Periodic solutions for p -Laplacian neutral Rayleigh equation with a deviating argument

Nonlinear Analysis 69 (2008) 1675–1685 www.elsevier.com/locate/na

Periodic solutions for p-Laplacian neutral Rayleigh equation with a deviating argument Shiguo Peng College of Automation, Guangdong University of Technology, Guangzhou 510090, PR China Received 18 December 2006; accepted 11 July 2007

Abstract By using topological degree theory and some analytical skill, some criteria to guarantee the existence of ω-periodic solutions are derived for p-Laplacian neutral Rayleigh equation with a deviating argument of the following form

0 φ p (x(t) − cx(t − σ ))0 + f (x 0 (t)) + g(x(t − τ (t))) = e(t). c 2008 Published by Elsevier Ltd

MSC: 34C25 Keywords: Periodic solutions; Neutral Rayleigh equation; Deviating argument; p-Laplacian; Topological degree

1. Introduction Throughout this paper, 1 < p < ∞ is a fixed real number. The conjugate exponent of p is denoted by q, i.e. + q1 = 1. Let φ p : R → R be the mapping defined by φ p (u) = |u| p−2 u. Then φ p is a homeomorphism of R with the inverse φq (u) = |u|q−2 u. In this paper, we will consider the existence of periodic solutions for the following neutral Rayleigh equation with a deviating argument

0 φ p (x(t) − cx(t − σ ))0 + f (x 0 (t)) + g(x(t − τ (t))) = e(t), (1.1) 1 p

where f, g, e and τ are real continuous functions on R, τ and e are periodic with period ω, ω > 0 is fixed, c, σ ∈ R are constants such that |c| 6= 1. In recent years, the existence of periodic solutions for second-order Rayleigh equations with a deviating argument x 00 (t) + f (x 0 (t)) + g(t, x(t − τ (t))) = e(t),

(1.2)

has been extensively studied in the literature, we refer the readers to [1–4] and the references cited therein. In [5–7], the problems on the existence of periodic solutions for Rayleigh equations of p-Laplacian type (φ p (x 0 (t)))0 + f (x 0 (t)) + g(x(t − τ (t))) = e(t), were also discussed by using Mahwin’s coincidence degree theory [8]. E-mail address: [email protected]. c 2008 Published by Elsevier Ltd 0362-546X/$ - see front matter doi:10.1016/j.na.2007.07.007

(1.3)

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In [9], Zonga and Liang consider the following equation (φ p (x 0 (t)))0 + f (t, x 0 (t − σ (t))) + g(t, x(t − τ (t))) = e(t),

(1.4)

where f, g ∈ C(R×R, R) and f, g are 2π -periodic with their first arguments, σ, τ and e are continuously 2π -periodic R 2π functions defined on R. Under the assumptions of f (t, 0) = 0 and 0 e(t)dt = 0, they obtained the following result. Theorem A ([9]). Suppose there exist positive constants K , d, M such that (H1 ) | f (t, x)| ≤ K for (t, x) ∈ R × R; (H2 ) xg(t, x) > 0 and |g(t, x)| > K for |x| > d and t ∈ R; (H3 ) g(t, x) ≥ −M, for x ≤ −d and t ∈ R. Then (1.4) has at least one solution with period 2π. For a kind of second-order neutral Rayleigh functional differential equation (x(t) + cx(t − r ))00 + f (x 0 (t)) + g(t, x(t − τ (t))) = p(t),

(1.5)

Lu and Ge [10] gave some existence theorems of 2π-periodic solutions for |c| 6= 1. Very recently, Zhu and Lu [11] discussed the existence of periodic solutions for p-Laplacian neutral functional differential equation with deviating argument when p > 2 (φ p (x(t) − cx(t − σ ))0 )0 + g(t, x(t − τ (t))) = p(t).

(1.6)

They obtained some results by translating (1.6) into a two-dimensional system to which Mawhin’s continuation theorem was applied. But for (1.1), the methods to obtain a priori bounds of periodic solutions in [11] cannot be applied to the present RT paper, since the crucial step 0 [g(t, x(t − τ (t))) − p(t)]dt = 0 is no longer valid for Eq. (1.1). The purpose of this paper is to establish some criteria to guarantee the existence of ω-periodic solutions of (1.1) for any p > 1 by using the following Lemma 2.6 in Section 2 which is similar to Theorem 3.1 in [12] obtained by using Leray–Schauder degree theory. The significance of this paper is that even if for c = 0, the results are different from the corresponding ones of [5–7]. 2. Preliminaries {x : x ∈ C(R, R), x(t + ω) ≡ x(t)} with norm kxk∞ = maxt∈[0,ω] |x(t)|, Cω1 = {x : x ∈ + ω) ≡ x(t)} with norm kxk = max{kxk∞ , kx 0 k∞ }. Clearly, Cω and Cω1 are Banach spaces. In Rω
1 what follows, we will use k · k p to denote the L p -norm in Cω , i.e. kxk p = 0 |x(t)| p dt p . We also define a linear operator A as follows

Let Cω = C 1 (R, R), x(t

A : Cω → Cω ,

(Ax)(t) = x(t) − cx(t − σ ).

Lemma 2.1 ([13–15]). If |c| 6= 1, then A has continuous bounded inverse on Cω , and kxk∞ (1) kA−1 xk∞ ≤ |1−|c|| , ∀x ∈ Cω , (2) X  c j x(t − jσ ), |c| < 1   j≥0 −1 X (A x)(t) =  − c− j x(t + jσ ), |c| > 1.   j≥1

(3)

Rω 0

|(A−1 x)(t)|dt ≤

Rω 1 |1−|c|| 0

Lemma 2.2. For ai , xi ≥ 0, and !p n n X X p ai xi ≤ ai xi . i=1

i=1

Pn

|x(t)|dt, ∀x ∈ Cω .

i=1 ai

= 1, the following inequality holds for any p > 1

(2.1)

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Proof. Since h(x) = x p is convex on [0, +∞) when p > 1, Lemma 2.2 is obtained immediately. Lemma 2.3. If |c| 6= 1 and p > 1, then Z ω Z ω 1 |(A−1 x)(t)| p dt ≤ |x(t)| p dt, |1 − |c|| p 0 0



∀x ∈ Cω .

(2.2)

Proof. From (2.1), we have !p |(A

−1

p

x)(t)| ≤

X

|c| |x(t − jσ )| j

,

|c| < 1.

j≥0 (1−|c|)|c| j 1−|c|n

P , then a j ≥ 0 and n−1 j=0 a j = 1, from Lemma 2.2, one can see !p !p  n−1  n−1 X 1 − |c|n p X j a j |x(t − jσ )| |c| |x(t − jσ )| = 1 − |c| j=0 j=0   n−1 1 − |c|n p X ≤ a j |x(t − jσ )| p 1 − |c| j=0  p−1 X  n−1 n 1 − |c| |c| j |x(t − jσ )| p . = 1 − |c| j=0

Let a j =

Let n → +∞, then we drive !p

X 1 |c| j |x(t − jσ )| p . (2.3) (1 − |c|) p−1 j≥0 j≥0 Rω Rω Noticing that the periodicity of x(t), for any j ≥ 0, 0 |x(t − jσ )| p dt = 0 |x(t)| p dt, from (2.3), we can obtain (2.2). If |c| > 1, from (2.1), we can also prove (2.2) is true in the same way.  |(A−1 x)(t)| p ≤

X

|c| j |x(t − jσ )|

≤

By elementary analysis, it is easy to prove the following lemma. Lemma 2.4. Suppose that a ≥ 0, b ≥ 0 and 0 < p ≤ 1. Then (a + b) p ≤ a p + b p . Now we consider the following equation in Cω1 of the form (φ p (y 0 ))0 = F(y),

(2.4)

where (φ p (y 0 ))0 (t) = (φ p (y 0 (t)))0 , F : Cω1 → Cω is continuous and takes bounded sets into bounded sets. Let us define Z 1 ω P : Cω1 → Cω1 , u 7→ u(0), Q : Cω → Cω , h 7→ h(s)ds, ω 0 and H (h)(t) =

t

Z

h(s)ds, 0

h ∈ Cω .

Then we can obtain that if y ∈ Cω1 solves (2.4), then y satisfies the abstract equation y = P y + Q F(y) + K(F(y)),

(2.5)

where the operator K : Cω → Cω1 is given by K(h)(t) = H {φq [a((I − Q)h) + H ((I − Q)h)]}(t), where a : Cω → R can be obtained from [12] for φ = φ p .

h ∈ Cω ,

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Lemma 2.5. Suppose that F : Cω1 → Cω is continuous and takes bounded sets into bounded sets, then the operator K ◦ F : Cω1 → Cω1 is completely continuous. Proof. From [12], we see that a : Cω → R is a continuous function which sends sets of Cω into bounded sets of R, so the continuity of K ◦ F in Cω1 follows by observing that this operator is a composition of continuous operators. Also, we have that [K ◦ F(y)]0 = φq [a((I − Q)F(y)) + H ((I − Q)F(y))],

(2.6)

which is also a composition of continuous operators and hence continuous. Now, let D be a nonempty bounded set in Cω1 . Then there exists a number b > 0 such that kF(y)k∞ ≤ b,

∀y ∈ D.

0 Suppose that {yn }∞ 1 ⊂ D, for t, t ∈ [0, ω], we have that Z t |H (I − Q)F(yn )(t) − H (I − Q)F(yn )(t 0 )| = H (I − Q)F(yn )(s)ds 0 t

≤ 2b|t − t 0 |. Hence the sequence {H (I − Q)F(yn )} is uniformly bounded and equicontinuous. By the Ascoli–Arzela theorem there is a subsequence of {H (I − Q)F(yn )}, which we rename the same, which is convergent. Then, passing to a subsequence if necessary, we obtain that the sequence {a((I − Q)F(yn )) + H ((I − Q)F(yn ))} is convergent. Using that φq is continuous it follows from [K ◦ F(yn )]0 = φq [a((I − Q)F(yn )) + H ((I − Q)F(yn ))] that the sequence {[K ◦ F(yn )]0 } is convergent and so is the sequence {K ◦ F(yn )}. Therefore, K ◦ F(D) is a compact set.  Lemma 2.6. Let Ω be an open bounded set in Cω1 . Suppose that the following conditions hold: (i) For each λ ∈ (0, 1) the problem (φ p (y 0 ))0 = λF(y),

(2.7)

has no solution on ∂Ω ; (ii) The equation Z 1 ω F(y)(t)dt = 0, F(y) := ω 0 has no solution on ∂Ω ∩ R; (iii) The Brouwer degree of F deg{F, Ω ∩ R, 0} 6= 0. Then (2.4) has at least one ω-periodic solution in Ω¯ . Proof. Let us embed problem (2.4) into the one parameter family of problems (φ p (y 0 ))0 = λF(y) + (1 − λ)Q F(y).

(2.8)

For λ ∈ (0, 1], observe that, in both cases, y is an ω-periodic solution of (2.7) or y is a solution of (2.8), we have necessarily Z 1 ω Q F(y) = F(y)(s)ds = 0. ω 0 It follows that, for λ ∈ (0, 1], (2.7) and (2.8) have the same solutions. Furthermore, given the property of F, it is easy to see that the operator N : Cω1 × [01] → Cω defined by N (y, λ) = λF(y) + (1 − λ)Q F(y)

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is continuous and takes bounded sets into bounded sets. Also problem (2.8) can be written in the equivalent form y = G(y, λ)

(2.9)

with G(y, λ) := P y + Q F(y) + (K ◦ [λF(y) + (1 − λ)Q F(y)])(y) = P y + Q F(y) + (K ◦ [λ(I − Q)F(y)])(y). From Lemma 2.5, we obtain that G : Cω1 × [0, 1] → Cω1 is a completely continuous operator. The remaining proof of this lemma is very similar to that of Theorem 3.1 in [12], we omit it here.  3. Existence of periodic solutions Rω Theorem 3.1. Suppose that f (0) = 0 e(t)dt = 0 and there exist constants r1 ≥ 0, r2 > 0, r3 > 0, K > 0 and d > 0 such that (A1 ) | f (x)| ≤ K + r1 |x| p−1 , ∀x ∈ R; (A2 ) g(x) < −K for x < −d and g(x) > K + r1r3 x p−1 for x > d; |g(x)| (A3 ) limx→−∞ |x| p−1 = r 1 r 2 . Then (1.1) has at least one ω-periodic solution if  1  1   p−1  p−1 p−1 2(1 + |c|)r1 r < |1 − |c|| p , + ω 1 + r2 r +ω where r = max{ r12 , r13 }. Proof. For any λ ∈ (0, 1), we consider the following parameter equation

0 φ p (x(t) − cx(t − σ ))0 + λ f (x 0 (t)) + λg(x(t − τ (t))) = λe(t).

(3.1)

Let x(t) be a possible ω-periodic solution of (3.1). Integrating both sides of (3.1) over [0, ω], we have Z ω [ f (x 0 (t)) + g(x(t − τ (t)))]dt = 0.

(3.2)

0

By the integral mean value theorem, there is η ∈ [0, ω] such that Z 1 ω f (x 0 (t))dt, g(x(ξ )) = − ω 0

(3.3)

where ξ = η − τ (η). If r1 > 0, let !p  1 p−1 1 r1 G(z) = 2(1 + |c|)(r1r2 + z) max , +ω r3 r1 r2 − z !    1 p−1 1 r1 + 2(1 + |c|)r1 max , +ω , r3 r1 r2 − z 



we can see G(z) is continuous on [0, 21 r1r2 ], and  1   1  p−1  G(0) = 2(1 + |c|)r1 r p−1 + ω 1 + r2 r p−1 + ω < |1 − |c|| p . So there exists a small number z 0 ∈ (0, 12 r1r2 ) such that G(z) < |1 − |c|| p ,

z ∈ (0, z 0 ].

(3.4)

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|1−|c|| p 1 Choose ε = min{z 0 , 4(1+|c|) ( r31+r ) }, then in either case r1 > 0 or r1 = 0, we can obtain 3ω

G(ε) < |1 − |c|| p .

(3.5)

For such ε, from (A3 ), there exists ρ > d, such that |g(x)| > (r1r2 − ε) > 0, for x < −ρ, r1 > 0 |x| p−1 |g(x)| < (r1r2 + ε), for x < −ρ, r1 ≥ 0. |x| p−1

and (3.6)

Now, we can assert that there exists t0 ∈ [0, ω) such that |x(t0 )| p−1 ≤ αω

− q1

p−1

kx 0 k p

+ β,

(3.7) n

1 r1 > 0, r1 where α = max{ r13 , r1 rr21−ε }, β = ρ p−1 + Kr1sgn r2 −ε , sgn r1 = 0 r1 = 0. Case 1. r1 = 0. If |x(ξ )| > d, from (A1 ), (A2 ) and (3.3), we can see K < |g(x(ξ ))| ≤ K , which is a contradiction. So

|x(ξ )| ≤ d < ρ.

(3.8)

Case 2. r1 > 0. (1) If x(ξ ) > ρ, then K + r1r3 (x(ξ )) p−1 < g(x(ξ )) ≤ K +

r1 ω

Rω 0

|x 0 (t)| p−1 dt, which yields that

1 − q1 0 p−1 ω kx k p . r3 (2) If x(ξ ) < −ρ, from (3.3), we have 0 < (x(ξ )) p−1 ≤

(3.9)

(r1r2 − ε)|x(ξ )| p−1 < |g(x(ξ ))| ≤ K + ≤ K + r1 ω

− q1

r1 ω

p−1

kx 0 k p

ω

Z

|x 0 (t)| p−1 dt

0

.

Hence K r1 p−1 −1 + ω q kx 0 k p . r1 r2 − ε r1 r2 − ε (3) If −ρ ≤ x(ξ ) ≤ ρ, then |x(ξ )| p−1 ≤

(3.10)

|x(ξ )| ≤ ρ.

(3.11)

From (3.8)–(3.11), we can see   1 −1 K r1 p−1 p−1 −1 |x(ξ )| p−1 ≤ max ρ p−1 , ω q kx 0 k p , + ω q kx 0 k p r3 r1 r2 − ε r1 r2 − ε ≤ αω

− q1

p−1

kx 0 k p

+ β.

Let ξ = kω + t0 , where k is an integer and t0 ∈ [0, ω), we get (3.7). Multiplying both sides of (3.1) by (Ax)(t) = x(t) − cx(t − σ ) and integrating them over [0, ω], we have Z ω p kAx 0 k p = λ (Ax)(t)[ f (x 0 (t)) + g(x(t − τ (t))) − e(t)]dt 0

≤ (1 + |c|)kxk∞

ω

Z

[| f (x 0 (t))| + |g(x(t − τ (t)))| + |e(t)|]dt.

(3.12)

0

Let E 1 = {t ∈ [0, ω] : x(t − τ (t)) < −ρ}, E 2 = {t ∈ [0, ω] : |x(t − τ (t))| ≤ ρ}, E 3 = {t ∈ [0, ω] : x(t − τ (t)) > ρ}. From (3.2), we know that Z Z |g(x(t − τ (t)))|dt = g(x(t − τ (t)))dt E3

E3

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S. Peng / Nonlinear Analysis 69 (2008) 1675–1685

Z 

Z =− Z ≤

+ Z 

|g(x(t − τ (t)))|dt +

+

Therefore, we obtain Z Z ω |g(x(t − τ (t)))|dt ≤ 2

Z  + E1

Z

f (x 0 (t))dt

| f (x 0 (t))|dt.

0

E2

E1

ω 0 ω

E2

E1

0

g(x(t − τ (t)))dt −

Z

0

E2

≤ 2(r1r2 + ε)

ω

Z

| f (x 0 (t))|dt Z ω | f (x 0 (t))|dt |x(t − τ (t))| p−1 dt + 2ωgρ +

|g(x(t − τ (t)))|dt +

Z E1

0

p−1

≤ 2(r1r2 + ε)ωkxk∞ + 2ωgρ +

ω

Z

| f (x 0 (t))|dt,

(3.13)

0

where gρ = max|x|≤ρ |g(x)|. Substituting (3.13) into (3.12), we have i h 1 p p p−1 kAx 0 k p ≤ 2(1 + |c|) (r1r2 + ε)ωkxk∞ + r1 ω p kxk∞ kx 0 k p + a1 kxk∞ (3.14) Rω   where a1 = (1 + |c|) 2ωgρ + 2K ω + 0 |e(t)|dt . We can claim that there exists a constant R1 > 0, such that kx 0 k p ≤ R1 .

(3.15)

(I) Case of p ≥ 2. Since 0 < 1

|x(t0 )| ≤ α p−1 ω

− 1p

1 p−1

≤ 1, from (3.7) and Lemma 2.4, we have 1

kx 0 k p + β p−1 .

So kxk∞ ≤ |x(t0 )| +

ω

Z

|x 0 (t)|dt

0

≤α

1 p−1

1

1

− 1p

ω kx 0 k p + ω q kx 0 k p + β p−1 h 1 i 1 −1 = ω p α p−1 + ω kx 0 k p + β p−1 := a0 kx 0 k p + b0 , −1

1

(3.16)

1

where a0 = ω p [α p−1 + ω], b0 = β p−1 . By elementary analysis, there is a constant h > 0, independent of λ, such that (1 + u) p < 1 + (1 + p)u,

∀u ∈ (0, h].

If kx 0 k p = 0, then kxk∞ ≤ b0 . Otherwise, if kx 0 k p ≤ If

b0 a0 kx 0 k p p kxk∞

b0 a0 kx 0 k p

(3.17) ≥ h, then

b0 . a0 h

(3.18)

≤ h, we have 0

≤ a0 kx k p + b0


p

=

p p a0 kx 0 k p

 p p ≤ a0 kx 0 k p 1 + ( p + 1)



p b0 1+ a0 kx 0 k p  p−1 p p p−1 = a0 kx 0 k p + ( p + 1)a0 b0 kx 0 k p .

b0 a0 kx 0 k p

(3.19)

Substituting (3.16) and (3.19) into (3.14), we get p

p

p−1

kAx 0 k p ≤ G(ε)kx 0 k p + α1 kx 0 k p

+ a2 kx 0 k p + b1 ,

(3.20)

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S. Peng / Nonlinear Analysis 69 (2008) 1675–1685 p−1

where α1 = 2(1 + |c|)b0 [(r1r2 + ε)( p + 1)a0 From Lemma 2.3, we can see that p

1

ω + r1 ω p ], a2 = a1 a0 , b1 = a1 b0 .

p

p

|1 − |c|| p · kx 0 k p = |1 − |c|| p kA−1 Ax 0 k p ≤ kAx 0 k p . So it follows from (3.20) that p

p

p−1

|1 − |c|| p · kx 0 k p ≤ G(ε)kx 0 k p + α1 kx 0 k p

+ a2 kx 0 k p + b1 .

(3.21)

In view of p > 1, from (3.5), it follows that there exists a constant R2 > 0, such that kx 0 k p ≤ R2 .

(3.22)

From (3.18) and (3.22), we can see that (3.15) is true. (II) Case of 1 < p < 2. Under this circumstance, we have 0 < p − 1 < 1 and such that   1 1 p−1 (1 + u) ≤1+ 1+ u = 1 + qu, u ∈ (0, h 1 ]. p−1

1 p−1

> 1. There is a number h 1 > 0,

(3.23)

1

From (3.7), if

0

βω q p−1 αkx 0 k p



kx k p ≤

β αh 1

> h 1 , then



1 p−1

1

ωp.

(3.24)

1

If

βω q p−1 αkx 0 k p

≤ h 1 , from (3.23), we obtain

1   p−1 p−1 −1 |x(t0 )| ≤ αω q kx 0 k p + β 1

≤α

1 p−1

ω

− 1p

1

qβα p−1 ω q

0

kx k p +

αω

1 p

2− p

kx 0 k p

.

Thus, kxk∞ ≤ ω

− 1p

h

1 1 i 1 qβα p−1 ω q 0 2− p 2− p kx k p := a0 kx 0 k p + b2 kx 0 k p , α p−1 + ω kx 0 k p + 1 αω p 1

where a0 is defined in (3.16), b2 = It is easy to see

(3.25)

1

qβα p−1 ω q 1

αω p

.

   p p−1 2− p 2− p p−1 kxk∞ = kxk∞ · kxk∞ ≤ a0 kx 0 k p + b2 kx 0 k p a0 kx 0 k p + b2 kx 0 k p    2− p p−1 p−1 p−1 (2− p)( p−1) ≤ a0 kx 0 k p + b2 kx 0 k p a0 kx 0 k p + b2 kx 0 k p p

p

p−1

= a0 kx 0 k p + b2 a0

p−1

kx 0 k p + a0 b2

(2− p)( p−1)+1

kx 0 k p

(2− p) p

p

+ b2 kx 0 k p

.

(3.26)

Substituting (3.25) and (3.26) into (3.14), we derive that p

p

p

(2− p)( p−1)+1

|1 − |c|| p kx 0 k p ≤ kAx 0 k p ≤ G(ε)kx 0 k p + a3 kx 0 k p (2− p) p

+ a4 kx 0 k p

p−1

where a2 = 2(1 + |c|)(r1r2 + ε)ωa0 b2 1 p

r1 ω + a1 a0 ], a6 = a1 b2 .

2− p

+ a5 kx 0 k p + a6 kx 0 k p

,

(3.27) p

p−1

, a4 = 2(1 + |c|)(r1r2 + ε)ωb2 , a5 = 2(1 + |c|)b2 [(r1r2 + ε)ωa0

+

S. Peng / Nonlinear Analysis 69 (2008) 1675–1685

1683

Since max{(2 − p)( p − 1) + 1, (2 − p) p, 1, 2 − p} < p, from (3.5), there is a number R3 > 0 such that kx 0 k p ≤ R3 .

(3.28)

From (3.24), (3.28) and (3.15) it is also true for 1 < p < 2. By (3.15), (3.16) and (3.25), we can see that there exists a positive number R4 > 0 such that kxk∞ ≤ R4 .

(3.29)

From (3.1), we have Z Z ω 0 0 |(φ p ((Ax )(t))) |dt ≤

ω

[| f (x 0 (t))| + |g(x(t − τ (t)))| + |e(t)|]dt Z ω p−1 |e(t)|dt ≤ K ω + r1 ω1/ p kx 0 k p + ωg R4 + 0

0

0

≤ Kω

p−1 + r1 ω1/ p R1

+ ωg R4 +

ω

Z

|e(t)|dt := R5 ,

(3.30)

0

where g R4 = max|s|≤R4 |g(s)|. Clearly, there exists t1 ∈ (0, ω) such that φ p ((Ax 0 )(t1 )) = 0. Thus, for any t ∈ [0, ω], we have Z t Z ω 0 0 0 |φ p ((Ax )(t))| = |φ p ((Ax )(s)) |ds ≤ |φ p ((Ax 0 )(s))0 |ds ≤ R5 , 0

t1

from which it follows that q−1

kAx 0 k∞ ≤ φq (R5 ) = R5

.

(3.31)

From Lemma 2.1, we derive q−1

kx 0 k∞ = kA−1 Ax 0 k∞ ≤

R5 kAx 0 k∞ ≤ := R6 . |1 − |c|| |1 − |c||

Now, let y(t) = (Ax)(t), we can see that (3.1) is equivalent to the following equation

0 φ p y 0 (t) + λ f ((A−1 y 0 )(t)) + λg((A−1 y)(t − τ (t))) = λe(t).

(3.32)

(3.33)

So, if y is an ω-periodic solution of (3.33), then x = A−1 y is an ω-periodic solution of (3.1). Let F(y)(t) := e(t) − f ((A−1 y)0 (t)) − g((A−1 y)(t − τ (t))).

(3.34)

Then (3.33) is (2.7). Since f, g are continuous functions and A has a continuous inverse, we can easily see the mapping F : Cω1 → Cω in (3.34) is continuous and takes bounded sets into bounded sets. Set R7 = 2(1 + |c|) max{R4 , R6 , d}, Ω = {y ∈ Cω1 : kyk < R7 }, we can see that (3.33) has no solution on ∂Ω for λ ∈ (0, 1). In fact, if y = Ax is a solution (3.33) on ∂Ω , then kyk = R7 , kyk∞ = R7 or ky 0 k∞ = R7 . If kyk∞ = R7 , then kyk∞ = 2 max{R4 , R6 , d} > R4 , kxk∞ ≥ 1 + |c| from (3.29), which is a contradiction. Similarly, ky 0 k∞ = R7 is also impossible. If y ∈ ∂Ω ∩ R, then y is a constant y and |y| = R7 , x = A−1 y = 1−c , |x| ≥ 2 max{R4 , R6 , d} > d, from (A2 ), we know Z ω 1 F(y) = [e(t) − f (0) − g(A−1 y)]dt = −g(A−1 y) 6= 0, (3.35) ω 0 on ∂Ω ∩ R. Let H (y, µ) = µ(−A−1 y) + (1 − µ)F(y), y ∈ ∂Ω ∩ R, µ ∈ [0, 1], then we have (A−1 y) · H (y, µ) = −µ(A−1 y)2 − (1 − µ)(A−1 y) · g(A−1 y) < 0.

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Thus H (y, µ) is a homotopic transformation and deg{F, Ω ∩ R, 0} = deg{−A−1 y, Ω ∩ R, 0} = sgn(c − 1) 6= 0. So, for (3.33), all of the conditions of Lemma 2.6 are satisfied. By Lemma 2.6, we conclude that

0 φ p y 0 (t) + f ((A−1 y 0 )(t)) + g((A−1 y)(t − τ (t))) = e(t) has at least one ω-periodic solution y¯ . Therefore, x¯ = A−1 y¯ is an ω-periodic solution of (1.1).

(3.36) 

A similar argument leads to Theorem 3.2. Suppose that f (0) = d > 0 such that

Rω 0

e(t)dt = 0 and there exist constants r1 ≥ 0, r2 > 0, r3 > 0, K > 0 and

(A∗1 ) | f (x)| ≤ K + r1 |x| p−1 , ∀x ∈ R; (A∗2 ) g(x) > K for x > d and g(x) < −K − r1r3 |x| p−1 for x < −d; (A∗3 ) limx→+∞ xg(x) p−1 = r 1 r 2 . Then (1.1) has at least one ω-periodic solution if  1   1  p−1  2(1 + |c|)r1 r p−1 + ω 1 + r2 r p−1 + ω < |1 − |c|| p , where r = max{ r12 , r13 }. Remark 3.1. If c = 0, (1.1) reduces to (1.3), we obtain some new results which are different from those in [5–7]. 1

1

Remark 3.2. It is clear that if c = 0 and r1 = 0, then 2(1 + |c|)r1 (r p−1 + ω)[1 + r2 (r p−1 + ω) p−1 ] = 0 < 1. For (1.4), when f (t, x) = f (x), g(t, x) = g(x) and σ (t) = 0, Theorem A(H2 ) can guarantee that Theorem 3.1(A2 ) holds. Furthermore, Theorem A(H3 ) leads to Theorem 3.1(A3 ) holds. By Theorem 3.1, (1.4) has at least one 2π-periodic solution. Clearly, assumption (A3 ) of Theorem 3.1 is weaker than (H3 ) of Theorem A. Rω Remark 3.3. If f (0) 6= 0 or 0 e(t)dt 6= 0, Eq. (1.1) can be converted to the following equation

0 φ p (x(t) − cx(t − σ ))0 + f 1 (x 0 (t)) + g1 (x(t − τ (t))) = e1 (t), (3.37) R R ω ω where f 1 (x) = f (x) − f (0), g1 (x) = g(x) + f (0) − ω1 0 e(t)dt, e1 (t) = e(t) − ω1 0 e(t)dt. As f 1 (0) = 0 and Rω 0 e1 (t)dt = 0, Eq. (3.37) can be discussed in the same way. As an application, we list the following example. Example 3.1. Consider equation (φ 3 ((x(t) − 5x(t − 4))0 ))0 + f (x 0 (t)) + g(x(t − sin(2πt))) = cos(2πt),

(3.38)

2

x 2√, x ≥ 0 √ where p = 32 , c = 5, σ = 4, ω = 1, τ (t) = sin(2π t), e(t) = cos(2π t), f (x) = 19 |x|, g(x) = 2 2 √3 p6 . x |x|, x < 0 9 √ √ Let d = 9, K = 2 3 2 , r1 = 91 , r2 = r3 = 2 2. We can see that (A1 ), (A2 ) and (A3 ) of Theorem 3.1 are satisfied.



1

1

Furthermore, we can compute 2(1 + |c|)r1 (r p−1 + ω)[1 +r2 (r p−1 + ω) p−1 ] = 6 < |1 − |c|| p = 8. From Theorem 3.1, (3.38) has at least one 1-periodic solution. Acknowledgments The author would like to thank an anonymous referee for a careful reading of the manuscript and helpful opinions. The author of this paper was supported by the National Natural Sciences Foundation of the People’s Republic of China under Grant 60572073.

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