cha1.dvi


CUBO A Mathematical Journal

Vol.12, No¯ 03, (153–165). October 2010

Existence of Periodic Solutions for a Class of

Second-Order Neutral Differential Equations with

Multiple Deviating Arguments1

CHENGJUN GUO

School of Applied Mathematics, Guangdong,

University of Technology 510006, P.R.China

email: guochj817@163.com

DONAL O’REGAN

Department of mathematics, National University of Ireland,

Galway, Ireland

email: donal.oregan@nuigalway.ie

AND

RAVI P. AGARWAL

Department of Mathematical Sciences, Florida Institute of Technology,

Melbourne, Florida 32901, USA

email: agarwal@fit.edu

ABSTRACT

Using Kranoselskii fixed point theorem and Mawhin’s continuation theorem we establish

the existence of periodic solutions for a second order neutral differential equation with

multiple deviating arguments.

1This project is supported by grant 10871213 from NNSF of China, by grant 06021578 from NSF of Guangdong.



154 Chengjun Guo, Donal O’Regan & Ravi P. Agarwal
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12, 3 (2010)

RESUMEN

Usando el teorema del punto fijo de Kranoselskii y el teorema de continuación de Mawhin

establecemos la existencia de soluciones periódicas de una ecuación diferencial neutral de

segundo orden con argumento de desviación multiple.

Key words and phrases: Periodic solution, Multiple deviating arguments, Neutral differen-

tial equation, Kranoselskii fixed point theorem, Mawhin’s continuation theorem.

Math. Subj. Class.: 34K15; 34C25.

1 Introduction

In this paper, we discuss the second-order neutral differential equation with multiple deviat-

ing arguments of the form

x
′′
(t) + cx

′′
(t −τ) + a(t)x(t) + g(t, x(t −τ1 (t)), x(t −τ2 (t))··· , x(t −τn (t))) = p(t), (1.1)

where |c| < 1, τ is a constant, τi (t)(i = 1, 2,··· , n), a(t) and p(t) are real continuous functions
defined on R with positive period T and g(t, x1 , x2,··· , xn) ∈ C(R × R × R × ··· × R, R) and is
T−periodic in t.

Periodic solutions for differential equations were studied in [2-4, 6-10, 12, 15] and we

note that most of the results in the literatue concern delay problems. There are only a few

papers[1, 5, 11, 13, 14] which discuss neutral problems.

For the sake of completeness, we first state Kranoselskii fixed point theorem and Mawhin’s

continuation theorem [3].

Theorem A (Kranoselskii). Suppose that Ω is a Banach space and X is a bounded, convex

and closed subset of Ω. Let U, S : X →Ω satisfy the following conditions:

(1) U x + S y ∈ X for any x, y ∈ X ;

(2) U is a contraction mapping;

(3) S is completely continuous.

Then U + S has a fixed point in X .

Let X and Y be two Banach space and L : D omL ⊂ X −→ Y is a linear mapping and
N : X −→ Y is a continuous mapping. The mapping L will be called a Fredholm mapping of
index zero if d imK erL = cod imI mL < +∞, and I mL is closed in Y. If L is a Fredholm map-
ping of index zero, there exist continuous projectors P : X −→ X and Q : Y −→ Y such that



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Existence of Periodic Solutions for a Class ... 155

I mP = K erL and I mL = K erQ = I m(I − Q). It follows that L|D omL∩K erP : (I − P)X −→ I mL
has an inverse which will be denoted by KP . If Ω is an open and bounded subset of X, the

mapping N will be called L−compact on Ω if Q N(Ω) is bounded and KP (I − Q)N(Ω) is com-
pact. Since I mQ is isomorphic to K erL, there exists an isomorphism J : I mQ −→ K erL.

Theorem B (Mawhin’s continuation theorem[3]). Let L be a Fredholm mapping of in-

dex zero, and let N be L−compact on Ω. Suppose

(1) for each λ ∈ (0, 1) and x ∈ ∂Ω, Lx 6= λN x and

(2) for each x ∈ ∂Ω∩ K er(L), Q N x 6= 0 and d e g(Q N,Ω∩ K er(L), 0) 6= 0.

Then the equation Lx = N x has at least one solution in Ω∩ D(L).

2 Main Results

Now we make the following assumption on a(t):

(H1) (
π
T

)2 > M = max t∈[0,T] a(t) ≥ a(t) ≥ m = mint∈[0,T] a(t) > 0.

Our main results are the following theorems.

Theorem 2.1 Suppose (H1) holds and also assume there exists a constant K1 > 0 such that
(H2)

‖g‖0 ≤ m − 3|c|M −‖p‖0 ,

where ‖g‖0 = max {t∈[0,T],|x1|≤K1 ,··· ,|xn|≤K1}|g(t, x1 , x2,··· , xn)| and ‖p‖0 = max t∈[0,T]|p(t)|.
Then Eq.(1.1) possesses a nontrivial T−periodic solution.

Theorem 2.2 Suppose (H1) holds and also assume

(H3)

|g(t, x1 , x2,··· , xn)| ≤ γ
∑

n
i=1 |xi|.

Then Eq.(1.1) has at least one T−periodic solution as 0 < γ < 1
n

[(1 −|c|)m −|c|M].

In order to prove the main theorems we need some preliminaries. Set

X := {x|x ∈ C2(R, R), x(t + T) = x(t),∀t ∈ R}

and x(0)(t) = x(t) and define the norm on X as follows

||x|| = max t∈[0,T]|x(t)|+ max t∈[0,T]|x
′
(t)|+ max t∈[0,T]|x

′′
(t)|.



156 Chengjun Guo, Donal O’Regan & Ravi P. Agarwal
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Remark 2.3 If x ∈ X , then it follows that x(i)(0) = x(i)(T)(i = 0, 1, 2).

In order to prove our main results, we need the following Lemma [10].

Lemma 2.4 ([10]). Suppose that M is a positive number and satisfies 0 < M < ( π
T

)2. Then

for any function ϕ defined in [0, T], the following equation





x
′′
(t) + M x(t) = ϕ(t),

x(0) = x(T), x
′
(0) = x

′
(T)

has a unique solution

x(t) =
´ T

0
G(t, s)ϕ(s)ds,

where

G(t, s)





w(t − s), (k − 1)T ≤ s ≤ t ≤ kT

w(T + t − s), (k − 1)T ≤ t ≤ s ≤ kT(k ∈ N),

w(t) =
cos α(t− T

2
)

2αsin αT
2

and α =
p

M. Here

max t∈[0,T]
´

T

0
|G(t, s)|ds = 1

M
.

Proof of Theorem 2.1: For ∀x ∈ X , define the operators U : X −→ X and S : X −→ X respec-
tively by

(U x)(t) = −cx(t −τ) (2.1)

and

(Sx)(t) = cx(t −τ) +
´ T

0
G(t, s)[−cx

′′
(s −τ)(M − a(s))x(s) + p(s)

−g(s, x(s −τ1 (s)), x(s −τ2 (s))··· , x(s −τn (s)))]ds.
(2.2)

It is clear that a fixed point of U + S is a T−periodic solution of Eq.(1.1).

We are going to demonstrate that U and S satisfy the conditions of Theorem A.

Let x, y ∈ X and |x| ≤ K1,|y| ≤ K1(here K1 is as in the statement of Theorem 2.1). Now we
prove that |U x + S y| ≤ K1 holds.

First, we have the following equality:

´

T

0
G(t, s)x

′′
(s −τ)ds = M

´

T

0
G(t, s)x(s −τ)ds. (2.3)



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Existence of Periodic Solutions for a Class ... 157

In fact, we have from Lemma 2.4

´ T

0
G(t, s)x

′′
(s −τ)ds =

´ t

0

cosα(t−s− T
2

)

2αsin Tα
2

d[x
′
(s −τ)] +

´ T

t

cos α(t−s+ T
2

)

2αsin Tα
2

d[x
′
(s −τ)]

=
cos α(t−s− T

2
)

2αsin Tα
2

x
′
(s −τ)|t

0
−α
´ t

0

sinα(t−s− T
2

)

2αsin Tα
2

d[x(s −τ)]

+
cos α(t−s+ T

2
)

2αsin Tα
2

x
′
(s −τ)|Tt −α

´

T

t

sin α(t−s+ T
2

)

2αsin Tα
2

d[x(s −τ)]

= −α[
sin α(t−s− T

2
)

2αsin Tα
2

x(s −τ)|t
0
+

sin α(t−s+ T
2

)

2αsin Tα
2

x(s −τ)|Tt ]

+α2[
´

t

0

cos α(t−s− T
2

)

2αsin Tα
2

x(s −τ)ds +
´

T

t

cos α(t−s+ T
2

)

2αsin Tα
2

x(s −τ)ds]

= M
´ T

0
G(t, s)x(s −τ)ds,

(2.4)

so (2.3) holds.

From (H1), (H2) and (2.1)-(2.3), we have

|(U y)(t) + (Sx)(t)| ≤ |(U y)(t)|+|(Sx)(t)|

≤ 2|c|K1 +|
´ T

0
G(t, s)(M − a(s))x(s) − cx

′′
(s −τ) + p(s)

−g(s, x(s −τ1 (s)), x(s −τ2 (s))··· , x(s −τn (s)))]ds|+|c|K1

≤ 2|c|K1 + M−mM K1 +
‖g‖0

M
+|c|M||

´ T

0
G(t, s)x(s −τ)ds|

≤ 3|c|K1 + M−mM K1 +
‖g‖0+‖p‖0

M

≤ K1, x, y ∈ X ,

(2.5)

where ‖g‖0 and ‖p‖0 are given in (H2).

Set

K2 =
ρ0[(M−m)K1+|c|K3+‖g‖0+‖p‖0]

1−2|c| , (2.6)

where ρ0 = T
2 sin Tα

2

,

K3 =
MK1+‖g‖0+‖p‖0

1−|c| (2.7)

and

G = {x ∈ X : |x(t)| ≤ K1,|x
′
(t)| ≤ K2,|x

′′
(t)| ≤ K3}.

It is clear that G is a bounded, convex and closed subset of X .

(1) For ∀x, y ∈ G, we will show that

| d
d t

[(U y)(t) + (Sx)(t)]| ≤ K2 (2.8)



158 Chengjun Guo, Donal O’Regan & Ravi P. Agarwal
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12, 3 (2010)

and

| d
2[(U y)(t)+(S x)(t)]

d t2
| ≤ K3. (2.9)

From (2.1) we have

d
d t

[(U x)(t)] = −cx
′
(t −τ) (2.10)

and
d2[(U x)(t)]

d t2
= −cx

′′
(t −τ). (2.11)

Also from Lemma 2.4 and (2.2) we have

d
d t

[(Sx)(t)] =
´

T

0
G t(t, s)[(M − a(s))x(s) − cx

′′
(s −τ) + p(s)

−g(s, x(s −τ1 (s)), x(s −τ2 (s))··· , x(s −τn (s)))]ds + cx
′′
(t −τ),

(2.12)

where

G t(t, s)





w̃(t − s), (k − 1)T ≤ s ≤ t ≤ kT

w̃(T + t − s), (k − 1)T ≤ t ≤ s ≤ kT(k ∈ N)

and

w̃(t) =
sin α(t− T

2
)

2 sin αT
2

,

since

d
d t

[(Sx)(t)] = {
´

T

0
G t(t, s)[(M − a(s))x(s) − cx

′′
(s −τ) + p(s)

−g(s, x(s −τ1 (s)), x(s −τ2 (s))··· , x(s −τn (s)))]ds + cx
′′
(t −τ)}

′

= {
´

t

0

cos α(t−s− T
2

)

2αsin Tα
2

[(M − a(s))x(s) − cx
′′
(s −τ) + p(s)

−g(s, x(s −τ1 (s)), x(s −τ2 (s))··· , x(s −τn (s)))]ds + cx
′′
(t −τ)}

′

+{
´

s

t

cos α(t−s+ T
2

)

2αsin Tα
2

[(M − a(s))x(s) − cx
′′
(s −τ) + p(s)

−g(s, x(s −τ1 (s)), x(s −τ2 (s))··· , x(s −τn (s)))]ds + cx
′′
(t −τ)}

′

= α{
´ t

0

cos α(t−s− T
2

)

2αsin Tα
2

[(M − a(s))x(s) − cx
′′
(s −τ) + p(s)

−g(s, x(s −τ1 (s)), x(s −τ2 (s))··· , x(s −τn (s)))]ds + cx
′′
(t −τ)}

+α{
´ s

t

cosα(t−s+ T
2

)

2αsin Tα
2

[(M − a(s))x(s) − cx
′′
(s −τ) + p(s)

−g(s, x(s −τ1 (s)), x(s −τ2 (s))··· , x(s −τn (s)))]ds + cx
′′
(t −τ)}.



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Note

´ T

0
|G t(t, s|ds ≤ T

2 sin αT
2

= ρ0

and

d2[(S x)(t)]

d t2
= p(t) − a(t)x(t) − g(t, x(t −τ1 (t)), x(t −τ2 (t))··· , x(t −τn (t))). (2.13)

From (2.6),(2.7) and (2.10)-(2.13), we have

| d
d t

[(U y)(t) + (Sx)(t)]| ≤ | d
d t

[(U y)(t)]|+| d
d t

[(Sx)(t)]|

≤ 2|c|K2 +ρ0[(M − m)K1 +|c|K3 +‖g‖0 +‖p‖0]

≤ K2

(2.14)

and

| d
2[(U y)(t)+(S x)(t)]

d t2
| = |(M − a(t))x(t) − c y

′′
(t −τ) + p(t)

−g(t, x(t −τ1 (t)), x(t −τ2 (t))··· , x(t −τn (t)))|

≤ (M − m)K1 +|c|K3 +‖g‖0 +‖p‖0

≤ K3.

(2.15)

From (2.5), (2.14) and (2.15), we have U x + S y ∈ G for ∀x, y ∈ G.

(2) U is a contraction mapping.

Let x, y ∈ G and we from (2.1) that

‖U x −U y‖ = max t∈[0,T]|cx(t −τ) − c y(t −τ)|+ max t∈[0,T]|cx
′
(t −τ) − c y

′
(t −τ)|

+max t∈[0,T]|cx
′′
(t −τ) − c y

′′
(t −τ)|

= |c|[max t∈[0,T]|x(t −τ) − y(t −τ)|+ max t∈[0,T]|x
′
(t −τ) − y

′
(t −τ)|

+max t∈[0,T]|x
′′
(t −τ) − y

′′
(t −τ)|]

= |c|‖x − y‖.

Since |c| < 1, U is a contraction mapping.

(3) S is completely continuous.

We can obtain the continuity of S from the continuity of a(t), p(t) and g(t, x(t−τ1 (t)), x(t−
τ2(t))··· , x(t − τn(t))) for t ∈ [0, T], x ∈ G. In fact, suppose that xk ∈ G and ‖xk − s‖ → 0 as



160 Chengjun Guo, Donal O’Regan & Ravi P. Agarwal
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k → +∞. Since G is closed convex subset of X , we have x ∈ G. Then

|Sxk − Sx| = c[xk(t −τ) − x(t −τ)] + c[xk (t −τ) − x(t −τ)]

+
´ T

0
G(t, s){(M − a(s))(xk (s) − x(s)) − c[x

′′

k
(s −τ) − x

′′
(s −τ)]

−[ g(s, xk (s −τ1(s)), xk (s −τ2(s))··· , xk (s −τn (s)))

−g(s, x(s −τ1 (s)), x(s −τ2 (s))··· , x(s −τn (s)))]}ds.

(2.16)

Using the Lebesgue dominated convergence theorem, we have from (2.12), (2.13) and (2.16)

that

limk→+∞ ‖Sxk − Sx‖ = 0.

Then S is continuous.

Next, we prove that Sx is relatively compact. It suffices to show that the family of

functions {Sx : x ∈ G} is uniformly bounded and equicontinuous on [0, T]. From (2.2), (2.12)
and(2.13), it is easy to see that {Sx : x ∈ G} is uniformly bounded and equicontinuity. Since S
is continuous and is relatively compact, S is completely continuous. By Theorem A (Kranosel-

skii fixed point theorem), we have a fixed point x of U + S. That means that x is a T−periodic
solution of Eq.(1.1).

In order to prove Theorem 2.2, we need some preliminaries. Set

Z := {x|x ∈ C1(R, R), x(t + T) = x(t),∀t ∈ R}

and x(0)(t) = x(t) and define the norm on Z as follows

||x|| = max {max t∈[0,T]|x(t)|, max t∈[0,T]|x
′
(t)|},

and set

Y := { y|y ∈ C(R, R), y(t + T) = y(t),∀t ∈ R}.

We define the norm on Y as follow ||y||0 = max t∈[0,T]|y(t)|. Thus both (Z,||·||) and (Y ,||·||0) are
Banach spaces.

Remark 2.5 If x ∈ Z, then it follows that x(i)(0) = x(i)(T)(i = 0, 1).

Define the operators L : Z −→ Y and N : Z −→ Y respectively by

Lx(t) = x
′′
(t), t ∈ R, (2.17)



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Existence of Periodic Solutions for a Class ... 161

and

N x(t) = −cx
′′
(t −τ) − a(t)x(t) + p(t)

−g(t, x(t −τ1 (t)), x(t −τ2 (t)),··· , x(t −τn (t))), t ∈ R.
(2.18)

Clearly,

K erL = {x ∈ Z : x(t) = c ∈ R} (2.19)

and

I mL = { y ∈ Y :
´

T

0
y(t)dt = 0} (2.20)

is closed in Y . Thus L is a Fredholm mapping of index zero.

Let us define P : Z → Z and Q : Y → Y /I m(L) respectively by

P x(t) = 1
T

´ T

0
x(t)dt = x(0), t ∈ R, (2.21)

for x = x(t) ∈ X and
Q y(t) = 1

T

´ T

0
y(t)dt, t ∈ R (2.22)

for y = y(t) ∈ Y . It is easy to see that I mP = K erL and I mL = K erQ = I m(I − Q). It follows
that L|D omL∩K erP : (I − P)Z −→ I mL has an inverse which will be denoted by KP .

Let Ω be an open and bounded subset of Z, we can easily see that Q N(Ω) is bounded and

KP (I − Q)N(Ω) is compact. Thus the mapping N is L−compact on Ω. That is, we have the
following result.

Lemma 2.6. Let L, N, P and Q be defined by (2.17), (2.18), (2.21) and (2.22) respectively.

Then L is a Fredholm mapping of index zero and N is L−compact on Ω, where Ω is any open
and bounded subset of Z.

In order to prove Theorem 2.2, we need the following Lemma [12].

Lemma 2.7 ([12 and Remark 2.5]). Let x(t) ∈ C(n)(R, R) ∩ CT . Then

||x(i)||0 ≤ 12
´ T

0
|x(i+1)(s)|ds, i = 1, 2,··· , n − 1,

where n ≥ 2 and CT := {x|x ∈ C(R, R), x(t + T) = x(t),∀t ∈ R}.

Now, we consider the following auxiliary equation

x
′′
(t) +cλx

′′
(t −τ) + a(t)λx(t) = λp(t)

−λg(t, x(t −τ1 (t)), x(t −τ2 (t)),··· , x(t −τn (t))),
(2.23)

where 0 < λ < 1.

Lemma 2.8. Suppose that conditions of Theorem 2.2 are satisfied. If x(t) is a T−periodic



162 Chengjun Guo, Donal O’Regan & Ravi P. Agarwal
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solution of Eq.(2.23), then there are positive constants D i (i = 0, 1), which are independent of
λ, such that

||x(i)||0 ≤ D i , t ∈ [0, T], i = 0, 1. (2.24)

Proof: Suppose that x(t) is a T−periodic solution of (2.23). We have from (H3) and (2.23) that

|x
′′
(t)| ≤ max t∈[0,T]|c||x

′′
(t)|+ M||x||0 +||p||0 +γn||x||0 . (2.25)

From (2.25), we have

max t∈[0,T]|x
′′
(t)| ≤ 1

1−|c| [(M +γn)||x||0 +||p||0 ]. (2.26)

On the other hand, from Lemma 2.4 and (2.23), we get

x(t) =
´

T

0
G̃(t, s)λ[(M − a(s))x(s) + p(s) − cx

′′
(s −τ)

−g(s, x(s −τ1 (s)), x(s −τ2 (s)),··· , x(s −τn (s))]ds,
(2.27)

where

G̃(t, s)





w̃(t − s), (k − 1)T ≤ s ≤ t ≤ kT

w̃(T + t − s), (k − 1)T ≤ t ≤ s ≤ kT(k ∈ N),
(2.28)

w̃(t) =
cosα1(t− T2 )

2α1 sin
α1 T

2

, (2.29)

α1 =
p
λM and

max t∈[0,T]
´

T

0
|G̃(t, s)|ds = 1

λM
. (2.30)

From (H3), (2.27) and (2.30), we have

‖x‖0 = max t∈[0,T]|
´ T

0
G̃(t, s)λ[(M − a(s))x(s) + p(s) − cx

′′
(s −τ)

−g(s, x(s −τ1 (s)), x(s −τ2 (s)),··· , x(s −τn (s))]ds|

≤ 1
M

[(M − m)‖x‖0 +‖p‖0 +|c|max t∈[0,T]|x
′′
(t)|+γn‖x‖0 ].

(2.31)

From (2.31), we have

‖x‖0 ≤
|c|max t∈[0,T]|x

′′
(t)|+‖p‖0

m−γn .
(2.32)

Thus combining (2.26) and (2.32), we see that

max t∈[0,T]|x
′′
(t)| ≤ M+m

m(1−|c|)−M|c|−γn = ξ (2.33)

and

‖x‖0 ≤
|c|ξ+‖p‖0

m−γn = D0. (2.34)



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Existence of Periodic Solutions for a Class ... 163

Finally from Lemma 2.4, (2.33) and (2.34), we get

||x
′
||0 ≤ D1. (2.35)

The proof of Lemma 2.8 is complete.

Proof of Theorem 2.2: Suppose that x(t) is a T-periodic solution of Eq.(2.23). By Lemma

2.8, there exist positive constants D i(i = 0, 1) which are independent of λ such that (2.24) is
true. Consider any positive constant D > max 0≤i≤1{D i } +‖p‖0 .

Set

Ω := {x ∈ Z : ||x|| < D}.

We know that L is a Fredholm mapping of index zero and N is L-compact on Ω(see [3]).

Recall

K er(L) = {x ∈ Z : x(t) = c ∈ R}

and the norm on Z is

||x|| = max {max t∈[0,T]|x(t)|, max t∈[0,T]|x
′
(t)|}.

Then we have

x = D or x = −D for x ∈ ∂Ω∩ K er(L). (2.36)

From (H3) and (2.36), we have(if D is chosen large enough)

a(t)D + g(t, D , D,··· , D) −‖p‖0 > 0 for t ∈ [0, T] (2.37)

and

x
′
(t) = 0 and x

′′
(t) = 0, for t ∈ [0, T]. (2.38)

Finally from (2.18), (2.22), (2.37) and (2.38), we have

(Q N x) = 1
T

´ T

0
[−cx

′′
(t −τ) − a(t)x(t) + p(t)]dt

−g(t, x(t −τ1 (t)), x(t −τ2 (t)),··· , x(t −τn (t)))]dt

6= 0, ∀x ∈ ∂Ω∩ K er(L).

Then, for any x ∈ K erL ∩∂Ω and η ∈ [0, 1], we have

xH(x,η) = −ηx2 − x
T

(1 −η)
´ T

0
[cx

′′
(t −τ) + a(t)x(t) − p(t)

+g(t, x(t −τ1 (t)), x(t −τ2 (t)),··· , x(t −τn (t)))dt]dt

6= 0.



164 Chengjun Guo, Donal O’Regan & Ravi P. Agarwal
CUBO

12, 3 (2010)

Thus

d e g{Q N, Ω∩ K er(L), 0}

= d e g{− 1
T

´

T

0
[cx

′′
(t −τ) + a(t)x(t) − p(t)

+g(t, x(t −τ1 (t)), x(t −τ2 (t)),··· , x(t −τn (t)))]dt,Ω∩ K er(L), 0}

= d e g{−x,Ω∩ K er(L), 0}

6= 0.

From Lemma 2.8 for any x ∈ ∂Ω∩ D om(L) and λ ∈ (0, 1) we have Lx 6= λN x. By Theorem B
(Mawhin’s continuation theorem), the equation Lx = N x has at least a solution in D om(L)∩Ω,
so there exists a T-periodic solution of Eq.(1.1). The proof is complete.

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