On the existence of periodic solutions to a p-Laplacian neutral differential equation in the critical case

Nonlinear Analysis: Real World Applications 10 (2009) 2884–2893

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On the existence of periodic solutions to a p-Laplacian neutral differential equation in the critical caseI Shiping Lu ∗,1 Department of Mathematics, Anhui Normal University, Wuhu 241000, Anhui, PR China

article

a b s t r a c t

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Article history: Received 23 July 2007 Accepted 26 September 2008

In this paper, the author studies the existence of periodic solutions for a second order neutral functional differential equation

(ϕp (x(t ) − cx(t − τ ))0 )0 = f (x(t ))x0 (t ) + g (t , x(t − µ(t ))) + e(t )

Keywords: Periodic solution Continuation theorem Deviating argument

in the critical case |c | = 1. By analyzing some properties of the linear difference operator A : [Ax](t ) = x(t ) − cx(t − τ ) and using Mawhin’s continuation theorem, some new results are obtained. © 2009 Published by Elsevier Ltd

1. Introduction In recent years, the existence of periodic solutions for the following types of neutral differential equations with deviating arguments

(ϕp (x(t ) − cx(t − τ ))0 )0 + g (t , x(t − τ (t ))) = p(t ) and

(ϕp (x(t ) − cx(t − τ ))0 )0 = f (x(t ))x0 (t ) +

n X

βi (t )g (x(t − τi (t ))) + p(t )

i=1

was studied by papers [1,2]. But the condition of constants |c | 6= 1 is required. For example, under the assumption |c | 6= 1, we obtained that A : C2π := {x : x ∈ C (R, R), x(t + 2π ) ≡ x(t )} → C2π , [Ax](t ) = x(t ) − cx(t − τ ) has a unique inverse A−1 : C2π → C2π defined by

X j c f (t − jτ ), |c | < 1,   j≥0X −1 [A f ](t ) =  c −j f (t + jτ ), |c | > 1, − j ≥1

and then 2π

Z 0

|[A−1 f ](t )|dt ≤

1

|1 − |c ||

2π

Z

|f (t )|dt ,

∀f ∈ C2π ,

(1.1)

0

which was crucial to obtaining estimation of a priori bounds of periodic solutions in the non-critical case |c | 6= 1. Under the critical condition |c | = 1, we studied a first order neutral differential equation I Sponsored by the NSF of Anhui Province of China (No. 2005kj031ZD and No. 050460103), the Teaching and Research Award Program for Excellent Teachers in Higher Education Institution of Anhui Province of China and the key NSF of Education Ministry of China (No. 207047). ∗ Tel.: +86 553 3828887. E-mail address: [email protected]. 1 Current address: College of Math. and Physics, Nanjing University of Information and Technology, Nanjing 210044, PR China.

1468-1218/$ – see front matter © 2009 Published by Elsevier Ltd doi:10.1016/j.nonrwa.2008.09.005

S. Lu / Nonlinear Analysis: Real World Applications 10 (2009) 2884–2893

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(x(t ) − cx(t − τ ))0 = g (t , x(t − µ(t ))) + e(t ), a Duffing differential equation of neutral type

(x(t ) − cx(t − τ ))00 = g (t , x(t − µ(t ))) + e(t ), and a Liénard differential equation of neutral type

(x(t ) − cx(t − τ ))00 = f (x(t ))x0 (t ) + g (t , x(t − µ(t ))) + e(t ) in [3–5], respectively. In this paper, we continue studying a p-Laplacian neutral differential equation as follows

(ϕp (x(t ) − cx(t − τ ))0 )0 = f (x(t ))x0 (t ) + g (t , x(t − µ(t ))) + e(t ), (1.2) where f ∈ C (R, R), g ∈ C (R × R, R) with g (t + 2π , x) ≡ g (t , x), ∀x ∈ R, e and µ are continuous periodic functions with period 2π , and e¯ = 0, c , τ ∈ R are two constants with |c | = 1. By employing Mawhin’s continuation theorem and topological degree theory, we obtain some new results on the existence of 2π -periodic solutions to Eq. (1.2). The significance of the present paper is that the approaches used for estimating a priori bounds of periodic solutions in [1–5] cannot be applied directly herein. This mainly due to the facts in this situation — on the one hand formula (1.1) which is crucial to obtaining a priori bounds of periodic solutions in [1,2] does not hold in the critical case |c | = 1, and on the other hand, Mawhin’s continuation theorem is not applicable directly since the p-Laplacian p(u) = |u|p−2 u is not linear with respect to u except when p = 2 in [3–5]. Throughout this paper, we will denote by Z the set of integers, Z1 the set of odd integers, Z2 the set of even integers, N the set of positive integers, N1 the set of odd positive integers and N2 the set of even positive integers. Let C2π = {x : x ∈ C (R, R), x(t + 2π ) ≡ x(t )} with the norm |ϕ|0 = maxt ∈[0,2π] |ϕ(t )|, C2+π = {x : x ∈ C2π , x(t + π ) ≡ x(t )},

R 2π

R 2π

C20π = {x : x ∈ C2π , 0 x(s)ds = 0}, C2π = {x : x ∈ C2+π , 0 x(s)ds = 0}, and C2−π = {x : x ∈ C2π , x(t + π ) ≡ −x(t )} equipped with the norm | · |0 ; C21π = {x : x ∈ C 1 (R, R), x(t + 2π ) ≡ x(t )} with the norm |ϕ|C 1 = max{|ϕ|0 , |ϕ 0 |0 }; +,0

a.e.

L22π = {x : x(t ) = x(t + 2π ), t ∈ R and

2π

a.e.

R 2π

2 |x(s)|2 ds < +∞}, L22+ π = {x : x ∈ L2π , x(t ) = x(t + π ), t ∈ R}, 0 R 1/2 a.e. 2π 2 |ϕ(s)|2 ds . Clearly, C21π , C2−π , C2π , L22− π = {x : x ∈ L2π , x(t ) = −x(t + π ), t ∈ R}, and the norm is defined by |ϕ|2 = 0 R 2π R 2π +,0 2− 1 2 ¯ C2+π , C20π , C2π , L22π , L22+ π and L2π are all Banach spaces. We also denote h = 2π 0 h(s)ds, |h|1 = 0 |h(s)|ds, ∀h ∈ L2π .

2. Main lemmas In this section, we will analyze some properties of the linear difference operator A : [Ax](t ) = x(t ) − cx(t − τ ), which will be used to estimate a priori bounds of periodic solutions in Section 3. Lemma 2.1 ([3]). Let |τ | =

q1 p1

π , where p1 and q1 are coprime positive integers. Then:

1. If c = −1 and q1 is even, then

σ1 := inf |1 − ce−kiτ | = inf [2(1 + cos kτ )]1/2 > 0. k∈N

k∈N

2. If c = −1, q1 is odd and p1 is even, then

σ2 := inf |1 − ce−kiτ | = inf [2(1 + cos kτ )]1/2 > 0. k∈N1

k∈N1

3. If c = −1, p1 and q1 are odd, then

σ3 := inf |1 − ce−kiτ | = inf [2(1 + cos kτ )]1/2 > 0. k∈N2

k∈N2

4. If c = 1 and q1 is odd, then

σ4 := inf |1 − ce−kiτ | = inf [2(1 − cos kτ )]1/2 > 0. k∈N1

k∈N1

5. If c = 1 and p1 = q1 = 1, then

σ5 := inf |1 − ce−kiτ | = inf [2(1 − cos kτ )]1/2 = 2 > 0. k∈N1

k∈N1

Lemma 2.2 ([5]). Suppose c = −1, |τ | = L22π

a unique inverse A−1 :

→

L22π

Lemma 2.3 ([5]). If c = −1, τ =

π , where p1 , q1 are coprime positive integers with q1 even, then A : L22π → L22π has satisfying kA−1 k ≤ σ1 . 1 q1 p1

q1 p1

π , where p1 , q1 are coprime odd positive integers, then

2+ A : L22+ π → L2π , [Ax](t ) = x(t ) − cx(t − τ ),

−1

has a unique inverse A

:

L22+ π

→

L22+ π

∀t ∈ [0, 2π] −1

satisfying kA

k≤

1

σ3

.

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S. Lu / Nonlinear Analysis: Real World Applications 10 (2009) 2884–2893

Lemma 2.4 ([5]). Suppose 2− A : L22− π → L2π : [Ax](t ) = x(t ) − cx(t − τ ),

∀t ∈ [0, 2π ].

Then the following propositions are true. 1. If c = −1, τ =

q1 p1

π , where p1 , q1 are coprime positive integers with q1 odd and p1 even, then A has a unique inverse

2− 1 −1 A−1 : L22− π → L2π satisfying kA k ≤ σ2 . q 2− 2. If c = 1, τ = p1 π , where p1 , q1 are coprime positive integers with q1 odd, then A has a unique inverse A−1 : L22− π → L2π 1

satisfying kA−1 k ≤ σ1 . 4 2− 1 −1 3. If c = 1, τ = π , then A has a unique inverse A−1 : L22− π → L2π satisfying kA k ≤ σ5 . Lemma 2.5 ([6]). Let X and Y be two Banach spaces, L : D(L) ⊂ X → Y be a Fredholm operator with index zero and Q : Y → Y /Im L be a projection. Furthermore, Ω ⊂ X is an open bounded set, and N : Ω → Y is L-compact on Ω . If all the following conditions: (1) Lx 6= λNx, ∀(x, λ) ∈ [(D(L) \ Ker L) ∩ ∂ Ω ] × (0, 1); (2) Nx ∈ 6 Im L, ∀x ∈ ∂ Ω ∩ Ker L; (3) deg {QN |Ker L , Ω ∩ Ker L, 0} 6= 0; hold, then equation Lx = Nx has a solution on Ω

T

D(L).

In order to use Lemma 2.5, now we define

 X =Y =

C2π × C2π , if τ and c satisfy the condition of Lemma 2.2, C2+π × C2+π , if τ and c satisfy the condition of Lemma 2.3,

and suppose

µ(t + π ) ≡ µ(t ), g (t + π , x) ≡ g (t , x),

∀x ∈ R, e(t + π ) ≡ e(t ).

(2.1)

Moreover, we should rewrite Eq. (1.2) in the following form



(Ax1 )0 (t ) = ϕq (x2 (t )) = |x2 (t )|q−2 x2 (t ) x02 (t ) = −f (x1 (t ))ϕq (x2 (t )) + g (x1 (t − τ (t ))) + e(t ),

(2.2)

where q > 1 is a constant with 1p + 1q = 1. Clearly, if x(t ) = (x1 (t ), x2 (t ))> is a T -periodic solution to System (2.2), then x1 (t ) must be a 2π -periodic solution to Eq. (1.2). Thus, in order to prove that Eq. (1.2) has a 2π -periodic solution, it suffices to show that System (2.2) has a 2π -periodic solution. Now, we set L : D(L) ⊂ X → Y , Lx =

N : X → Y, N x =





(Ax1 )0



0

x2

.

(2.3)

 ϕq (x2 ) f (x1 (t ))ϕq (x2 (t )) + g (x1 (t − τ (t ))) + e(t ),

(2.4)

R 2π

where D(L) = {x ∈ C 1 (R, R2 ) : x(t + 2π ) ≡ x(t )}. It is easy to see that Ker L = R2 , Im L = {x : x ∈ Y , 0 x(s)ds = 0}. So L is a Fredholm operator with index zero. Also let projectors P : X → Ker L and Q : Y → Im Q be defined by Px = ((Ax1 )(0), x2 (0))> ; Qy =

1 2π

T

Z

y(s)ds,

0

and let K represent the inverse of L|Ker P ∩D(L) . Obviously, Ker L = Im Q = R2 .

[Ky]−1 (t ) = ((A−1 Fy1 )(t ), (Fy2 )(t ))> , (2.5) Rt Rt > where (Fy1 )(t ) = 0 y1 (s)ds, (Fy2 )(t ) = 0 y2 (s)ds, y(t ) = (y1 (t ), y2 (t )) . From (2.5), one can easily see that N is Lcompact on Ω , where Ω is an open, bounded subset of X . The following Lemma is crucial for us to investigate the relation between the existence of periodic solutions to Eq. (1.2) and the deviating argument τ (t ). Lemma 2.6 ([7]). Let n1 > 1, α ∈ [0, +∞) be constants, s ∈ C (R, R) with s(t + T ) ≡ s(t ), and s(t ) ∈ [−α, α], ∀t ∈ [0, T ]. Then ∀x ∈ C 1 (R, R) with x(t + T ) ≡ x(t ), we have T

Z

|x(t ) − x(t − s(t ))|n1 dt ≤ 2α n1 0

Z 0

T

|x0 (t )|n1 dt .

S. Lu / Nonlinear Analysis: Real World Applications 10 (2009) 2884–2893

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Throughout this paper, we suppose e¯ = 0 and set F (x) =

x

Z

f (s)ds,

E (t ) =

t

Z

e(s)ds.

(2.6)

0

0

From e¯ = 0, we see E (t + 2π ) ≡ E (t ). Meanwhile, for constant ω > 0, we set Fω = max |F (x)|, |x|≤ω

fω = max |f (x)|, |x|≤ω

gω =

max

|x|≤ω,t ∈[0,2π ]

|g (t , x)|.

Furthermore, we assume that there are integers k and l such that max |µ(t ) − 2kπ | ≤ 2π

and

t ∈[0,2π]

max |µ(t + τ ) − τ − 2lπ | ≤ 2π .

t ∈[0,2π]

Let

γ (t ) = µ(t ) − 2kπ ,

γ1 (t ) = µ(t + τ ) − τ − 2lπ .

(2.7)

Then

|γ |0 ≤ 2π ,

|γ1 |0 ≤ 2π .

(2.8)

3. Main results In this section, we will study the existence of periodic solutions for Eq. (1.1) in the critical case |c | = 1. Theorem 3.1. Suppose p ∈ [2, +∞), and there are constants r1 ≥ 0, r0 ≥ 0 and d ≥ 0 such that:

(A1 ) limx→+∞ sup ||gx(|tp,−x1)| ≤ r1 and limx→−∞ sup ||gx(|tp,−x1)| ≤ r2 uniformly for t ∈ R, and lim|x|→+∞ sup ||xF|(px−)|1 ≤ r0 ; (A2 ) xg (t , x) > 0, for (t , x) ∈ R × ∆, where ∆ = (−∞, −d) ∪ (d, +∞). Then Eq. (1.2) has at least one 2π -periodic solution, if one of the following conditions holds: q 1. c = −1, |τ | = p1 π with p1 , q1 coprime positive integers and q1 even, and 1  1 (p−1)/p p−1 1 (2π ) p r0p−1 + 2 2(p−1) (r1 + r2 )1/(p−1) (|γ |0 + |γ1 |0 )1/(p−1) < 1; σ1 q 2. Condition (2.2) holds, c = −1, |τ | = p1 π , where p1 , q1 are coprime positive odd integers, and 1  1 (p−1)/p p−1 1 (2π ) p r0(p−1) + 2 2(p−1) (r1 + r2 ]1/(p−1) (|γ |0 + |γ1 |0 )1/(p−1) < 1; σ3 where σ1 , σ3 are defined by Lemma 2.1; F (x) are defined by (2.6); γ (t ) and γ1 (t ) are defined by (2.7). Proof. As the proof of case 2 works almost exactly as the proof of case 1, we should prove case 1 only. Consider the operator equation as follows Lx = λN x,

λ ∈ (0, 1),

(3.1)

where L and N are defined by (2.3) and (2.4), respectively. Let x ∈ D(L) ⊂ X be an arbitrary 2π -periodic solution of Eq. (3.1), then



(Ax1 )0 (t ) = λϕq (x2 (t )) = λ|x2 (t )|q−2 x2 (t ) x02 (t ) = λf (x1 (t ))[A−1 ϕq (x2 )](t ) + λe(t ).

(3.2)

From the first formula of (3.1), we have x2 (t ) = ϕp ( λ1 [Ax1 ]0 (t )), which together with the second formula of (3.2) yields

(x(t ) − cx(t − τ ))00 = λp−1 f (x(t ))x0 (t ) + λp g (t , x(t − µ(t ))) + λp e(t ), Integrating the two sides of (3.3), we get 2π

Z

λp−1 f (x(t ))x0 (t )dt +

0

0

Since 2π

Z 0

2π

Z

f (x(t ))x0 (t )dt = 0,

λp g (t , x(t − µ(t )))dt = 2π e¯ = 0.

λ ∈ (0, 1).

(3.3)

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S. Lu / Nonlinear Analysis: Real World Applications 10 (2009) 2884–2893

it follows that 2π

Z

g (t , x(t − µ(t )))dt = 0.

(3.4)

0

By using the integral mean value theorem, we have that there is a point t0 ∈ [0, 2π ] such that 2π g (t0 , x(t0 − µ(t0 ))) = 0 i.e., g (t0 , x(t0 − µ(t0 ))) = 0, which together with assumption (A2 ) leads to

|x(t0 − µ(t0 ))| ≤ d. Since t0 − µ(t0 ) ∈ R, there must be an integer m and a constant t1 ∈ [0, 2π ) such that t0 − µ(t0 ) = 2mπ + t1 . Therefore, |x(t1 )| = |x(t0 − µ(t0 ))| ≤ d. So

|x|0 =

max

t ∈[t1 ,t1 +2π ]

|x(t )| ≤ d +

Z

t1 + 2 π

|x0 (s)|ds = d +

t1

2π

Z

|x0 (s)|ds.

(3.5)

0

From the assumption

(2π )

p−1 p



1 p−1

r0

1

+ 2 2(p−1) (r1 + r2 )1/(p−1) (|γ |0 + |γ1 |0 )1/(p−1)

(p−1)/p < 1,

σ1 we know that there is a small constant ε > 0 such that

(2π )

p−1 p

h

1

1

(r0 + ε) p−1 + 2 2(p−1) (r1 + r2 + 2ε)1/(p−1) (|γ |0 + |γ1 |0 )1/(p−1) σ1

i(p−1)/p < 1.

(3.6)

For such an ε > 0, in view of assumption (A1 ), there must be a constant ρ > d, which is independent of λ and x, such that

|g (t , u)| < r1 + ε, |u|p−1

for u > ρ, t ∈ [0, 2π ]

|g (t , u)| < r2 + ε, |u|p−1

for u < −ρ, t ∈ [0, 2π ]

(3.7)

and

|F (x)| < r0 + ε, |x|p−1

for |u| > ρ.

(3.8)

Take E1 = {t : t ∈ [0, 2π ], x(t − µ(t )) > ρ}, E2 = {t : t ∈ [0, 2π ], |x(t − µ(t ))| ≤ ρ} and E3 = {t : t ∈ [0, 2π], x(t − µ(t )) < −ρ}, E4 = {t : t ∈ [0, 2π ], |x(t )| ≤ ρ} and E5 = {t : t ∈ [0, 2π ], |x(t )| > ρ}, then we have from (3.4) and assumption (A1 ) that

Z

Z 

Z + E1

+ E2

g (t , x(t − µ(t )))dt = 0.

E3

Hence,

Z

|g (t , x(t − µ(t )))|dt ≤ (r1 + ε)

Z

|g (t , x(t − µ(t )))|dt ≤ (r2 + ε)

Z

E1

Z

|x1 ((t − µ(t )))|p−1 dt

0

E3

Z

2π

2π

|x1 ((t − µ(t )))|p−1 dt

0

|g (t , x1 (t − µ(t )))|dt ≤ 2π gρ . E2

Meanwhile, by (3.8) we have

|F (x1 (t ))| ≤ (r0 + ε)|x1 |p0−1 , |x1 (t )|

for t ∈ E5 .

(3.9)

S. Lu / Nonlinear Analysis: Real World Applications 10 (2009) 2884–2893

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Multiplying the two sides of Eq. (3.3) by (Ax1 )(t ) and integrating them on the interval [0, 2π ], we obtain 2π

Z

|(Ax1 )0 (t )|p dt = −

2π

Z

(Ax1 )00 (t )(Ax1 )(t )dt

0

0

2π

Z = −λ

f (x1 (t ))x01 (t )(Ax1 )(t )dt − λ

−λ

g (t , x1 (t − µ(t )))(Ax1 )(t )dt

0

0 2π

Z

2π

Z

e(t )(Ax1 )(t )dt .

(3.10)

0

Since 2π

Z

f (x1 (t ))x1 (t )(Ax1 )(t )dt = − 0

−

2π

Z

f (x1 (t ))x1 (t )x1 (t )dt + c 0

f (x1 (t ))x01 (t )x1 (t − τ )dt

0

0

0

2π

Z

2π

Z = −c

F (x1 (t ))x01 (t − τ )dt

0

Z Z ≤ |c | |F (x1 (t ))x01 (t − τ )|dt + |c | |F (x1 (t ))x01 (t − τ )|dt E4 E5 Z Z 0 = |F (x1 (t ))x1 (t − τ )|dt + |F (x1 (t ))x01 (t − τ )|dt , E4

E5

it follows from (3.9) that 2π

Z

f (x1 (t ))x01 (t )(Ax1 )(t )dt ≤ Fρ

2π

Z

0

Z

|x01 (t − τ )|dt + (r0 + ε)|x1 |p0−1

0

2π

|x01 (t − τ )|dt

0 1/2

= Fρ (2π )

2π

Z

|x1 (t )| dt 0

2

1/2

1/2

+ (2π )

p−1 x1 0

(r0 + ε)| |

0

2π

Z

|x1 (t )| dt 0

2

1/2

.

0

(3.11) In view of 2π

Z

g (t , x1 (t − µ(t )))(Ax1 )(t )dt

− 0

= (c − 1)

Z

g (t , x1 (t − µ(t )))x1 (t − µ(t ))dt + (c − 1) E1 ∪E3

Z +

E2

E

3 Z 

Z

+c

g (t , x1 (t − µ(t )))[x1 (t ) − x1 (t − µ(t ))]dt

+

E

Z1 +

+

E1

g (t , x1 (t − µ(t )))x1 (t − µ(t ))dt E2

Z 

Z

−

Z

E2

g (t , x1 (t − µ(t )))[x1 (t − τ ) − x1 (t − µ(t ))]dt ,

E3

we have from assumptions (A1 ) and (A2 ) that 2π

Z −

g (x1 (t − µ(t )))(Ax1 )(t )dt ≤ |c − 1|

0

Z

|g (t , x1 (t − µ(t )))x1 (t − µ(t ))|dt E2

Z

Z 

Z

+

+ E1

E2

Z

E3

Z 

Z

+ |c |

+ E1

|g (x1 (t − µ(t )))[x1 (t ) − x1 (t − µ(t ))]|dt

+ + E2

|g (t , x1 (t − µ(t )))[x1 (t − τ ) − x1 (t − µ(t ))]|dt

E3

≤ 2π |c − 1|ρ gρ + gρ

2π

Z

|x1 (t ) − x1 (t − µ(t ))|dt

0

+ (r1 + r2 + 2ε)

2π

Z

|x1 (t − µ(t )))|p−1 |x1 (t ) − x1 (t − µ(t ))|dt + gρ

0

+ (r1 + r2 + 2ε)

0 2π

Z 0

2π

Z

|x1 (t − µ(t )))|p−1 |x1 (t − τ ) − x1 (t − µ(t ))|dt

|x1 (t − τ ) − x1 (t − µ(t ))|dt

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S. Lu / Nonlinear Analysis: Real World Applications 10 (2009) 2884–2893

≤ 2π |c − 1|ρ gρ + (2π )

1/2

2π

Z

|x1 (t ) − x1 (t − µ(t ))| dt 2

gρ

1/2

+ (2π )1/2 (r1 + r2 + 2ε)|x1 |p0−1

0 2π

Z

|x1 (t ) − x1 (t − µ(t ))|2 dt

×

1/2

+ (2π)1/2 gρ

2π

Z

|x1 (t − τ ) − x1 (t − µ(t ))|2 dt

1/2

0

0 2π

Z

+ (2π )1/2 (r1 + r2 + 2ε)|x1 |p0−1

|x1 (t − τ ) − x1 (t − µ(t ))2 |dt

1/2

.

(3.12)

0

By using Lemma 2.6, 2π

Z

|x1 (t ) − x1 (t − µ(t ))|dt

1/2

√ 2|γ |0

1/2

√

0

|x01 (s)|2 ds

1/2

0

2π

Z

2π

Z

≤

|x1 (t − τ ) − x1 (t − µ(t ))|dt

≤

Z

2π

2|γ1 |0

|x01 (s)|2 ds

1/2

.

(3.13)

0

0

Substituting (3.13) into (3.12), we have 2π

Z −

g (x1 (t − µ(t )))(Ax1 )(t )dt ≤ 2π |c − 1|ρ gρ + 2π

1/2

gρ (|γ |0 + |γ1 |0 )

2π

Z

0

|x1 (s)| ds 0

2

1/2

0 1/2

+ 2(π )

p−1 x1 0

(r1 + r2 + 2ε)| |

2π

Z

(|γ |0 + γ1 |0 )

|x1 (s)| ds 0

2

1/2

.

(3.14)

0

Furthermore, 2π

Z −

e(t )(Ax1 )(t )dt =

2π

Z

0

E (t )(Ax1 )0 (t )dt

0

≤ (1 + |c |)|E |2

2π

Z

|x01 (t )|2 dt

1/2

,

(3.15)

0

where |E |2 = 2π

Z

R 2π

|E (t )|2 dt

0

1/2

. Substituting (3.15), (3.14) and (3.11) into (3.10), we have

|(Ax1 )0 (t )|p dt ≤ Fρ (2π )1/2

2π

Z

0

|x01 (t )|2 dt

1/2

+ (2π )1/2 (r0 + ε)|x1 |p0−1

Z

0

2π

|x01 (t )|2 dt

1/2

0

+ 2π |c − 1|ρ gρ + [2π

1/2

gρ (|γ |0 + |γ1 |0 ) + |E |2 ]

2π

Z

|x1 (s)| ds 0

2

1/2

0

+ 2π 1/2 (r1 + r2 + 2ε)|x1 |p0−1 (|γ |0 + γ1 |0 )

2π

Z

1/2 |x01 (s)|2 ds ,

0

which yields 2π

Z

|(Ax1 ) (t )| dt 0

p

1/(p−1)

0

h

1 2(p−1)

1/(p−1)

1/2

1/(p−1)

+ [2π gρ (|γ |0 + |γ1 |0 ) + |E |2 ] ≤ Fρ (2π ) h 1 1 + (2π ) 2(p−1) (r0 + ε) (p−1) + [2π 1/2 (r1 + r2 + 2ε)]1/(p−1) 2π

Z i × (|γ |0 + |γ1 |0 )1/(p−1) |x1 |0

|x01 (t )|2 dt

 2(p1−1)

i Z

2π

|x1 (t )| dt 0

2

 2(p1−1)

0

+ (2π |c − 1|ρ gρ )1/(p−1) .

0

It follows from (3.5) that 2π

Z 0

|(Ax1 ) (t )| dt 0

p

1/(p−1)

≤α

2π

Z 0

|x1 (t )| dt 0

2

 2(p1−1)

+β

2π

Z

|x1 (t )| dt 0

2

 2(pp−1)

+ (2π |c − 1|ρ gρ )1/(p−1) ,

0

(3.16)

S. Lu / Nonlinear Analysis: Real World Applications 10 (2009) 2884–2893

2891

where

i h 1 α = Fρ1/(p−1) (2π ) 2(p−1) + [2(π )1/2 gρ (|γ |0 + |γ1 |0 ) + |E |2 ]1/(p−1) h i 1 1 + d (2π ) 2(p−1) (r0 + ε) (p−1) + [2π 1/2 (r1 + r2 + 2ε)]1/(p−1) (|γ |0 + |γ1 |0 )1/(p−1) i h 1 1 β = (2π )1/2 (2π ) 2(p−1) (r0 + ε) (p−1) + [2(π )1/2 (r1 + r2 + 2ε)]1/(p−1) (|γ |0 + |γ1 |0 )1/(p−1) i h p 1 1 = (2π ) 2(p−1) (r0 + ε) (p−1) + 2 2(p−1) (r1 + r2 + 2ε)1/(p−1) (|γ |0 + |γ1 |0 )1/(p−1) . By using Lemma 2.2 and the Hölder inequality, 2π

Z

|x1 (t )| dt 0

2

 12 ≤

0

1

2π

Z

σ1

|(Ax1 ) (t )| dt 0

2

 12

0

p−2

(2π ) 2p ≤ σ1

2π

Z

|(Ax1 ) (t )| dt 0

p

 1p

.

0

So it follows from (3.16) that 2π

Z

|x1 (t )| dt 0

2

 12 ≤

(2π )

+ (2π)

p−2 2p

σ1

β

2π

Z

p−1 p

< 1 and

|x01 (t )|2 dt

 12

1 2p

α

p−1 p

(2π )

2π

Z

σ1

0

Since

p−2 2p

|x1 (t )| dt 0

2

 2p1

0

p−2 2p

β

p−1 p

2π

Z

σ1

|x1 (t )| dt 0

2

 12 +

0

(2π ) σ1

p−2 p

(2π |c − 1|ρ gρ )1/p) .

(3.17)

< 1, it follows from (3.17) that there is a constant M > 0 such that

≤ M,

(3.18)

0

which together with (3.5) yields

|x1 |0 ≤ d + (2π )1/2 |x01 |2 ≤ d + (2π )1/2 M := M .

(3.19)

Again from the first equation of (3.2), we have 2π

Z

|x2 (s)|q−2 x2 (s)ds = 0,

0

which implies that there is a constant η ∈ [0, 2π ] such that x2 (η) = 0. Let fM maxt ∈[0,2π],u∈[−M ,M ] |g (t , u)|, and from the second equation of (3.2), we have

= maxu∈[−M ,M ] |f (u)|, gM =

Z t Z t Z t Z t |x2 (t )| = x02 (s)ds ≤ |f (x1 (s))(Ax1 )0 (s)|ds + |g (s, x1 (s − µ(s)))|ds + |e(s)|ds η η η η Z 2π ≤ 2fM |x01 (s)|ds + 2π gM + |e|1 , ∀t ∈ [η, η + 2π ], 0

R 2π

|x01 (s)|ds, which together with (3.18) gives Z 2π 1/2 1 |x2 (t )| ≤ 2fM 2π 2 |x01 (s)|2 ds + 2π gM + |e|1

where |e|1 =

0

0 1

≤ 2fM (2π ) 2 M + 2π gM + |e|1 ,

∀t ∈ [η, η + T ],

i.e., 1

|x2 |0 ≤ 2fM (2π ) 2 M1 + 2π gM + |e|1 := M1 .

(3.20)

Let Ω2 = {x : x ∈ Ker L, QNx = 0}. If x ∈ Ω2 , then x = (u, v)

>

2

∈ R is a constant vector with

 q −2 |v| Z v = 0 2π

1



2π

[f (u)|v|q−2 v + g (s, u)]ds = 0.

0

So v = 0 and by assumption [H1 ], we see |u| ≤ d, which implies Ω2 ⊂ Ω1 .

2892

S. Lu / Nonlinear Analysis: Real World Applications 10 (2009) 2884–2893

Now, if we set Ω = {x : x = (u, v)> ∈ X , |u|0 < M + 1, |v|0 < M1 + 1}, then Ω ⊃ Ω1 ∪ Ω2 . So from (3.16) and (3.17), we see that condition (1) and condition (2) of Lemma 2.3 are satisfied. Condition (3) of Lemma 2.3 can be proved in a similar way as that used for proving Theorem 3.1 of [8]. Therefore, by using Lemma 2.3, we see that the equation Lx = N x has a solution x∗ (t ) = (u∗ (t ), v ∗ (t ))> on Ω , i.e., Eq. (1.2) has a T -periodic solution u∗ (t ) with |u∗ |0 ≤ M + 1. In what follows, we will discuss the existence of odd periodic solutions to Eq. (1.1) under the condition of Lemma 2.4. By [8], we know that if x is an odd 2π -periodic solution to Eq. (1.2), then x1 (t + π ) ≡ −x1 (t ). So we need to impose some symmetric conditions on g (t , x1 ), µ(t ) and e(t ) in the following form f (−x) ≡ f (x),

g (t + π , −x) ≡ −g (t , x),

µ(t + π ) ≡ µ(t ),

e(t + π ) ≡ −e(t ),

(3.21)

and let L : D(L) ⊂ C2−π × C2−π → C2−π × C2−π , Lx =

N : C2−π × C2−π → C2−π × C2−π , N x =



(Ax1 )0 0

x2



,

 ϕq (x2 ) , f (x1 (t ))ϕq (x2 (t )) + g (t , x1 (t − µ(t ))) + e(t )



where D(L) = {x ∈ C 1 (R, R2 ) : x(t + π ) ≡ −x(t )}. Theorem 3.2. Suppose that (3.21) holds; and there are constants r0 , r1 , r2 ≥ 0 and d ≥ 0 such that assumptions (A1 ) and (A2 ) of Theorem 3.1 hold. Then Eq. (1.2) has at least one 2π -odd periodic solution, if one of the following conditions holds: 1. c = −1, |τ | =

q1 p1

p−1 p



(2π )

2. c = 1, |τ | =

(2π )

q1 p1

p−1 p

π , where p1 , q1 are coprime positive integers with q1 odd p1 even, and (p−1)/p 1 1 (p−1) r0 + 2 2(p−1) (r1 + r2 ]1/(p−1) (|γ |0 + |γ1 |0 )1/(p−1) < 1; σ2

π , where p1 , q1 are coprime positive integers with q1 odd, and  1 (p−1)/p 1 (p−1) r0 + 2 2(p−1) (r1 + r2 ]1/(p−1) (|γ |0 + |γ1 |0 )1/(p−1) < 1;

σ4 3. If c = 1, |τ | = π , and

(2π )

p−1 p



1

(p−1) r0

+2

1 2(p−1)

(r1 + r2 ]

1/(p−1)

1/(p−1)

(|γ |0 + |γ1 |0 )

(p−1)/p < 1;

σ5 where σ2 , σ4 and σ5 are constants defined by Lemma 2.1.

The proof of Theorem 3.2 works almost exactly as the proof of Theorem 3.1. So it is omitted. As an application, we consider the neutral functional differential as follows







ϕ4 x(t ) + x t −

3π θ

0 0

p =

5

p 1 − cos π5 0 1 − cos π5 x (t )x(t ) sin x(t ) + h(x(t − θ sin t )) + cos t 2π 2π 2

where h(x) =

x3 , x > 0 x, x ≤ 0,



θ ∈ (−1, 1) is a constant. Corresponding to Eq. (1.2), we have p 1 − cos π5 3π θ c = −1, p = 4, τ= , f ( x) = x sin x, 5 2π p 1 − cos π5 g ( t , x) = h(x), µ(t ) = θ sin t , e(t ) = cos t . 2π 2 √ √ 1−cos π5 1−cos π5 So e¯ = 0, r0 = r2 = 0, r1 = and we can chose constants K = 1, r = , r2 = 1 2π 2π 2 h  π i1/2 σ3 := inf |1 − ce−kiτ | = inf [2(1 + cos kτ )]1/2 = 2 1 − cos . k∈N2

k∈N2

5

1

π

, r3 = π2 and

(3.22)

S. Lu / Nonlinear Analysis: Real World Applications 10 (2009) 2884–2893

2893

Meanwhile,

π r1 2 (1 + 2π r2 + r2 max{r2−1 , r3−1 }) = √ < 1. √ 3σ3 6 Thus, if |θ| <

5(1−cos π5 ) , 24π 4/3

by applying Theorem 3.1 we see that Eq. (3.22) has at least one 2π -periodic solution.

References [1] Yanling Zhu, Shiping Lu, Periodic solutions for p-Laplacian neutral functional differential equation with deviating arguments, J. Math. Anal. Appl. 325 (2007) 377–385. [2] Yanling Zhu, Shiping Lu, Periodic solutions for p-Laplacian neutral functional differential equation with multiple deviating arguments, J. Math. Anal. Appl. 236 (2) (2007) 1357–1367. [3] Shiping Lu, Weigao Ge, Periodic solutions to to a kind of neutral functional differential equations in the critical case, J. Math. Anal. Appl. 293 (2) (2004) 462–475. [4] Shiping Lu, Zhanjie Gui, Weigao Ge, Periodic solutions to a second order nonlinear neutral functional differential equation with multiple deviating arguments in the critical case, Nonlinear Anal. TMA. 64 (2006) 1587–1607. [5] Shiping Lu, On the existence of periodic solutions to rayleigh differential equation of neutral type in the critical case, Nonlinear Anal. TMA (in press). [6] J. Mawhin, Topological degree and boundary value problems for nonlinear differential equations, in: Topological Methods for Ordinary Differential Equations, in: Lecture Notes in Mathematics, vol. 1537, Springer-Verlag, New York, 1991, pp. 74–142. [7] Shiping Lu, Weigao Ge, Periodic solutions for a kind of Liénard equation with a deviating argument, J. Math. Anal. Appl. 289 (2004) 231–243. [8] Shiping Lu, Zhanjie Gui, On the existence of periodic solutions to p-Laplacian Rayleigh differential equation with a delay, J. Math. Anal. Appl. 325 (2007) 685–C702.