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Existence of periodic solutions for a p-Laplacian neutral functional differential equation with multiple variable parametersI Zhengxin Wang a,∗ , Shiping Lu b , Jinde Cao a a

Department of Mathematics, Southeast University, Nanjing 210096, Jiangsu, China

b

Department of Mathematics, Anhui Normal University, Wuhu 241000, Anhui, China

article

abstract

info

Article history: Received 24 July 2008 Accepted 6 July 2009

By means of Mawhin’s continuation theorem, a kind of p-Laplacian neutral functional differential equation with multiple variable parameters as follows:

ϕp

MSC: 34B15 34K13

u(t ) −

n X

!0 !!0 ci (t )u(t − ri )

= f (u(t ))u0 (t ) + α(t )g (u(t ))

i=1

+

Keywords: Periodic solutions Neutral functional differential equation Mawhin’s continuation theorem Variable parameter arguments

n X

βj (t )g (u(t − γj (t ))) + e(t ),

j=1

is studied. The interesting thing is that the coefficient c isP not a constant, by analyzing some n properties of operator A : CT → CT , [Ax](t ) = x(t ) − i=1 ci (t )x(t − ri ) and then using Mawhin’s continuation theorem, some new results on the existence of periodic solutions are obtained. Crown Copyright © 2009 Published by Elsevier Ltd. All rights reserved.

1. Introduction In this paper, we investigate the existence of periodic solutions for a p-Laplacian neutral functional differential equation with multiple variable parameters ci (t )(i = 1, 2, . . . , n) as follows:

ϕp

u(t ) −

n X

!0 !!0 ci (t )u(t − ri )

= f (u(t ))u0 (t ) + α(t )g (u(t )) +

i =1

n X

βj (t )g (u(t − γj (t ))) + e(t ),

(1.1)

j =1

where ϕp (u) = |u|p−2 u, u ∈ R, p > 1; f , g ∈ C (R, R), α(t ), e(t ), βj (t ), γj (t ), (j = 1, 2, . . . , n) are continuous periodic functions defined on R with period T > 0, ci (t ) ∈ C 1 (R, R) and ci (t + T ) = ci (t ), (i = 1, 2, . . . , n); T , ri are given constants, (i = 1, 2, . . . , n). In recent years, when c (t ) is a constant c, the problem of existence of periodic solutions to some types of neutral functional differential equations (NFDEs) were studied in some papers, see [1–5] and the references therein. But the work to study the existence of periodic solutions to NFDEs with variable parameter rarely appeared. Recently, Du, Guo, Ge and Lu in [6] studied the following NFDEs with variable parameter as follows:

(x(t ) − c (t )x(t − τ ))00 + f (x(t ))x0 (t ) + g (x(t − γ (t ))) = e(t ).

(1.2)

I This work was sponsored by the NSF of Anhui Province of China (No. 2005kj031ZD; No. 050460103), the Teaching and Research Award Program for Excellent Teachers in Higher Education Institutions of Anhui Province of China and the key NSF of Education Ministry of China (No. 207047). ∗ Corresponding author. E-mail address: [email protected] (Z. Wang).

0362-546X/$ – see front matter Crown Copyright © 2009 Published by Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.07.014

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

735

But as far as we know, the problem of periodic solutions for a p-Laplacian NFDEs with multiple variable parameters has not been studied until now. Thus, it is worth investigating the existence of periodic solutions of Eq. (1.1). The main purpose of this paper is to establish sufficient conditions for guaranteeing theP existence of periodic solutions n to Eq. (1.1). At first, by analyzing some properties of operator A : CT → CT , [Ax](t ) = x(t ) − i=1 ci (t )x(t − ri ), we establish some new inequalities and other important results, which are crucial for estimating a priori bounds of periodic solutions, and then by using the continuation theorem of coincidence degree theory, we obtain some new results on the existence of periodic solution to Eq. (1.1). Moreover, some examples are given to show that the results of this paper are easily applicable. 2. Main lemmas In this section, we give some lemmas which will be used in this paper. Let CT = {x : x ∈ C (R, R), x(t + T ) ≡ x(t ), for all t ∈ R}, with norm |ϕ|0 = maxt ∈[0,T ] |ϕ(t )|, for all ϕ ∈ CT , and CT1 = {x : x ∈ C 1 (R, R), x(t + T ) ≡ x(t ), for all t ∈ R}, with norm kϕk = max{|ϕ|0 , |ϕ 0 |0 }, for all ϕ ∈ CT1 . Clearly, CT and CT1 are Banach spaces. Also, for ci ∈ CT1 , let ci0 = max |ci (t )|,

ci1 = max |ci0 (t )|,

t ∈[0,T ]

(i = 1, 2, . . . , n).

t ∈[0,T ]

Define linear operators: A : CT → CT , [Ax](t ) = x(t ) −

n X

ci (t )x(t − ri ),

for all t ∈ [0, T ],

(2.1)

i =1

and n X

B : CT → CT , [Bx](t ) =

ci (t )x(t − ri ),

for all t ∈ [0, T ].

(2.2)

i=1

Throughout this paper, let Lemma 2.1. If

Pn

i=1 0 i=1 ci m

Pm

i =n r i

= 0 if n > m.

ci0 < 1, then B satisfies the following conditions:

(1) kBk ≤ . P∞ (2) m=0 [B f ](t ) ≤

Pn

(3)

R T 0

|[Bj f ](t )|p dt

1p

1−

≤

|f | Pn0

0 i=1 ci

.

Pn

0 i=1 ci

j R T 0

|f (t )|p dt

1p

, for all f ∈ CT , p > 1, j ≥ 1.

Proof. (1) The conclusion of case 1 is obviously true. (2) From the definition of operator B, we have

[B f ](t ) = m

n X n X

···

i1 =1 i2 =1

n Y m X

t−

cij

im =1 j=1

m X

! rik

f

t−

k=j+1

m X

! ,

ris

(2.3)

s=1

for all m ≥ 1. Therefore ∞ X

[B f ](t ) = f (t ) + m

m=0

∞ X n X n X

···

m=1 i1 =1 i2 =1

n Y m X

cij

t−

im =1 j=1

m X

! rik

f

t−

k=j+1

m X

! ris

,

s=1

which yields

! ∞ ∞ X n X n n Y m m X X X X m ··· rik f [B f ](t ) ≤ |f (t )| + cij t − m=0 m=1 i =1 i =1 i =1 j =1 k=j+1 1

≤ |f |0 +

m

2

∞ X n X n X

n

···

m=1 i1 =1 i2 =1

≤ |f |0 +

∞ n X X m=1

≤

|f |0 . n P 1− ci0 i=1

i=1

XY im =1 j=1

!m ci0

m

|f |0

ci0j |f |0

t−

m X s =1

! ris

736

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

(3) From (2.3) we have T

Z

|[B f ](t )| dt m

p

1p

n X n X

T

Z ≤

0

0

≤

n Y m X

···

i1 =1 i2 =1

n X n X

im =1 j=1

Z n X

···

i1 =1 i2 =1

T

Z n X n n Y m X X 0 ≤ ··· cij n

≤

im =1 j=1

n

n

XX

···

i1 =1 i2 =1 n X

≤

!m Z

Pn

i =1

(3)

RT 0

1 1−

Pn

s=1

! !p ! 1p ris dt 1p

ris

|f (t )|p dt

ris

T

|f (t )| dt p

1p

0

T

|f (t )| dt p

1p

,

for all f ∈ CT , p > 1, m ≥ 1.

ci0 < 1, then A has continuous inverse A−1 on CT with the following properties, where A is defined by (2.1).

(1) [A−1 f ](t ) = f (t ) + (2) kA−1 k ≤

m X

! !p ! 1p ris dt

0

i =1

Lemma 2.2. If

ci0j

t−

s=1 m P

−

Z

im =1 j=1

ci0

T−

m X s=1

m P

s=1

m

XY

t−

m Y 0 cij f j=1

0

im =1

i1 =1 i2 =1

f

ci0j

0 i=1 ci

P∞ Pn m=1

.

|[A−1 f ](t )|p dt ≤

1

(4) [Af 0 ](t ) = [Af ]0 (t ) +

1−

Pn

i 1 =1

Pn

0 i=1 ci

Pn

i=1

i 2 =1

RT

p

0

Pn

···

im =1

Qm

j =1

cij t −

Pm

k=j+1 rik

f t−

Pm

s=1 ris

.

|f (t )|p dt , for all f ∈ CT , p > 1.

ci0 (t )f (t − ri ), for all f ∈ CT1 .

Pn

Proof. (1) From Lemma 2.1, we see that kBk ≤ A−1 : CT → CT , A−1 = (I − B)−1 = I +

i =1

∞ X

ci0 < 1, we obtain that A has bounded inverse

Bm ,

m=1

then for all f ∈ CT , together with (2.3) we have

[A−1 f ](t ) = f (t ) +

∞ X

[Bm f ](t )

m=1

= f (t ) +

∞ X n X n X

···

m=1 i1 =1 i2 =1

n Y m X

cij

t−

im =1 j=1

!

m X

rik

t−

f

k=j+1

m X

! ris

.

s =1

(2) It follows that A−1 = (I − B)−1 and kA−1 k = k(I − B)−1 k ≤

1 1−

Pn

0 i=1 ci

.

(3) From the proof of Lemma 2.1, we obtain T

Z

|[A−1 f ](t )|p dt

1p

Z

T

=

0

0

! ∞ X n X n n Y m m X X X ··· cij t − rik f f (t ) + m=1 i =1 i =1 i =1 j =1 k=j+1 1

T

Z

|f (t )|p dt

≤

t−

s=1

m

2

m X

! p ! 1p ris dt

1p

0

Z + 0

T

! ∞ X n X n n Y m m X X X ··· cij t − rik f m=1 i =1 i =1 i =1 j =1 k=j+1 1

T

Z

|f (t )|p dt

≤

1p

T

Z

|f (t )|p dt 0

p ! 1p ∞ T X [Bm f ](t ) dt 0 m=1

Z +

0

≤

m

2

1p +

∞ Z X m=1

T

|[Bm f ](t )|p dt 0

1p

t−

m X s =1

! p ! 1p ris dt

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747 T

Z

|f (t )|p dt

≤

1 p

+

0

∞ n X X m=1

n P

1−

T

Z

1

≤

ci0

|f (t )|p dt

ci0

!m Z

|f (t )|p dt

0

i=1 1 p

T

737 1 p

.

0

i =1

(4) The conclusion of case (4) follows from the continuity of A−1 and the definition of A immediately. Now suppose that there is an integer k ∈ {1, 2, . . . , n} such that ck = mint ∈[0,T ] |ck (t )| > 1 and δ := From the definition of A we have

[Ax](t ) = x(t ) −

n X i=1

"

1 ck

+

ci0 i6=k ck

P

< 1.

# X ci (t ) ci (t )x(t − ri ) = −ck (t ) x(t − rk ) − + x(t − ri ) . ck (t ) c (t ) i6=k k x(t )

Let [Ax](t ) = f (t ), that is

# X ci (t ) + x(t − ri ) = f (t ), −ck (t ) x(t − rk ) − ck (t ) c (t ) i6=k k "

x(t )

so x(t − rk ) −

x(t ) ck (t )

+

X ci (t ) f (t ) x(t − ri ) = − . c ( t ) c k (t ) i6=k k

Doing variable transformation: x(t ) −

1 ck (t + rk )

x(t + rk ) −

X ci (t + rk ) f ( t + rk ) − x(t − ri + rk ) = − := f1 (t ), ck (t + rk ) ck (t + rk ) i6=k

(2.4)

and f1 (t ) ∈ CT as well. Let E : CT → CT , [Ex](t ) = x(t ) −

1 ck (t + rk )

x(t + rk ) −

1 ci (t + rk ) , di ( t ) = − , ck (t + rk ) ck (t + rk ) i = 1, 2, . . . , k − 1, k + 1, . . . , n,

dk (t ) =

X ci (t + rk ) − x(t − ri + rk ), ck (t + rk ) i6=k

e rk = −rk ,

e ri = ri − rk ,

and d0i = max |di (t )|, t ∈[0,T ]

i = 1 , 2 , . . . , n.

Then we have [Ex](t ) = x(t ) −

Pn

i =1

di (t )x(t −e ri ) and

Pn

i =1

d0i ≤

1 ck

+

P

ci0 i6=k ck

= δ < 1. Therefore, by Lemma 2.2 E has

continuous inverse E −1 : CT → CT and it follows from Eq. (2.4) which is equivalent to [Ex](t ) = f1 (t ) that x(t ) = [E −1 f1 ](t ), and [E −1 f1 ](t ) ∈ CT also holds, this implies that A has continuous inverse A−1 : CT → CT , and

[A−1 f (t )] = x(t ) = [E −1 f1 ](t ). Pn In view of i=1 d0i ≤ δ < 1, it follows from Lemma 2.2 that the E −1 satisfies the following property: Pm P∞ Pn Pn Pn Qm Pm (1) [E −1 f ](t ) = f (t ) + m=1 i1 =1 i2 =1 · · · im =1 j=1 dij t − k=j+1 e rik f t − s=1 e ris , where di ,e ri (i = 1, 2, . . . , n) are defined above and for all f ∈ CT . 1 (2) kE −1 k ≤ 1−δ . R T −1 p R T 1 (3) 0 |[E f ](t )|p dt ≤ 1−δ |f (t )|p dt , for all f ∈ CT , p > 1. 0 Furthermore, we can obtain the following result easily.

738

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

Lemma 2.3. If there is an integer k ∈ {1, 2, . . . , n} such that ck = mint ∈[0,T ] |ck (t )| > 1 and δ = has continuous inverse A−1 : CT → CT satisfying: f (t +r )

(1) [A−1 f (t )] = − c (t +rk ) + m=1 i1 =1 i2 =1 · · · im =1 j=1 dij t − k k where di ,e ri (i = 1, 2, . . . , n) are defined above and for all f ∈ CT . 1 P (2) kA−1 k ≤ c −1c δ = 0. k

(3)

RT 0

ck −1−

k

|[A f ](t )| dt ≤ −1

P∞ Pn

p

Pn

Pn

Qm

Pm

e k=j+1 rik

1 ck

+

ci0 i6=k ck

P

< 1, then A

P f t− m e r +r − c ( t −Psm=1 eris +rk ) , k( k) s=1 is

i6=k ci

p

1 ck −1−

0 i6=k ci

RT 0

P

|f (t )|p dt , for all f ∈ CT , p > 1.

Proof. (1) From Lemma 2.2,

[A−1 f (t )] = x(t ) = [E −1 f1 ](t ) ! ! m m ∞ X n X n n Y m X X X X e e ··· dij t − rik f1 t − ris = f1 ( t ) + m=1 i1 =1 i2 =1

im =1 j=1

k=j+1

s=1

=−

f (t + rk ) ck (t + rk )

+

∞ X n X n X

···

m=1 i1 =1 i2 =1

n Y m X

dij

t−

im =1 j=1

m X k=j+1

! e rik −

f

t−

m P s=1

ck

t−

e ris + rk

m P s=1

e ris + rk

,

where di ,e ri (i = 1, 2, . . . , n) are defined above and for all f ∈ CT . (2) By the definition of the operator’s norm, we have

−1 kE (f1 )k kA−1 (f )k kA k = sup = sup k f k kf k f ∈CT f ∈CT −1

= sup f ∈CT

≤ sup f1 ∈CT

=

1 ck

1 kE −1 (f1 )k kf 1 k kf1 k kf k

kE −1 (f1 )k kf k 1 kf 1 k ck kf k

k E −1 k ≤

1 ck − ck δ

=

1 ck − 1 −

P

ci0

.

i6=k

(3) T

Z

|[A−1 f ](t )|p dt = 0

T

Z

|[E −1 f1 ](t )|p dt p Z T 1 |f1 (t )|p dt 1−δ 0 p Z T f (t + rk ) p 1 c (t + r ) dt 1−δ k k 0 p Z T 1 1 |f (t + rk )|p dt p 1−δ ck 0 p Z T 1 P 0 |f (t )|p dt , ck − 1 − ci 0 0

≤ = ≤

=

for all f ∈ CT , p > 1.

i6=k

Remark 2.1. If c1 (t ) ≡ c2 (t ) ≡ · · · ≡ ck−1 (t ) ≡ ck+1 (t ) ≡ · · · ≡ cn (t ) ≡ 0, furthermore, ck0 = maxt ∈[0,T ] |ck (t )| < 1 or ck = mint ∈[0,T ] |ck (t )| > 1 and p = 1, then one can easily find that Lemmas 2.2 and 2.3 generalized the corresponding results of [6]. Lemma 2.4. If

Pn

i =1

ci0 <

1 2

and f (t ) ≡ 1, then [A−1 f ](t ) > 0, for all t ∈ R.

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

Proof. If

Pn

i=1

ci0 <

1 2

739

and f (t ) ≡ 1, then from Lemma 2.2(1) we obtain

[A f ](t ) = f (t ) + −1

∞ X n X n X

···

m=1 i1 =1 i2 =1

= 1+

∞ X n X n X

= 1+

n X

ci1 (t ) +

≥ 1−

ci0

t−

im =1 j=1 n X n X

n X

−

i=1

m X

ci0

1−

n P

rik n X n X n X

ci1 (t − ri2 − ri3 )ci2 (t − ri3 )ci3 (t ) + · · ·

n X

!3 ci0

− ···

i =1 n P

ci0

i=1 n

= ci0

ris

s=1

i1 =1 i2 =1 i3 =1

−

1−2

i=1

!

!

ci1 (t − ri2 )ci2 (t ) +

!2 ci0

t−

f

m X

k=j+1

i =1

n P

= 1−

cij

rik

k=j+1

i1 =1 i2 =1

i1 =1 n X

n Y m X

!

m X

t−

cij

im =1 j=1

···

m=1 i1 =1 i2 =1

n Y m X

1−

i=1

P

> 0.

ci0

i =1

Lemma 2.5. If there is an integer k ∈ {1, 2, . . . , n} such that ck = mint ∈[0,T ] |ck (t )| > 1 and δ ck0 < (1 − δ)ck where δ = 1 ck

+

ci0 i6=k ck ,

P

furthermore f (t ) ≡ 1, then [A−1 f ](t ) 6= 0, for all t ∈ R.

Proof. From assumptions and Lemma 2.3(1) we obtain

[A−1 f (t )] = −

f ( t + rk ) ck (t + rk )

+

n n X ∞ X X

···

m=1 i1 =1 i2 =1

n Y m X

dij

t−

im =1 j=1

m X k=j+1

! e rik −

t−

f

m P s=1

t−

ck

e ris + rk

m P s=1

e ris + rk

=−

1 ck (t + rk )

+

∞ X n X n X

n

···

m=1 i1 =1 i2 =1

m

XY

dij

t−

im =1 j=1

!

m

X k=j+1

e rik −

1

ck

t−

m P s=1

e ris + rk

.

Similar to Lemma 2.4, we obtain

|[A f (t )]| ≥ −1

≥ =

1 ck0 1 ck0 1 ck0

− − −

n 1 X

d0i

ck i=1 1 1 ck 1

δ−

ck

δ

ck 1 − δ

−

1

n X

ck

i =1

δ2 −

1 ck

!2 d0i

−

1

n X

ck

i =1

!3 d0i

− ···

δ3 − · · ·

> 0,

for all t ∈ R by δ ck0 < (1 − δ)ck . Thus for f (t ) ≡ 1, [A−1 f ](t ) 6= 0, for all t ∈ R. Remark 2.2. From the inequality δ ck0 < (1 − δ)ck one obtains δ <

1 2

easily.

Lemma 2.6 ([5]). Let g ∈ CT , τ ∈ CT1 with τ 0 (t ) < 1, for all t ∈ [0, T ], then g (µ(s)) ∈ CT , where µ(t ) is the inverse function of t − τ (t ). Now, We recall Mawhin’s continuation theorem which our study is based upon. Let X and Y be real Banach spaces and L : D(L) ⊂ X → Y be a Fredholm operator with index zero; here D(L) denotes the domain of L. This means that Im L is closed in Y and L dim ker L = dim(Y L /Im L) < +∞. Consider the supplementary subspaces X1 , Y1 of X , Y respectively, such that X = ker L X1 and Y = Im L Y1 , and let P : X → ker L and Q : Y → Y1 be natural projectors. Clearly, ker L ∩ (D(L) ∩ X1 ) = {0}, thus the restriction LP := L|D(L)∩X1 is invertible. Denote by K the inverse of LP . Now, let Ω be an open bounded subset of X with D(L) ∩ Ω 6= ∅. A map N : Ω → Y is said to be L-compact in Ω , if QN (Ω ) is bounded and the operator K (I − Q )N : Ω → X is compact.

740

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

Lemma 2.7 (Gaines and Mawhin [7]). Suppose that X and Y are two Banach spaces, and L : D(L) ⊂ X → Y is a Fredholm operator with index zero. Furthermore, Ω ⊂ X is an open bounded set and N (λ, x) : [0, 1] × Ω → Y is L-compact on [0, 1] × Ω . If: (1) Lx 6= λN (λ, x), for all x ∈ ∂ Ω ∩ D(L), λ ∈ (0, 1); (2) N (0, x) 6∈ Im L, for all x ∈ ∂ Ω ∩ ker L; (3) deg(JQN (0, ·), Ω ∩ ker L, 0) 6= 0, where J : Im Q → ker L is an isomorphism; then the equation Lx = N (1, x) has a solution in Ω ∩ D(L). In order to use Lemma 2.7 to study the existence of T -periodic solutions for Eq. (1.1), suppose that

Pn

i=1

ci0 < 1 or there c0

i is an integer k ∈ {1, 2, . . . , n} such that ck = mint ∈[0,T ] |xk (t )| > 1 and δ ck0 < (1 − δ)ck where δ = c1 + i= 6 k ck , so the k problem of existence of T -periodic solutions to Eq. (1.1) can be converted to the corresponding problem of the following system:

P

0 q −2 [Ax1 ] (t ) = ϕq (x2"(t )) = |x2 (t )| x2 (t ), !# n n X X 0 −1 0 x ( t ) = f ( x ( t )) A ϕ ( x (·)) + c (·) x (· − r ) ( t ) + α( t ) g ( x ( t )) + βj (t )g (x1 (t − γj (t ))) + e(t ). 1 q 2 1 i 1 i 2 i=1

j =1

(2.5) where q > 1 is a constant with + = 1. In fact, if x(·) = (x1 (·), x2 (·)) is a T -periodic solution to Eqs. (2.5), then x1 (t ) must be a T -periodic solution to Eq. (1.1). Thus, in order to prove that Eq. (1.1) has a T -periodic solution, it suffices to show that Eqs. (2.5) has a T -periodic solution. Now, we set 1 p

1 q

>

X = Y = {x = (x1 (·), x2 (·)) ∈ C (R, R2 ) : x1 ∈ CT , x2 ∈ CT } with norm kϕk = max{|ϕ1 |0 , |ϕ2 |0 }, for all ϕ ∈ X or Y . Clearly, X and Y are two Banach spaces. Meanwhile, define a linear operator

L : D(L) ⊂ X → Y ,

Lx =

(Ax1 )0

0

x2

,

(2.6)

where D(L) = {x = (x1 (·), x2 (·)) ∈ C 1 (R, R2 ) : x1 ∈ CT1 , x2 ∈ CT1 }, and a nonlinear operator N (λ, x) : [0, 1] × X → Y by setting

ϕq (x2 (t )) !# n X f (x1 (t )) A−1 λ2 ϕq (x2 (·)) + λ ci0 (·)x1 (· − ri ) (t ) [N (λ, x)](t ) = . i=1 n X +α(t )g (x1 (t )) + βj (t )g (x1 (t − γj (t ))) + e(t )

"

(2.7)

j =1

It is easy to see that Eqs. (2.5) can be converted to the abstract equation Lx = N (1, x). Moreover, for all x = (x1 (·), x2 (·)) ∈ ker L, we have

n [Ax ](t ) = x (t ) − X c (t )x (t − r ) ≡ a , 1 1 j 1 j 1 j=1 x ( t ) ≡ a , 2 2 where a1 , a2 ∈ R are constants. Let ϕ(t ) be a solution of x1 (t ) − j=1 cj (t )x1 (t − rj ) ≡ 1, then from Lemma 2.4 or Lemma 2.5 we know that ϕ(t ) 6= 0 for all t ∈ R. Therefore ker L = {(a1 ϕ(t ), a2 ) : a1 , a2 ∈ R}. Clearly

Pn

Im L =

Z

y∈Y :

T

y(s)ds = 0 .

0

So Im L is closed in CT and dim kerL = codim ImL = 2, then the operator L is a Fredholm operator with index zero. Also let projectors P : X → ker L and Q : Y → ImQ be defined by

ϕ(t )

Z

T Px = 1Z T

0 T

T

x2 (t )dt 0

x1 (t )dt

,

Qy =

1 T

T

Z

y(s)ds. 0

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

741

Obviously, ker L = Im P , ker Q = Im L. Set operator Kp represents the inverse of L|D(L)∩ker P , then T

Z Kp : Im L → D(L) ∩ ker P ,

[Kp y](t ) =

0

[A−1 (G(t , ·)y1 (·))](s)ds , Z T G(t , s)y2 (s)ds

(2.8)

0

From (2.7) and (2.8), one can easily see that N (λ, ·) is L-compact on [0, 1] × Ω , where Ω is an open bounded subset of X . 3. Main results Throughout this paper, we assume that γj ∈ CT1 and γj0 (t ) < 1, for all t ∈ [0, T ], (j = 1, 2, . . . , n). So the function t − γj (t ) has a unique inverse denoted by µj (t ), (j = 1, 2, . . . , n). We also denote h=

1

T

Z

T

|h(s)|ds,

e h=

for all h ∈ CT .

0

0

Γ (t ) = α(t ) +

T

Z

h(s)ds,

β(µj (t ))

n X

1 − γj0 (µj (t )) j=1

,

n X β(µj (t )) Γ1 (t ) = |α(t )| + . 1 − γj0 (µj (t )) j =1 For the sake of convenience, we list the following conditions which will be used to study Eq. (1.1) in this section.

[H1 ] eΓ (t ) < 0, t ∈ [0, T ]. [H2 ] There is a constant d > 0 such that e x g (x) + > 0, for all x ∈ R with |x| > d. Γ [H3 ] There are nonnegative constants a and b such that |g (x)| |F (x)| lim sup p−1 ≤ b, lim sup p−1 ≤ a, and x→−∞ |x|→+∞ |x| |x| Rx where F (x) = 0 f (s)ds. Pn Theorem 3.1. Suppose that assumptions [H1 ]–[H3 ] hold, and i=1 ci0 < 21 . Then Eq. (1.1) has at least one T-periodic solution, if

n n p n P P 1 P eb 1 + ci0 + T (2T )p−1 a (ci0 + ci1 T ) + 2 ΓΓ1 0 Γ ci i=1 i=1 i=1 < 1. p n P 1− ci0 i =1

Proof. Let Ω1 = {x : Lx = λN (λ, x), λ ∈ (0, 1)}. If x(·) = (u(·), v(·))> ∈ Ω1 , then from (2.6) and (2.7)

[Au]0 (t ) = λϕq (v("t )) = λ|v(t )|q−2 v(t ), !# n X 0 v 0 (t ) = λf (u(t )) A−1 λ2 ϕq (v(·)) + λ ci (·)u(· − ri ) (t ) + λα(t )g (u(t )) i =1

(3.1)

n X βj (t )g (u(t − γj (t ))) + λe(t ). +λ j =1

From the first equation of (3.1), we have v(t ) = ϕp ( λ1 [Au]0 (t )), which together with the second equation of (3.1) yields

ϕp

1

λ

[Au]0 (t )

0

= λ2 f (u(t ))u0 (t ) + λα(t )g (u(t )) + λ

n X

βj (t )g (u(t − γj (t ))) + λe(t ).

(3.2)

j =1

Integrating two sides of Eq. (3.2) on the interval [0, T ], we obtain T

Z

α(t )g (u(t ))dt + 0

Z 0

T

n X j=1

βj (t )g (u(t − γj (t )))dt +

T

Z

e(t )dt = 0. 0

(3.3)

742

As

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

RT 0

R T −γj (T )

βj (t )g (u(t − γj (t )))dt =

−γj (0)

T

Z

T

Z

βj (t )g (u(t − γj (t )))dt =

0

0

β(µj (s)) g 1−γj0 (µj (s))

(u(s))ds, by applying Lemma 2.6, we know that

β(µj (s)) g (u(s))ds, 1 − γj0 (µj (s))

β(µj (t )) 1−γj0 (µj (t ))

∈ CT , So

(j = 1, 2, . . . , n).

So it follows from (3.3) that T

Z

Γ (t )g (u(t ))dt = −eT .

(3.4)

0

By using integral mean value theorem, we have that there is t1 ∈ [0, T ] such that g (u(t1 ))Γ T = −eT , i.e., g (u(t1 )) = −

e

Γ

,

and then by using assumption [H2 ], we obtain |u(t1 )| ≤ d. Therefore T

Z

|u0 (t )|dt .

|u|0 ≤ d +

(3.5)

0

On the other hand, by multiplying two sides of Eq. (3.2) with [Au](t ) and integrating them on [0, T ], we obtain T

Z

|[Au] (t )| dt = −λ 0

p

p+1

0

"

T

Z

f (u(t ))u (t ) u(t ) − 0

0

−λ

p

" α(t )g (u(t )) u(t ) −

0

−λ

p

n Z X

−λ

T

# ci (t )u(t − ri ) dt

βj (t )g (u(t − γj (t ))) u(t ) − 0

n X

# ci (t )u(t − ri ) dt

i=1

" e(t ) u(t ) −

n X

0

Z ≤

n X

"

T

T

Z

ci (t )u(t − ri ) dt

i=1

j=1 p

#

i =1

T

Z

n X

# ci (t )u(t − ri ) dt

i=1

X n Z T 0 f (u(t ))u (t )u(t )dt + f (u(t ))u (t )ci (t )u(t − ri )dt 0 i =1 ! "Z # n n Z T T X X 0 1+ ci |u|0 |α(t )||g (u(t ))|dt + |βj (t )||g (u(t − γj (t )))|dt 0

0

+

0

i =1

+

1+

n X

j =1

0

! ci0

|u|0e e.

(3.6)

i =1

Clearly,

RT

f (u(t ))u0 (t )u(t )dt = 0. In view of

0

(2T )

p−1

a

n P

(

i =1

ci0

+

ci1 T

n p n Γ P P 1 0 1 e ) + 2 Γ 0 Γ b 1 + ci + T ci i=1 i =1 < 1, p n P 1− ci0 i=1

it is easy to see that there is a small constant ε > 0 such that

n p n n P P P 1 e(b + ε) 1 + ci0 + T (2T )p−1 (a + ε) (ci0 + ci1 T ) + 2 ΓΓ1 0 Γ ci i=1 i =1 i=1 < 1. p n P 0 1− ci i =1

(3.7)

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

743

For such an ε , from assumption [H3 ], one can find that there is a constant ρ > d such that

|F (x)| ≤ (a + ε)|x|p−1 ,

for all |x| > ρ,

(3.8)

|g (x)| ≤ (b + ε)|x|

for all x < −ρ.

(3.9)

p−1

,

Let E1 = {t : t ∈ [0, T ], |u(t )| > ρ} and E2 = {t : t ∈ [0, T ], |u(t )| ≤ ρ}. As T

Z

0

T

Z

f (u(t ))u0 (t )ci (t )u(t − ri )dt =

F (u(t ))[ci (t )u0 (t − ri ) + ci0 (t )u(t − ri )]dt

0

T

Z

|F (u(t ))|[ci0 |u0 (t − ri )| + ci1 |u(t − ri )|]dt Z Z = + |F (u(t ))|[ci0 |u0 (t − ri )| + ci1 |u(t − ri )|]dt ≤

0

E1

E2

Z T [ci0 |u0 (t − ri )| + ci1 |u(t − ri )|]dt |u(t )|p−1 [ci0 |u0 (t − ri )| + ci1 |u(t − ri )|]dt + Fρ 0 0 Z T Z T p−1 p 0 0 1 0 ≤ (a + ε) ci |u|0 |u (t )|dt + ci |u|0 T + Fρ ci |u0 (t )|dt + ci1 Fρ |u|0 T 0 0 " p−1 Z p # Z Z ≤ (a + ε)

T

Z

T

≤ (a + ε) ci0 d +

T

|u0 (t )|dt

0

+ Fρ ci0

T

Z

|u0 (t )|dt

0

0

T

Z |u0 (t )|dt + ci1 Fρ T d +

0

≤ (a + ε)(

T

|u0 (t )|dt + ci1 T d +

|u0 (t )|dt

0

ci0

+

ci1 T

) d+

T

Z

|u (t )|dt 0

p

+ Fρ (ci0 + ci1 T )

T

Z

0

|u0 (t )|dt + ci1 Fρ dT ,

(3.10)

0

where Fρ = max|x|≤ρ |F (x)|, (i = 1, 2, . . . , n). Let

∆1 = {t : t ∈ [0, T ], u(t ) < −ρ},

∆2 = {t : t ∈ [0, T ], |u(t )| ≤ ρ},

∆3 = {t : t ∈ [0, T ], u(t ) > ρ}.

Then by (3.4) we obtain

Z

Z

Z + ∆1

+ ∆2

∆3

Γ (t )g (u(t ))dt = −

T

Z

e(t )dt . 0

Since

Z ∆3

Z |Γ (t )||g (u(t ))|dt =

∆3

Z Γ (t )g (u(t ))dt = −

Z + ∆1

∆2

Γ (t )g (u(t ))dt −

Z 0

T

e(t )dt .

It follows from (3.9) that T

Z

Z

|Γ (t )||g (u(t ))|dt =

Z

Z

+ |Γ (t )||g (u(t ))|dt ∆ Z 3 + |Γ (t )||g (u(t ))|dt +e e

+ ∆

0

∆2

Z1 ≤2 ∆1

∆2 p−1 e

e +e e, ≤ 2(b + ε)|u|0 Γ + 2gρ Γ where gρ = max|x|≤ρ |g (x)|. Therefore T

Z

|α(t )||g (u(t ))|dt + 0

n Z X j =1

T

|βj (t )||g (u(t − γj (t )))|dt = 0

T

Z 0

"

# n X |βj (µj (t ))| |α(t )| + |g (u(t ))|dt 1 − γj0 (µ(t )) j =1

T

Z

Γ1 (t )|g (u(t ))|dt

= 0

Γ1 ( t ) |Γ (t )||g (u(t ))|dt 0 |Γ (t )| Γ1 Γ1 Γ1 p−1 e e ≤ 2 Γ (b + ε)|u|0 + 2gρ Γ +e e . Γ 0 Γ 0 Γ 0

Z

T

=

(3.11)

744

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

Substituting (3.10) and (3.11) into (3.6), we have

Z n X 0 1 |[Au] (t )| dt ≤ (ci + ci T )(a + ε) d +

T

Z

0

T

|u (t )|dt 0

p

0

T

Z

|u (t )|dt + 1 + 0

Pn

i =1

|[Au ](t )| dt ≤ p

0

|[Au] (t )| + 0

|[Au] (t )| dt + 2 0

≤2

p−1

p

n X

n X

T

0

p−1

|[Au] (t )| dt + 2 0

≤2

p−1

p

n X

T

0

|u|0

ci1

i =1

p−1

+2 RT

Case 2. If

RT

0 0

ci0

! Z e e d+

T

|u0 (t )|dt .

(3.12)

0

Pn

i=1

ci0 (t )u(t − ri ). Then by (3.12) we

!p ci1

|u|p0 !p

Z

T

d+

|u (t )|dt 0

p

0

i=1

"

Case 1. If

Fρ ci1 T d

i =1

dt

n n X X (a + ε) (ci0 + ci1 T ) + 2 1 + ci0

≤2

n X

!p ci1

" p−1

|u0 (t )|dt +

i=1 T

Z

)

i=1

T

Z

T

Z 0

ci (t )u(t − ri ), u ∈ CT1 , we have [Au0 ](t ) = [Au]0 (t ) +

0

p−1

n X i=1

T

Z

+

ci1 T

i=1

0

× d+

0

(

Fρ ci0

! p Z T Γ1 0 0 e |u (t )|dt 1+ ci 2 Γ (b + ε) d + Γ 0 0 i=1 ! n X Γ1 Γ1 0 e +e 1+ ci 2gρ Γ e Γ 0 Γ 0 i=1

+

T

n X

n X

+

Z

+

0

i=1

From [Au](t ) = u(t ) − obtain

p

i =1

!p # ! p Z T n X Γ1 1 0 Γ e(b + ε) + T c d + | u ( t )| dt i Γ 0 0 i=1

! # Z T n n X X Γ1 Γ1 0 1 0 e +e e +e e d+ |u0 (t )|dt . Fρ (ci + ci T ) + 1 + ci 2gρ Γ Γ 0 Γ 0 0 i=1 i=1

(3.13)

|u0 (t )|dt = 0, from (3.5) we have |u|0 ≤ d. |u0 (t )|dt > 0, then we know that T

Z

|u0 (t )|dt

d+

p

T

Z

|u0 (t )|dt

=

0

p 1+ RT

0

0

!p

d

.

|u0 (t )|dt

By the knowledge of mathematical analysis, there is a σ > 0, such that

If R T 0

(1 + x)p < 1 + (p + 1)x, for all x ∈ (0, σ ]. RT d > σ , then 0 |u0 (t )|dt < σd , so from (3.5) we obtain |u|0 ≤ d + σd . 0

(3.14)

|u (t )|dt

If R T 0

d

|u0 (t )|dt

≤ σ , then by (3.14) we have T

Z

|u (t )|dt 0

d+ 0

p

T

Z

|u (t )|dt 0

≤

p

!

(p + 1)d

1+ RT

|u0 (t )|dt Z T p Z T p−1 = |u0 (t )|dt + (p + 1)d |u0 (t )|dt 0

0

0

≤T

p−1

0

Z 0

T

|u (t )| dt + (p + 1)dT 0

p

(p−1)2

T

Z

|u (t )| dt 0

p

0

p

p−p 1

,

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

745

which together with (3.13), we obtain

"

T

Z

|[Au ](t )| dt ≤ 2 0

(a + ε)

p−1

p

n X

0

(

ci0

+

ci1

)+2 1+

n X

i=1

× T p−1

i=1

T

Z

|u0 (t )|p dt + (p + 1)dT

(p−1)2 p

! !p # n X Γ1 1 Γ e(b + ε) + T ci Γ 0 i=1 p−p 1 Z T |u0 (t )|p dt

ci0

0

0

"

n n X X Fρ (ci0 + ci1 T ) + 1 + ci0

+ 2p−1

i =1

× ( 1 + σ )T

!

i=1 T

Z

p−1 p

|u0 (t )|p dt

1p

# Γ1 Γ1 e +e 2gρ Γ e +e e Γ 0 Γ 0

.

(3.15)

0

By applying the third part of Lemma 2.2, we get T

Z

|u (t )| dt = 0

p

0

RT

T

Z

0 |[A Au ](t )| dt ≤

−1

0

0

p

|[Au0 ](t )|p dt p . n P 0 1− ci i=1

So it follows from (3.15) that T

Z

T

Z

|u (t )| dt ≤ C1 0

p

0

|u (t )| dt + C2 0

p

0

T

Z

|u (t )| dt 0

p

1− 1p

T

Z

|u (t )| dt 0

+ C3

p

1p

,

(3.16)

0

0

where

n p n n P P P 1 e(b + ε) + T (2T )p−1 (a + ε) (ci0 + ci1 T ) + 2 1 + ci0 ΓΓ1 0 Γ ci i =1 i=1 i=1 , C1 = p n P 1− ci0 i =1

n p n n Γ (p−1)2 P 1 P P e(b + ε) + T ci (p + 1)dT p 2p−1 (a + ε) (ci0 + ci1 T ) + 2 1 + ci0 Γ1 0 Γ i =1 i=1 i =1 , C2 = p n P 1− ci0 i=1 n n Γ P P 1− 1 e +e e Γ1 0 +e e (1 + σ )T p 2p−1 Fρ (ci0 + ci1 T ) + 1 + ci0 2gρ ΓΓ1 0 Γ i =1 i=1 C3 = . p n P 0 1− ci i =1

From (3.7), we see that C1 < 1, which together with p > 1 and (3.16) implies that there is a constant M > 0 such that T

Z

|u0 (t )|p dt ≤ M , 0

then it follows from (3.5) that

|u|0 ≤ d + T

p−1 p

T

Z

|u (t )| dt 0

p

1p ≤d+T

p−1 p

1

M p := M1 .

0

RT

Again from the first equation of (3.1), we have 0 |v(t )|q−2 v(t )dt = 0, which implies that there is a constant t2 ∈ [0, T ] such that v(t2 ) = 0, then from the second equation of (3.1), we have

Z t Z T 0 |v(t )| = v (s)ds ≤ |v 0 (s)|ds t2 0 Z T Z T ≤ |f (u(t ))||u0 (t )|dt + Γ1 (t )|g (u(t ))|dt +e e 0

0

746

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

Z

T

e1 +e e |u0 (t )|dt + gM1 Γ

≤ f M1 0 p−1

≤ f M1 T

p

1

e1 +e e := M2 , M p + gM1 Γ

for all t ∈ [t2 , t2 + T ].

i.e., |v|0 ≤ M2 . Let Ω2 = {x : x ∈ ker L, QN (0, x) = 0}. If x(·) = (x1 (·), x2 (·))> ∈ Ω2 , then

x1 (t ) = a1 ϕ(t ), x2 (t ) ≡ a2 .

for all a1 , a2 ∈ R;

and

1 T

QN (0, x) = 1 Z

T

T

Z

T

|a2 |q−2 a2 dt 0

[Γ (t )g (a1 ϕ(t )) + e(t )]dt

= 0,

0

i.e.,

a2 Z≡ 0, T

1

T

[Γ (t )g (a1 ϕ(t )) + e(t )]dt = 0.

0

Therefore, there is a point ξ ∈ [0, T ] such that g (a1 ϕ(ξ ))Γ = −e, then from assumption [H2 ], we have |a1 ϕ(ξ )| ≤ d, so |a1 | ≤ |ϕ(ξd )| ≤ min d |ϕ(t )| := D1 . Thus |x1 |0 ≤ maxt ∈[0,T ] |a1 ϕ(t )| ≤ |ϕ|0 D1 := D2 , |x2 |0 ≡ 0 < M2 . t ∈[0,T ]

Now, if we set Ω = {x : x = (u, v)> ∈ X , |u|0 < M1 + D2 + 1, |v|0 < M2 + 1}, then Ω1 ∪ Ω2 ⊂ Ω . So conditions (1) and (2) of Lemma 2.7 are satisfied. Now we prove that condition (3) of Lemma 2.7 is satisfied. For all x ∈ Ω ∩ ker L, λ ∈ [0, 1], define H (x, λ) = λx + (1 − λ)JQN (0, x), where J : Im Q → ker L is a homeomorphism with J (x1 , x2 ) = ( H (x, λ) 6= 0,

ϕ(t ) x2 , x1 ). Γ

Hence

for all (x, λ) ∈ ∂ Ω ∩ ker L × [0, 1].

And then by the degree theory deg{JQN (0, ·), Ω ∩ ker L, 0} = deg{H (·, 0), Ω ∩ ker L, 0}

= deg{H (·, 1), Ω ∩ ker L, 0} = deg{I , Ω ∩ ker L, 0} 6= 0. Applying Lemma 2.7, we reach the conclusion. By using Lemmas 2.3 and 2.5, we can obtain the following result by a similar way as in the proof of Theorem 3.1.

Theorem 3.2. Suppose that assumptions [H1 ] – [H3 ] hold, and there is an integer k ∈ {1, 2, . . . , n} such that ck = mint ∈[0,T ]

P c0 |ck (t )| > 1 and δ ck0 < (1 − δ)ck where δ = c1k + i6=k cik . Then Eq. (1.1) has at least one T-periodic solution, if n n p n P P P 1 eb 1 + ci0 + T (2T )p−1 a (ci0 + ci1 T ) + 2 ΓΓ1 0 Γ ci i =1 i=1 i =1 < 1. !p P 0 ck − 1 − ci i6=k

As application, we consider the following examples. Example 3.1. Considering the following equation:

0 0 1 1 ϕ3 u( t ) − sin 8π t u(t − 10) − cos 8π t u(t − 5) 64π 64π cos 8π t 2 0 = [u (t ) sin u(t )]u (t ) + (cos 8π t )g (u(t )) + (sin 8π t − 2)g u t + 16π 1 sin 8π t + cos 8π t − 1 g u t − + sin 8π t + 1. 3 24π

(3.17)

Z. Wang et al. / Nonlinear Analysis 72 (2010) 734–747

Corresponding to Eq. (1.1), we have p = 3, T =

, r1 = 10, r2 = 5, c1 (t ) = 8π t cos 8π t , β1 (t ) = sin 8π t − 2, β2 (t ) = cos 8π t − 1, γ1 (t ) = − cos , γ2 (t ) = 16π 2 so F (x) = −x cos x + 2x sin x + 2 cos x − 2 and a = 1. Furthermore, x2 ex , x ≥ 0; g (x) = x2 sin x − e , x < 0. 1 3

1 4

747

π t , c2 (t ) = 641π cos 8π t , α(t ) = , ( ) = sin 8π t + 1, f (x) = x2 sin x,

1 sin 8 64π sin 8π t e t 24π

2

So b =

1 2

, e = 1 and

Γ (t ) = cos 8π t +

sin(8π µ1 (t )) − 2 1−

1 2

sin(8π µ1 (t ))

+

1 3

cos(8π µ2 (t )) − 1

1−

1 3

cos(8π µ2 (t ))

= cos 8π t − 3 < 0,

8π t where µ1 (t ) and µ2 (t ) are the inverse of γ1 (t ) = − cos and γ2 (t ) = 16π

x g (x) + Γ¯

for all t ∈ [0, T ],

Γ

sin 8π t . 24π

e = 1 + 3, Therefore Γ = −3, Γ1 0 = 4, Γ 2π 4

> 0, for |x| > 1 and n p n n Γ P P 1 P 0 0 1 p−1 1 e ci + T ci (2T ) a (ci + ci T ) + 2 Γ 0 Γ b 1 + i=1 i=1 i=1 = 0.9658 < 1. p n P 0 1− ci e¯

i =1

Thus by applying Theorem 3.1, we know that Eq. (3.17) has at least one 14 -periodic solution. Example 3.2. Considering the following equation:

ϕ3

u(t ) −

1 4π

sin 8π t + 4 u(t − 10) −

1 4π

cos 8π t

0 0

u( t − 5 )

cos 8π t = [u2 (t ) sin u(t )]u0 (t ) + (cos 8π t )g (u(t )) + (sin 8π t − 2)g u t + 16π 1 sin 8π t + cos 8π t − 1 g u t − + sin 8π t + 1. 3 24π Observing c1 (t ) = 41π sin 8π t + 4, c2 (t ) = 41π cos 8π t , T = (1 − δ)ck , moreover,

1 , and c1 4

(3.18)

= mint ∈[0,T ] |c1 (t )| = 4 − 41π > 1, δ = 0.3697, δ ck0 <

n n p n P P P 1 eb 1 + ci0 + T (2T )p−1 a (ci0 + ci1 T ) + 2 ΓΓ1 0 Γ ci i=1 i=1 i=1 = 0.4353 < 1. !p P 0 ck − 1 − ci i6=k

Similar to Example 3.1 we know that Eq. (3.18) has at least one 14 -periodic solution by applying Theorem 3.2. Acknowledgment The authors are very grateful to the referee for the valuable suggestions concerning the improvement of this paper. References [1] M.R. Zhang, Periodic solutions of linear and quasilinear neutral functional differential equations, J. Math. Anal. Appl. 189 (1995) 378–392. [2] Shiping Lu, Weigao Ge, On the existence of periodic solutions for neutral functional differential equation, Nonlinear Anal. 54 (2003) 1285–1306. [3] Shiping Lu, Weigao Ge, Periodic solutions for a kind of second-order neutral functional differential equational differential equation, Appl. Math. Comput. 157 (2004) 443–448. [4] Shiping Lu, Existence of periodic solutions for a p-Laplacian neutral functional differential equation, Nonlinear Anal. 70 (2009) 231–243. [5] Shiping Lu, On the existence of positive periodic solutions for neutral functional differential equation with multiple deviating arguments, J. Math. Anal. Appl. 280 (2003) 321–333. [6] Bo Du, Liliang Guo, Weigao Ge, Shiping Lu, Periodic solutions for generalized Liénard neutral equation with variable parameter, Nonlinear Anal. 70 (2009) 2387–2394. [7] R.E. Gaines, J.L. Mawhin, Coincidence Degree and Nonlinear Differential Equations, Springer Verlag, Berlin, 1977.

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