pðtÞ; x00 ðtÞ ¼ f ðt; xðtÞ; xðt  s0 ðtÞÞx0 ðtÞ þ bðtÞgðxðt  s1 ðtÞÞÞ ¼ pðtÞ and x00 ðtÞ þ f ðxðtÞÞx0 ðtÞ þ

n X

bj ðtÞgðxðt  cj ðtÞÞÞ ¼ pðtÞ

j¼1

were studied. But the study of corresponding problem for second-order neutral differential system with deviating arguments, as far as we know, appeared rarely. In this paper, we investigate the existence of periodic solutions of system (1). By use of the generalized continuation theorem of coincidence degree theory, we obtain a new result to guarantee the existence of periodic solutions. The interesting is that our result has a bearing on the deviating argument sðtÞ and r, which is useful in the theory and application of control system [14]. 2. Main lemmas Let In ¼ f1; 2; . . . ; ng, X , Y are two Banach spaces. Considering an abstract equation as follows: Lx ¼ kN ðk; xÞ; where L : Dom L  X ! Y is a linear operator, N ðk; xÞ : ½0; 1  X ! Y is a nonlinear operator and k 2 ½0; 1 is a parameter.

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721

Lemma 1 [8,15]. Let X , Y be two Banach spaces and L a Fredholm operator with index zero. Assume that N ðk; xÞ : ½0; 1  X ! Y is L-compact on ½0; 1  X with X open bounded in X . Furthermore, suppose (1) For each k 2 ð0; 1Þ, x 2 oX \ Dom L, Lx 6¼ kN ðk; xÞ. (2) For each x 2 oX \ ker L, QN ð0; xÞ 6¼ 0. (3) deg fQN ð0; Þ; X \ ker L; 0g 6¼ 0. Then equation Lx ¼ N ð1; xÞ has at least one solution in X. In order to use Lemma 1, we set X ¼ fx : x 2 C 1 ðR; Rn Þ; xðt þ T Þ  xðtÞg qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 with the norm kxk ¼ max½0;T  jxðtÞj þ jx0 ðtÞj and Y ¼ fx : x 2 CðR; Rn Þ; pffiffiffiffiffiffiffiffiffiffiffi xðt þ T Þ  xðtÞg with the norm kyk0 ¼ max½0;T  jyðtÞj, where jxj ¼ ðx; xÞ for x 2 Rn , ð; Þ is the inner Rproduct in Rn . Then X and Y are two Banach spaces. T We also denote ½x; y ¼ 0 ðxðtÞ; yðtÞÞ dt 8x; y 2 Y and A : Y ! Y;

½AxðtÞ ¼ xðtÞ þ Cxðt  rÞ:

ð2Þ

Clearly, A is a continuous linear operator with kAk < þ1. Lemma 2. Let k1 ; k2 ; . . . ; kn be the eigenvalues of matrix C. If jki j 6¼ 1 8i 2 In , then A has continuous bounded inverse with following relationships: RT RT Pn 1 (1) 0 j½A1 f jðtÞ dt 6 i¼1 j1jk jf ðtÞj dt 8f 2 Y ; 0 ik 00 (2) Ax00 ¼ ðAxÞ 8x 2 CT2 :¼ fx : x 2 C 2 ðR; Rn Þ; xðt þ T Þ  xðtÞg. Proof. (1) As C is symmetrical, we find that there is a orthogonal matrix U such that UCU T ¼ Ek ¼ diagðk1 ; k2 ; . . . ; kn Þ. In view of xðtÞ þ Cxðt  rÞ ¼ f ðtÞ; we have yðtÞ þ Ek yðt  rÞ ¼ f~ ðtÞ;

ð3Þ

where f~ ðtÞ ¼ Uf ðtÞ; yðtÞ :¼ ðy1 ðtÞ; y2 ðtÞ; . . . ; yn ðtÞÞ ¼ UxðtÞ. That is yi ðtÞ þ ki yi ðt  rÞ ¼ f~i ðtÞ;

i 2 In :

By [16], we obtain from (4) that 8P > ðki Þj f~i ðt  jrÞ; < jP0 P yi ðtÞ ¼ j > :  ðki Þ f~i ðt þ jrÞ; jP0

ð4Þ

jki j < 1 jki j > 1:

ð5Þ

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L. Shiping, G. Weigao / Appl. Math. Comput. 156 (2004) 719–732

Hence A has continuous inverse A1 with ½A1 f ðtÞ ¼ xðtÞ ¼ U T yðtÞ. If jki j < 1, we have from (5) that Z T X jZ T X jZ T ~ jyi ðtÞj dt 6 jki j jfi ðt  jrÞj dt ¼ jki j jf~i ðtÞj dt 0

¼ Similarly, Z

0

jP0

RT 0

1 1  jki j

jyi ðtÞj dt 6

T

jyðtÞj dt ¼

0

Z

1 jki j1

0

6

i¼1

0

jP0 T

jf~i ðtÞj dt 6

0

1 1  jki j

Z

T

jf ðtÞj dt:

0

RT

jf ðtÞj dt, if jki j > 1. So !1=2 n n Z T X X 2 yi ðtÞ dt 6 jyi ðtÞj dt

T

n X

Z

0

i¼1

1 j1  jki k

i¼1

Z

0

T

jf ðtÞj dt: 0

(2) Clearly, the statement of case (2) follows from the conclusion of case (1) immediately. Now, we define L : DðLÞ  X ! Y ;

00

Lx ¼ ðAxÞ ; d N : ½0; 1  X !; N ðk; xÞ ¼  grad F ðxðtÞÞ  grad Gðxðt  sðtÞÞÞ þ kpðtÞ; dt ð6Þ where DðLÞ :¼ fx : x 2 X ; x00 2 RCðR; RÞg. From Lemma 2, we know that T ker L ¼ Rn , Im L ¼ fx : x 2 CT ; 0 xðsÞ ds ¼ 0g. Hence, L is a Fredholm operator with index zero. Let project operator P ; Q as follows, respectively, Z 1 T P : X ! ker L; Px ¼ xðsÞ ds; Q : Y ! Y =Im L; T 0 Z T 1 yðsÞ ds Qy ¼ T 0 then Im P ¼ ker L, ker Q ¼ Im L. Set LP ¼ LjDðLÞ\ker p and L1 P : Im L ! DðLÞ denote the (unique) right-inverse LP , then  Z t 1 2 Z T 1 1 t  2 T syðsÞ ds þ ðt  sÞyðsÞ ds ðLP yÞðtÞ ¼ A T 0 0  Z Z u 1 T  ðu  sÞyðsÞ ds du : T 0 0 From the restriction on F ; G and p, it is easy to see that N is L-compact on ½0; 1  X, for any X open in X . Since system

L. Shiping, G. Weigao / Appl. Math. Comput. 156 (2004) 719–732

723

d2 d ðxðtÞ þ Cxðt  rÞÞ þ k grad F ðxðtÞÞ þ kgrad Gðxðt  sðtÞÞÞ ¼ k2 pðtÞ 2 dt dt ð7Þ can be represented by Lx ¼ kN ðk; xÞ

ð8Þ

then the question about the existence of periodic solution to system (1) is now translated to one about the existence of solutions to Eq. (8) in X for k ¼ 1. For the sake of convenience, we list following assumptions which are needed later for us to study system (1). 2

(H1) There is 2a constant a > 0 such that jðH ðvÞx; xÞj P ajxj 8v; x 2 Rn where H ðxÞ ¼ o oxF ðxÞ 2 . GðxÞÞ (H2) There is a constant d > 0 such that limjxj!þ1 ðx;grad > d or jxj2 ðx;grad GðxÞÞ limjxj!þ1 < d. jxj2 (H3) There is a constant L > 0, such that jgrad GðxÞ  grad GðyÞj 6 Ljx  yj 8x; y 2 Rn . (H4) There are constants l1 2 ½0; T , l2 2 ½0; T  and two positive integers m; k such that jsðtÞ  mT j 6 l1 , jsðtÞ  r  kT j 6 l2 8t 2 ½0; T . h The following lemma is important for us to study system (1). Lemma 3. Let 0 6 a 6 T be a constant, s 2 CðR; RÞ with sðt þ T Þ  sðtÞ and maxt2½0;T  jsðtÞj 6 a. Then for 8x 2 X , we have Z T Z T jxðtÞ  xðt  sðtÞÞj2 dt 6 2a2 jx0 ðtÞj2 dt: ð9Þ 0

0

Proof. Let D1 ¼ ft : t 2 ½0; T ; sðtÞ P 0g and D2 ¼ ft : t 2 ½0; T ; sðtÞ < 0g. Then for 8t 2 D1 , we have 2 Z !2 Z t Z t   t   0 0 x ðrÞ dr 6 jx ðrÞj dr 6 jsðtÞj jx0 ðrÞj2 dr     tsðtÞ tsðtÞ tsðtÞ Z t 6a jx0 ðrÞj2 dr ð10Þ ta

and for 8t 2 D2 , we have Z 2  Z 2 Z tsðtÞ  t  tsðtÞ   2 0 0 x ðrÞ dr 6 jx ðrÞj dr 6 jsðtÞj jx0 ðrÞj dr   tsðtÞ  t t Z tþa 6a jx0 ðrÞj2 dr: t

ð11Þ

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From (10) and (11), we find Z

T

2 2 Z  Z t Z  Z t       0 0 jxðtÞ  xðt  sðtÞÞj dt 6 x ðrÞdr dt þ x ðrÞdr dt       tsðtÞ tsðtÞ D1 D2 Z Z t Z Z tþa 2 2 6a jx0 ðrÞj drdt þ a jx0 ðrÞj drdt 2

0

D1

Z

D2

ta

Z

T

2

jx0 ðrÞj drdt:

6a 0

ð12Þ

ta

Step 1. If a 2 ½0; T =2, then we have Z T Z tþa Z a Z rþa Z 2 2 0 0 jx ðrÞj dr dt ¼ jx ðrÞj dt dr þ 0

a

ta

þ ¼

0

Z

T a

ra 2

ra

Z

a

T þa

2

jx0 ðrÞj dr

2

jx0 ðrÞj ðT  r þ aÞ dr

T a

Z

a 2

0

jx ðrÞj dr þ 2a

a

Z

Z a

T a

2

0

jx ðrÞj dr ¼ 2a

T a

Z

a

0

jx0 ðrÞj2 dr

T 2

jx0 ðrÞj dr: 0

So it follows from (12) that Z T Z jxðtÞ  xðt  sðtÞÞj2 dt 6 2a2

T

jx0 ðrÞj2 dr:

ð13Þ

0

Step 2. If a 2 ðT =2; T , then Z T Z tþa Z 2 0 jx ðrÞj dr dt ¼ 0

T a a

Z

¼ 2a

2

jx0 ðrÞj dt dr

jx0 ðrÞj2 dt dr

jx0 ðrÞj ðr þ aÞ dr þ 2a

¼ 2a

rþa

T

a

þ

Z

a

Z

T þa

T a

Z

t

tþa

T a

Z

a

ta

þ ¼

Z

0

T þa

2

jx ðrÞj dt dr þ

0

a

Z

rþa

Z

a

T 2

jx0 ðrÞj dt dr

0

2

jx0 ðrÞj dt dr 2

a

Z

Z

T

jx0 ðrÞj ðr þ aÞ dr þ T

þ

a T a

ra

T a

Z

Z

a 2

jx0 ðrÞj dr T a

T þa

2

jx0 ðrÞj ðT  r þ aÞ dr:

ð14Þ

L. Shiping, G. Weigao / Appl. Math. Comput. 156 (2004) 719–732

725

As Z

T þa

2

jx0 ðrÞj ðT  r þ aÞ dr ¼

Z

a

a

þ þ

2

jx0 ðrÞj ðT  r þ aÞ dr

a

Z Z

T a

2

jx0 ðrÞj ðT  r þ aÞ dr

a T þa

2

jx0 ðrÞj ðT  r þ aÞ dr T a

it follows from (14) that Z T Z tþa Z T a Z a 2 2 2 jx0 ðrÞj dr dt ¼ ðT þ 2aÞ jx0 ðrÞj dr þ T jx0 ðrÞj dr 0 ta a T a Z a 2 0 þ jx ðrÞj ðT  r þ aÞ dr a Z T þa 2 þ jx0 ðrÞj ðT  r þ aÞ dr T a Z T a Z a 2 2 ¼ ðT þ 2aÞ jx0 ðrÞj dr þ T jx0 ðrÞj dr a T a Z a 2 0 þ jx ðrÞj ðT  r þ aÞ dr Za a 2 þ jx0 ðrÞj ðr þ aÞ dr a Z T a Z T a 2 0 ¼ ðT þ 2aÞ jx ðrÞj dr  T jx0 ðrÞj2 dr a a Z a T jx0 ðrÞj2 dr a Z T a Z T a 2 2 0 ¼ ðT þ 2aÞ jx ðrÞj dr  T jx0 ðrÞj dr a a Z T 2 ¼ 2a jx0 ðrÞj dr: 0

Thus, from (12) we get that Z 0

T 2

jxðtÞ  xðt  sðtÞÞj dt 6 2a2

Z

T 2

jx0 ðrÞj dr:

ð15Þ

0

Therefore, the conclusion of Lemma 3 immediately follows from (13) and (15). h

726

L. Shiping, G. Weigao / Appl. Math. Comput. 156 (2004) 719–732

From the symmetric of matrix H ðxÞ and C, one can find that H ðxÞCCH ðxÞ is symmetric for all x 2 Rn . So the matrix H ðxÞCCH ðxÞ exist n eigenvalues l1 ; l2 ; . . . ; ln . Throughout this paper, we denote kM ¼ maxi2In fjki jg. Furthermore, we assume jki j 6¼ 1 8i 2 In and lM :¼ maxi2In fmaxx2Rn jli jg < þ1.

3. Main result Theorem. Suppose that assumption (H1 )–(H ) hold, then system (1) has at least pffiffiffiffiffiffi pffiffiffi 4 one T -periodic solution for a > lM þ 2Lðl1 þ kM l2 Þ. Proof. By considering the explanation preceding the Lemma 3, we need only to prove that there is a domain X satisfying all the requirements in Lemma 1. Without loss of generality, we may suppose from (H2 ) that ðx; grad GðxÞÞ > d=2;

8x 2 Rn

with jxj > A1 > 0;

ð16Þ

where A1 is a constant. Assume x ¼ xðtÞ is a solution of Eq. (8) for a constant k 2 ð0; 1Þ, then we have  2 d ðxðtÞ þ Cxðt  rÞÞ þ kH ðxðtÞÞx0 ðtÞ þ k grad Gðxðt  sðtÞÞÞ; dt2    d d 2 ðxðtÞ þ Cxðt  rÞÞ ¼ k pðtÞ; ðxðtÞ þ Cxðt  rÞÞ : dt dt As ½x00 ðtÞ; x0 ðtÞ ¼ 0, ½grad GðxðtÞÞ; x0 ðtÞ ¼ 0 and ½grad Gðxðt  rÞÞ;Cx0 ðt  rÞ ¼ 0, it follows from above formula that Z T a jx0 ðtÞj2 dt 6 j½H ðxðtÞÞx0 ðtÞ; x0 ðtÞj 0

6 j½H ðxðtÞÞx0 ðtÞ; Cx0 ðt  rÞj þ j½grad Gðxðt  sðtÞÞÞ  grad GðxðtÞÞ; x0 ðtÞj þ kj½pðtÞ; x0 ðtÞj þ j½grad Gðxðt  sðtÞÞÞ  grad Gðxðt  rÞÞ; Cx0 ðt  rÞj 6 j½H ðxðtÞÞx0 ðtÞ; Cx0 ðt  rÞj þ j½grad Gðxðt  sðtÞÞÞ  grad GðxðtÞÞ; x0 ðtÞj þ kj½pðtÞ; x0 ðtÞj þ j½grad Gðxðt þ r  sðt þ rÞÞÞ  grad GðxðtÞÞ; Cx0 ðtÞj: ð17Þ In view of j½H ðxðtÞÞx0 ðtÞ; Cx0 ðt  rÞj ¼ j½C T H ðxðtÞÞx0 ðtÞ; x0 ðt  rÞj ¼ j½CH ðxðtÞÞx0 ðtÞ; x0 ðt  rÞj 6 j½CH ðxðtÞÞx0 ðtÞ; CH ðxðtÞÞx0 ðtÞj1=2  j½x0 ðtÞ; x0 ðtÞj1=2 Z pffiffiffiffiffiffi T 0 2 6 lM jx ðsÞj ds: ð18Þ 0

L. Shiping, G. Weigao / Appl. Math. Comput. 156 (2004) 719–732

727

From (H4 ), we find that sðtÞ 2 ½mT  l1 ; mT þ l1 , 8t 2 ½0; T  and sðtÞ  r 2 ½kT  l2 ; kT þ l2 , 8t 2 ½0; T . Let sðtÞ  mT ¼ sðtÞ and sðt þ rÞ  r  kT ¼ s1 ðtÞ, then maxt2½0;T  jsðtÞj 6 l1 and maxt2½0;T  js1 ðtÞj 6 l2 . Substituting (18) into (17), we have Z Z T Z T pffiffiffiffiffiffi T 0 2 jx0 ðtÞj2 dt 6 lM jx ðsÞj ds þ L jx0 ðtÞkxðtÞ  xðt  sðtÞÞjdt a 0 0 0 Z T Z T jCx0 ðtÞkxðtÞ  xðt  s1 ðtÞÞjdt þ k jpðtÞkAx0 ðtÞjdt þL 0 0 Z pffiffiffiffiffiffi T 0 2 6 lM jx ðsÞj ds Z þL

0 T

0

þL

Z

1=2  Z jx ðtÞj dt 

1=2

T

2

0

2

jxðtÞ  xðt  sðtÞÞj dt 0

T

1=2  Z jCx ðtÞj dt 

0

þ kmkAk

Z

2

0

jx ðtÞj dt

2

jxðtÞ  xðt  s1 ðtÞÞj dt 0

1=2

T

1=2

T

2

0

ð19Þ

;

0

where m :¼ kpk0 T 1=2 . As Z T 1=2 Z jCx0 ðtÞj2 dt ¼ ð½Cx0 ðtÞ; Cx0 ðtÞÞ1=2 6 kM 0

T

jx0 ðsÞj2 ds

1=2

0

and by applying Lemma 3, we get from (19) that Z T Z pffiffiffi Z pffiffiffiffiffiffi T 0 2 jx0 ðtÞj2 dt 6 lM jx ðsÞj ds þ 2Ll1 a 0

0

Z pffiffiffi þ 2LkM l2 0

T

jx0 ðtÞj2 dt

0 T 0

2

jx ðsÞj ds þ kmkAk

Z

T 2

jxðtÞj dt

1=2 :

0

pffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffi pffiffiffi As a > lM þ 2Lðl1 þ l2 kM Þ, it follows that r1 ¼: a  lM þ 2Lðl1 þ l2 kM Þ > 0. So Z T 2 k2 m2 kAk 2 jx0 ðtÞj dt 6 ; ð20Þ r21 0 i.e., Z

2

T

m2 kAk ¼: R1 ; ð21Þ r21 0 RT pffiffiffiffiffiffiffiffi which yields that 0 jx0 ðtÞj dt 6 R1 T . Thus there is a t1 2 ½0; T  such that pffiffiffiffiffiffiffiffiffiffiffiffiffi ð22Þ jx0 ðt1 Þj < R1 T 1 : 2

jx0 ðtÞj dt 6

728

L. Shiping, G. Weigao / Appl. Math. Comput. 156 (2004) 719–732

In addition to this, we have 

d2 ðxðtÞ þ Cxðt  rÞÞ þ kH ðxðtÞÞx0 ðtÞ þ kgrad Gðxðt  sðtÞÞÞ; xðtÞ dt2



¼ k2 ½pðtÞ; xðtÞ: RT 2 As ½x00 ðtÞ; xðtÞ ¼  0 jx0 ðtÞj dt, and ½H ðxðtÞÞx0 ðtÞ; xðtÞ ¼ 0, it follows from (20) and above formula that kj½grad Gðxðt  sðtÞÞÞ; xðtÞj Z T Z 2 2 0 ¼ jk ½pðtÞ; xðtÞ þ jx ðtÞj dt þ 0

6 k2 m

Z

T

jxðtÞj2 dt

T

x0T ðtÞCx0 ðt  rÞ dtj 0

1=2

2

þ

0

k2 ð1 þ kM Þm2 kAk : r21

Thus j½grad GðxðtÞÞ; xðtÞj ¼ j½grad GðxðtÞÞ  grad Gðxðt  sðtÞÞÞ; xðtÞ þ ½grad Gðxðt  sðtÞÞÞ; xðtÞj Z T 6L jxðtÞkxðtÞ  xðt  sðtÞÞj dt 0

Z þ km

T

jxðtÞj dt

0

pffiffiffi 6 l1 L 2

1=2

km2 kAk2 ð1 þ kM Þ r21 1=2  Z T 1=2 jxðtÞj2 dt  jx0 ðtÞj2 dt 2

Z

T

þ

0

Z þ km

0

T 2

1=2

jxðtÞj dt

2

2

km kAk r21 1=2 2 m2 kAk ð1 þ kM Þ 2 jxðtÞj dt þ : r21 þ

0

Z pffiffiffi 1=2 6 ð 2Ll1 R1 þ mÞ

T 0

ð23Þ 2

From (16) we find there is a constant B > A1 such that ðx; grad GðxÞÞ > d2 jxj , pffiffiffiffiffiffiffiffiffi dþ d 2 þ4e 1=2 8x 2 Rn with jxj > B. It is easy to see that B can be defined by B > T , 2 pffiffi 1=2 2ð 2l1 LR1 þmÞ 2m2 kAk2 ð1þkM Þ where d ¼ ,e¼ . d dr2 1

Now we show that there is a t 2 ½0; T  such that jxðt Þj < B. Otherwise, jxðtÞj P B 8t 2 ½0; T . Then for t 2 ½0; T , we have from (23) that

L. Shiping, G. Weigao / Appl. Math. Comput. 156 (2004) 719–732

d 2

Z

729

T 2

jxðtÞj dt 6 ½grad GðxðtÞÞ; xðtÞ 0

6

 pffiffiffi  Z 1=2 2Ll1 R1 þ m

1=2

T 2

jxðtÞj dt

2

þ

0

m2 kAk ð1 þ kM Þ r21

i.e., Z

T 2

jxðtÞj dt 6 d

Z

0

1=2

T 2

jxðtÞj dt

þ e:

0

pffiffiffiffiffiffiffiffiffi R 1=2 T dþ d 2 þ4e 2 It follows that 0 jxðtÞj dt < . On the other hand, we have from 2 the assumption jxðtÞj > B, 8t 2 ½0; T  that Z

1=2

T 2

jxðtÞj dt

> BT

1=2

¼



0

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d 2 þ 4e ; 2

which yields a contradiction. This contradiction implies that there is a t 2 ½0; T  such that jxðt Þj 6 B. Hence

kxk0 6 B þ

Z

T

jx0 ðtÞj dt 6 B þ

pffiffiffiffiffiffiffiffi R1 T :

ð24Þ

0

According to the restriction on F ðxÞ and GðxÞ, it is easy to see that there are pffiffiffiffiffiffiffiffi constants M1 , M2 such that jH ðxÞj < M1 , forpffiffiffiffiffiffiffiffi x 2 Rn with jxj 6 B þ R1 T and jgrad GðxÞj < M2 , for x 2 Rn with jxj 6 B þ R1 T . By using Lemma 2, we have jx0 ðtÞj 6 jx0 ðt1 Þj þ

Z

T

jx00 ðsÞj ds

0

0

6 jx ðt1 Þj þ

Z

T 1

00

0

j½A Ax ðsÞj ds 6 jx ðt1 Þj þ

0

6 jx0 ðt1 Þj þ

n X i¼1

þ

Z 0

n X i¼1

1 j1  jki k

Z

1 j1  jki k

T

jH ðxðtÞÞx0 ðtÞj dt

0

T

jgrad Gðxðt  sðtÞÞÞj dt þ

Z 0

T

 jpðtÞj dt :

Z 0

T

j½Ax00 ðsÞj ds

730

L. Shiping, G. Weigao / Appl. Math. Comput. 156 (2004) 719–732

It follows from (21) and (22) that max jx0 ðtÞj 6

t2½0;T 

n pffiffiffiffiffiffiffiffiffiffiffiffiffi X R1 T 1 þ i¼1 n X

1 j1  jki k i¼1 n pffiffiffiffiffiffiffiffiffiffiffiffiffi X 6 R1 T 1 þ þ

i¼1

Z T 1=2  Z T 1=2 1 2 2 jH ðxðtÞÞj dt  jx0 ðtÞj dt j1  jki k 0 0  Z T jgrad Gðxðt  sðtÞÞÞj dt þ mT 1=2 0

1 1=2 ðM1 T 1=2 R1 þ M2 T þ mT 1=2 Þ :¼ R2 : j1  jki k ð25Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1=2 2 Let R3 ¼ ðB þ ðR1 T Þ Þ þ R22 þ 1 and X ¼ fx j x 2 X ; kxk < R3 g, then from (24) and (25) we can verify that X satisfies the conditions (1) and (2) in Lemma 1. Let Z 1l T Uðl; xÞ ¼ lx  grad Gðxðt  sðtÞÞÞ dt: T 0 It is easy to verify that Uðl; xÞ 6¼ 0 for all x 2 oX \ ker L and deg fQN ð0; Þ; X \ ker L; 0g ¼ deg fUð0; Þ; X \ ker L; 0g ¼ deg fUð1; Þ; X \ ker L; 0g ¼ deg fI; X \ ker L; 0g 6¼ 0: Therefore, the system (1) has at least one T -periodic solution.

h

As for application, we consider a kind of scalar neutral differential equation as follows: d2 ðxðtÞ þ bxðt  rÞÞ þ f ðxðtÞÞx0 ðtÞ þ gðxðt  sðtÞÞÞ ¼ pðtÞ; dt2

ð26Þ

where f ; g : R ! R are continuous with f ðxÞ is bounded, s; p 2 CðR; RÞ with sðt þ T Þ  sðtÞ; pðt þ T Þ  pðtÞ, b and r are two constants. corresponding to system 1, one can find l1 ¼ b2 f 2 ðxÞ, k1 ¼ b. So, by using the theorem, we have following result. Corollary. Suppose that following conditions hold. (H01 ) There are two constants a1 > 0, a2 > 0 such that jf ðxÞj P a1 , jbf ðxÞj 6 a2 8x 2 R; (H02 ) There is a constant d0 > 0 such that lim

jxj!1

xgðxÞ > d0 x2

or

limjxj!1

xgðxÞ < d0 ; x2

L. Shiping, G. Weigao / Appl. Math. Comput. 156 (2004) 719–732

731

(H03 ) There is a constant L0 > 0 such that jgðxÞ  gðyÞj 6 L0 jx  yj;

8x; y 2 R;

(H04 ) There are two constants l1 2 ½0; T , l2 2 ½0; T  and two positive integers m; k such that jsðtÞ  mT j 6 l01 ; jsðtÞ  r  kT j 6 l02 . Then Eq. (26) exists at least one T -periodic solutions, if a1 > ja2 j þ jbjl02 Þ.

pffiffiffi 0 0 2L ðl1 þ

For example, let us consider the neutral differential equation as follows:

00  x t  12 sin t 1 0 xðtÞ þ xðt  1Þ þ 2x ðtÞ þ pffiffiffi

ð27Þ

¼ cos2 t: 2 2 2 1 þ x2 t  12 sin t Corresponding to the corollary, we have b ¼ 1=2, a1 ¼ 2, a2 ¼ 1, d0 ¼ 2p1 ffiffi2 > 0, L0 ¼ 2p1 ffiffi2, l01 ¼ maxt2½0;2p j 12 sin tj ¼ 1=2 and l02 ¼ maxt2½0;2p j 12 sin t  1j ¼ 3=2. So pffiffiffi a1 ¼ 2 > 13=8 ¼ ja2 j þ 2L0 ðl01 þ jbjl02 Þ: By using the corollary, we obtain that Eq. (27) exists at least one 2p-periodic solution.

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