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Periodic solutions for Rayleigh type p -Laplacian equation with deviating arguments
Applied Mathematics Letters 20 (2007) 43–47 www.elsevier.com/locate/aml
Periodic solutions for Rayleigh type p-Laplacian equation with deviating arguments Minggang Zong a,b,∗ , Hongzhen Liang a a Faculty of Science, Jiangsu University, Zhenjiang 212013, Jiangsu, PR China b School of Statistics, Renmin University of China, Beijing, 100872, PR China
Received 18 June 2005; received in revised form 15 February 2006; accepted 22 February 2006
Abstract By using topological degree theory and some analysis skill, we obtain some sufficient conditions for the existence of periodic solutions for Rayleigh type p-Laplacian differential equation with deviating arguments. c 2006 Elsevier Ltd. All rights reserved. Keywords: p-Laplacian; Periodic solutions; Rayleigh equation; Deviating argument; Topological degree
In the last few years, the existence of periodic solutions for delay differential equations has received a lot of attention, For example, in [1–4], the following second order differentials equations with delay: x (t) + f (x (t)) + g(x(t − τ )) = p(t), x (t) + m 2 x(t) + g(x(t − τ )) = p(t), x (t) + f (t, x (t − τ )) + g(x(t − σ )) = p(t), were studied. However, so far as we are aware, fewer papers discuss the Rayleigh type equation with deviating arguments. In [5] Lu and Ge studied the following Rayleigh equation with a deviating argument: x (t) + f (x (t)) + g(x(t − τ (t))) = p(t),
(1)
and by using coincidence degree theory, the authors obtained sufficient conditions for the existence of periodic solutions for Eq. (1). In this work we deal with the existence of periodic solutions for the Rayleigh type p-Laplacian differential equation with deviating arguments of the form (ϕ p (x (t))) + f (t, x (t − σ (t))) + g(t, x(t − τ (t))) = e(t),
(2)
where p > 1 and ϕ p : R → R is given by ϕ p (s) = |s| p−2 s for s = 0 and ϕ p (0) = 0, f, g are continuous functions defined on R 2 and are periodic about t with f (t, ·) = f (t + 2π, ·), g(t, ·) = g(t + 2π, ·), f (t, 0) = 0 for ∗ Corresponding author at: Faculty of Science, Jiangsu University, Zhenjiang 212013, Jiangsu, PR China.
E-mail address: [email protected] (M. Zong). c 2006 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2006.02.021
44
M. Zong, H. Liang / Applied Mathematics Letters 20 (2007) 43–47
2π t ∈ R. e, σ and τ are continuous periodic functions defined on R with period 2π and 0 e(t) dt = 0. Obviously, as p = 2, ϕ2 (x) = x is a linear function and when σ (t) = τ (t) = 0, f (t, x) = f (x), g(t, x) = g(x), Eq. (2) reduces to Eq. (1). So the results in [4] are contained in our work. However, because the p-Laplacian operator ϕ p : ϕ p (x) = |x| p−2 x is nonlinear, the continuation theorem of Mawhin [6] is not applicable, which leads to difficulty for solving the problem (2). By using topological degree theory and some analysis skill, we obtain the sufficient conditions for the existence of periodic solutions for Eq. (2). For convenience, let us define C T1 := C T1 [0, T ] = {x ∈ C 1 [0, T ], x(0) = x(T ), x (0) = x (T ), T > 0}. For the periodic boundary value problem (ϕ p (x (t))) = f (t, x, x ),
x(0) = x(T ), x (0) = x (T )
(3)
where f ∈ C(R 3 , R), we have the following lemma: Lemma ([7]). Let Ω be an open bounded set in C T1 ; suppose that the following conditions hold: (i) For each λ ∈ (0, 1) the problem (ϕ p (x (t))) = λ f (t, x, x ), x(0) = x(T ), x (0) = x (T ) has no solution on ∂Ω . (ii) The equation 1 T F(a) := f (t, a, 0) dt = 0, T 0 has no solution on ∂Ω R. (iii) The Brouwer degree of F deg F, Ω R, 0 = 0. Then the periodic boundary value problem (3) has at least one periodic solution on Ω . By using this lemma, we obtain our main results: Theorem 1. 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 Eq. (2) has at least one solution with period 2π. Proof. Consider the homotopic equation of Eq. (2) as follows: (ϕ p (x (t))) + λ f (t, x (t − σ (t))) + λg(t, x(t − τ (t))) = λe(t),
λ ∈ (0, 1).
(4) x (t)
be an arbitrary solution of Eq. (4) with period 2π. Firstly, we give a priori bound which is Let x(t) ∈ independent of λ. By integrating both sides of Eq. (4) over [0, 2π], and noticing that x (0) = x (2π), we have 2π [ f (t, x (t − σ (t))) + g(t, x(t − τ (t)))] dt = 0. (5) 1 C2π
0
As x(0) = x(2π), there exists t0 ∈ [0, 2π] such that x (t0 ) = 0, while ϕ p (0) = 0; we see t |ϕ p (x (t))| = (ϕ p (x (s))) ds t0
2π
≤λ
2π
| f (t, x (t − σ (t)))| dt + λ
0
0
2π
≤ 2π K + 0
2π
|g(t, x(t − τ (t)))| dt + λ
|e(t)| dt
0
|g(t, x(t − τ (t)))| dt + 2π max |e(t)|. t ∈[0,2π]
(6)
M. Zong, H. Liang / Applied Mathematics Letters 20 (2007) 43–47
and then we can assert that there exists some positive constant D1 such that 2π |g(t, x(t − τ (t)))| dt ≤ 4π D1 + 2π K .
45
(7)
0
In fact, in view of condition (H1 ) and Eq. (5) we have 2π 2π {g(t, x(t − τ (t))) − K } dt ≤ {g(t, x(t − τ (t))) − | f (t, x (t − σ (t)))|} dt 0
0 2π
≤
{g(t, x(t − τ (t))) + f (t, x (t − σ (t)))} dt
0
= 0.
(8)
Defining E 1 = {t : t ∈ [0, 2π], x(t − τ (t)) > D} E 2 = {t : t ∈ [0, 2π], |x(t − τ (t))| ≤ D} ∪ {t : t ∈ [0, 2π], x(t − τ (t)) < −D} we can get |g(t, x(t − τ (t)))| dt ≤ 2π max{M,
E2
{|g(t, x(t − τ (t)))| − K } dt =
E1
sup
t ∈[0,2π],|x|≤D
|g(t, x)|}
{g(t, x(t − τ (t))) − K } dt
E 1
≤ − {g(t, x(t − τ (t))) − K } dt E2 {|g(t, x(t − τ (t)))| + K } dt ≤ E2
which yields |g(t, x(t − τ (t)))| dt ≤ |g(t, x(t − τ (t)))| dt + K dt E1 E E 1 +E 2 2 = |g(t, x(t − τ (t)))| dt + 2π K , E2
that is
2π
|g(t, x(t − τ (t)))| dt =
0
|g(t, x(t − τ (t)))| dt + E1
≤2
|g(t, x(t − τ (t)))| dt E2
|g(t, x(t − τ (t)))| dt + 2π K E2
≤ 4π max{M,
sup
t ∈[0,2π],|x|≤D
|g(t, x)|} + 2π K
= 4π D1 + 2π K where D1 = max{M, supt ∈[0,2π],|x|≤D |g(t, x)|}. Now substituting inequality (7) into (6) we obtain |ϕ p (x (t))| ≤ 4π D1 + 4π K + 2π max |e(t)| := M1 . t ∈[0,2π]
By using the monotonicity of ϕ p , we can get some positive constant M2 such that for all t ∈ R, |x (t)| ≤ M2 . Since (8) is satisfied, there must exist t1 ∈ (0, 2π) such that f (t1 , x (t1 − σ (t1 ))) + g(t1 , x(t1 − τ (t1 ))) = 0.
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M. Zong, H. Liang / Applied Mathematics Letters 20 (2007) 43–47
By applying condition (H1 ) we have |g(t1 , x(t1 − τ (t1 )))| = | − f (t1 , x (t1 − σ (t1 )))| ≤ K , and from condition (H2 ) we can get that |x(t1 − τ (t1 ))| ≤ D. Since x(t) is periodic with period 2π, so t1 − τ (t1 ) = 2nπ + t2 , t2 ∈ (0, 2π) where n is some integer; therefore |x(t2 )| ≤ D. So for all t ∈ R, t |x(t)| = |x(t2 ) + x (s) ds| ≤ D + 2π M2 := M3 . (9) t2
Setting 1 (R, R) x| ≤ M3 + 1, |x | ≤ M2 + 1} Ω = {x ∈ C2π
we know that Eq. (4) has no solution on ∂Ω as λ ∈ (0, 1) and when x(t) ∈ ∂Ω from (9) we know that D + 2π M2 = M3 , so M3 + 1 > D; we see that 2π 2π 1 1 {g(t, M3 + 1) + e(t)} dt = g(t, M3 + 1) dt > 0 2π 0 2π 0 2π 2π 1 1 {g(t, −M3 − 1) + e(t)} dt = g(t, −M3 − 1) dt < 0 2π 0 2π 0
R, x(t) = M3 +1 or x(t) = −M3 −1,
so condition (ii) is also satisfied. Setting 2π 1 H (x, μ) = μx + (1 − μ) g(t, x) dt, 2π 0 when x ∈ ∂Ω R, μ ∈ [0, 1] we have 2π 1 2 x H (x, μ) = μx + (1 − μ)x g(t, x) dt > 0 2π 0 and thus H (x, μ) is a homotopic transformation and
2π
1 g(t, x) dt, Ω R, 0 deg F, Ω R, 0 = deg 2π 0
= deg x, Ω R, 0 = 0 so condition (iii) is satisfied. In view of the previous lemma, there exists at least one solution with period 2π. This completes the proof. By using a similar argument, we can obtain the following theorem: Theorem 2. 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 t ∈ R and x ≥ D. Then Eq. (2) has at least one solution with period 2π. As an application, let us consider the following equation: x(t − sin t) (ϕ p x (t)) + cos t sin x (t − sin t) + arctan = cos t 1 + cos2 t
(10)
x where we have f (t, x) = cos t sin x, g(t, x) = arctan 1+cos 2 t , σ (t) = τ (t) = sin t and e(t) = cos t. We can easily π π choose K = 1, D > 2 and M = 2 such that (H1 )–(H3 ) hold. By Theorem 1, Eq. (10) has at lest one solution.
M. Zong, H. Liang / Applied Mathematics Letters 20 (2007) 43–47
47
2π Remark 1. If 0 e(t) dt = 0 and f (t, 0) = 0, the problem of existence of 2π-periodic solutions to Eq. (2) can be converted to the existence of 2π-periodic solutions to the equation (11) (ϕ p (x (t))) + f 1 (t, x (t − σ (t))) + g1 (t, x(t − τ (t))) = e1 (t) 2π 2π where f 1 (t, x) = f (t, x) − f (t, 0), g1 (t, x) = g(t, x) − 0 e(t) dt + f (t, 0) and e1 (t) = e(t) − 0 e(t) dt. Clearly, 2π 0 e1 (t) dt = 0 and f 1 (t, 0) = 0, Eq. (11) can be discussed by using Theorem 1 (or Theorem 2). Remark 2. When f (t, x) ≡ 0, Eq. (2) reduces to a Duffing type equation; the results also hold. Acknowledgement The authors are grateful to the referee for his or her suggestions on the first draft of the work. References [1] X. Huang, Z. Xiang, On the existence of 2π -periodic solution for delay Duffing equation x (t) + g(t, x(t − τ )) = p(t), Chinese Sci. Bull. 39 (3) (1994) 201–203. [2] Y. Li, Periodic solutions of the Li´enard equation with deviating arguments, J. Math. Res. Exposition 18 (4) (1998) 565–570 (in Chinese). [3] G.Q. Wang, S.S. Cheng, A priori bounds for periodic solutions of a delay Rayleigh equation, Appl. Math. Lett. 12 (1999) 41–44. [4] G. Wang, J. Yan, On existence of periodic solutions of the Rayleigh equation of retarded type, Int. J. Math. Math. Sci. 23 (1) (2000) 65–68. [5] S. Lu, W. Ge, Some new results on the existence of periodic solutions to a kind of Rayleigh equation with a deviating argument, Nonlinear Anal. 56 (2004) 501–514. [6] R.E. Gaines, J. Mawhin, Coincidence Degree, and Nonlinear Differential Equations, in: Lecture Notes in Mathematics, vol. 568, SpringerVerlag, Berlin, New York, 1977. [7] R. Man´asevich, J. Mawhin, Periodic solutions for nonlinear systems with p-Laplacian-like operators, J. Differential Equations 145 (1998) 367–393.
Periodic solutions for Rayleigh type p-Laplacian equation with deviating arguments Minggang Zong a,b,∗ , Hongzhen Liang a a Faculty of Science, Jiangsu University, Zhenjiang 212013, Jiangsu, PR China b School of Statistics, Renmin University of China, Beijing, 100872, PR China
Received 18 June 2005; received in revised form 15 February 2006; accepted 22 February 2006
Abstract By using topological degree theory and some analysis skill, we obtain some sufficient conditions for the existence of periodic solutions for Rayleigh type p-Laplacian differential equation with deviating arguments. c 2006 Elsevier Ltd. All rights reserved. Keywords: p-Laplacian; Periodic solutions; Rayleigh equation; Deviating argument; Topological degree
In the last few years, the existence of periodic solutions for delay differential equations has received a lot of attention, For example, in [1–4], the following second order differentials equations with delay: x (t) + f (x (t)) + g(x(t − τ )) = p(t), x (t) + m 2 x(t) + g(x(t − τ )) = p(t), x (t) + f (t, x (t − τ )) + g(x(t − σ )) = p(t), were studied. However, so far as we are aware, fewer papers discuss the Rayleigh type equation with deviating arguments. In [5] Lu and Ge studied the following Rayleigh equation with a deviating argument: x (t) + f (x (t)) + g(x(t − τ (t))) = p(t),
(1)
and by using coincidence degree theory, the authors obtained sufficient conditions for the existence of periodic solutions for Eq. (1). In this work we deal with the existence of periodic solutions for the Rayleigh type p-Laplacian differential equation with deviating arguments of the form (ϕ p (x (t))) + f (t, x (t − σ (t))) + g(t, x(t − τ (t))) = e(t),
(2)
where p > 1 and ϕ p : R → R is given by ϕ p (s) = |s| p−2 s for s = 0 and ϕ p (0) = 0, f, g are continuous functions defined on R 2 and are periodic about t with f (t, ·) = f (t + 2π, ·), g(t, ·) = g(t + 2π, ·), f (t, 0) = 0 for ∗ Corresponding author at: Faculty of Science, Jiangsu University, Zhenjiang 212013, Jiangsu, PR China.
E-mail address: [email protected] (M. Zong). c 2006 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2006.02.021
44
M. Zong, H. Liang / Applied Mathematics Letters 20 (2007) 43–47
2π t ∈ R. e, σ and τ are continuous periodic functions defined on R with period 2π and 0 e(t) dt = 0. Obviously, as p = 2, ϕ2 (x) = x is a linear function and when σ (t) = τ (t) = 0, f (t, x) = f (x), g(t, x) = g(x), Eq. (2) reduces to Eq. (1). So the results in [4] are contained in our work. However, because the p-Laplacian operator ϕ p : ϕ p (x) = |x| p−2 x is nonlinear, the continuation theorem of Mawhin [6] is not applicable, which leads to difficulty for solving the problem (2). By using topological degree theory and some analysis skill, we obtain the sufficient conditions for the existence of periodic solutions for Eq. (2). For convenience, let us define C T1 := C T1 [0, T ] = {x ∈ C 1 [0, T ], x(0) = x(T ), x (0) = x (T ), T > 0}. For the periodic boundary value problem (ϕ p (x (t))) = f (t, x, x ),
x(0) = x(T ), x (0) = x (T )
(3)
where f ∈ C(R 3 , R), we have the following lemma: Lemma ([7]). Let Ω be an open bounded set in C T1 ; suppose that the following conditions hold: (i) For each λ ∈ (0, 1) the problem (ϕ p (x (t))) = λ f (t, x, x ), x(0) = x(T ), x (0) = x (T ) has no solution on ∂Ω . (ii) The equation 1 T F(a) := f (t, a, 0) dt = 0, T 0 has no solution on ∂Ω R. (iii) The Brouwer degree of F deg F, Ω R, 0 = 0. Then the periodic boundary value problem (3) has at least one periodic solution on Ω . By using this lemma, we obtain our main results: Theorem 1. 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 Eq. (2) has at least one solution with period 2π. Proof. Consider the homotopic equation of Eq. (2) as follows: (ϕ p (x (t))) + λ f (t, x (t − σ (t))) + λg(t, x(t − τ (t))) = λe(t),
λ ∈ (0, 1).
(4) x (t)
be an arbitrary solution of Eq. (4) with period 2π. Firstly, we give a priori bound which is Let x(t) ∈ independent of λ. By integrating both sides of Eq. (4) over [0, 2π], and noticing that x (0) = x (2π), we have 2π [ f (t, x (t − σ (t))) + g(t, x(t − τ (t)))] dt = 0. (5) 1 C2π
0
As x(0) = x(2π), there exists t0 ∈ [0, 2π] such that x (t0 ) = 0, while ϕ p (0) = 0; we see t |ϕ p (x (t))| = (ϕ p (x (s))) ds t0
2π
≤λ
2π
| f (t, x (t − σ (t)))| dt + λ
0
0
2π
≤ 2π K + 0
2π
|g(t, x(t − τ (t)))| dt + λ
|e(t)| dt
0
|g(t, x(t − τ (t)))| dt + 2π max |e(t)|. t ∈[0,2π]
(6)
M. Zong, H. Liang / Applied Mathematics Letters 20 (2007) 43–47
and then we can assert that there exists some positive constant D1 such that 2π |g(t, x(t − τ (t)))| dt ≤ 4π D1 + 2π K .
45
(7)
0
In fact, in view of condition (H1 ) and Eq. (5) we have 2π 2π {g(t, x(t − τ (t))) − K } dt ≤ {g(t, x(t − τ (t))) − | f (t, x (t − σ (t)))|} dt 0
0 2π
≤
{g(t, x(t − τ (t))) + f (t, x (t − σ (t)))} dt
0
= 0.
(8)
Defining E 1 = {t : t ∈ [0, 2π], x(t − τ (t)) > D} E 2 = {t : t ∈ [0, 2π], |x(t − τ (t))| ≤ D} ∪ {t : t ∈ [0, 2π], x(t − τ (t)) < −D} we can get |g(t, x(t − τ (t)))| dt ≤ 2π max{M,
E2
{|g(t, x(t − τ (t)))| − K } dt =
E1
sup
t ∈[0,2π],|x|≤D
|g(t, x)|}
{g(t, x(t − τ (t))) − K } dt
E 1
≤ − {g(t, x(t − τ (t))) − K } dt E2 {|g(t, x(t − τ (t)))| + K } dt ≤ E2
which yields |g(t, x(t − τ (t)))| dt ≤ |g(t, x(t − τ (t)))| dt + K dt E1 E E 1 +E 2 2 = |g(t, x(t − τ (t)))| dt + 2π K , E2
that is
2π
|g(t, x(t − τ (t)))| dt =
0
|g(t, x(t − τ (t)))| dt + E1
≤2
|g(t, x(t − τ (t)))| dt E2
|g(t, x(t − τ (t)))| dt + 2π K E2
≤ 4π max{M,
sup
t ∈[0,2π],|x|≤D
|g(t, x)|} + 2π K
= 4π D1 + 2π K where D1 = max{M, supt ∈[0,2π],|x|≤D |g(t, x)|}. Now substituting inequality (7) into (6) we obtain |ϕ p (x (t))| ≤ 4π D1 + 4π K + 2π max |e(t)| := M1 . t ∈[0,2π]
By using the monotonicity of ϕ p , we can get some positive constant M2 such that for all t ∈ R, |x (t)| ≤ M2 . Since (8) is satisfied, there must exist t1 ∈ (0, 2π) such that f (t1 , x (t1 − σ (t1 ))) + g(t1 , x(t1 − τ (t1 ))) = 0.
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M. Zong, H. Liang / Applied Mathematics Letters 20 (2007) 43–47
By applying condition (H1 ) we have |g(t1 , x(t1 − τ (t1 )))| = | − f (t1 , x (t1 − σ (t1 )))| ≤ K , and from condition (H2 ) we can get that |x(t1 − τ (t1 ))| ≤ D. Since x(t) is periodic with period 2π, so t1 − τ (t1 ) = 2nπ + t2 , t2 ∈ (0, 2π) where n is some integer; therefore |x(t2 )| ≤ D. So for all t ∈ R, t |x(t)| = |x(t2 ) + x (s) ds| ≤ D + 2π M2 := M3 . (9) t2
Setting 1 (R, R) x| ≤ M3 + 1, |x | ≤ M2 + 1} Ω = {x ∈ C2π
we know that Eq. (4) has no solution on ∂Ω as λ ∈ (0, 1) and when x(t) ∈ ∂Ω from (9) we know that D + 2π M2 = M3 , so M3 + 1 > D; we see that 2π 2π 1 1 {g(t, M3 + 1) + e(t)} dt = g(t, M3 + 1) dt > 0 2π 0 2π 0 2π 2π 1 1 {g(t, −M3 − 1) + e(t)} dt = g(t, −M3 − 1) dt < 0 2π 0 2π 0
R, x(t) = M3 +1 or x(t) = −M3 −1,
so condition (ii) is also satisfied. Setting 2π 1 H (x, μ) = μx + (1 − μ) g(t, x) dt, 2π 0 when x ∈ ∂Ω R, μ ∈ [0, 1] we have 2π 1 2 x H (x, μ) = μx + (1 − μ)x g(t, x) dt > 0 2π 0 and thus H (x, μ) is a homotopic transformation and
2π
1 g(t, x) dt, Ω R, 0 deg F, Ω R, 0 = deg 2π 0
= deg x, Ω R, 0 = 0 so condition (iii) is satisfied. In view of the previous lemma, there exists at least one solution with period 2π. This completes the proof. By using a similar argument, we can obtain the following theorem: Theorem 2. 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 t ∈ R and x ≥ D. Then Eq. (2) has at least one solution with period 2π. As an application, let us consider the following equation: x(t − sin t) (ϕ p x (t)) + cos t sin x (t − sin t) + arctan = cos t 1 + cos2 t
(10)
x where we have f (t, x) = cos t sin x, g(t, x) = arctan 1+cos 2 t , σ (t) = τ (t) = sin t and e(t) = cos t. We can easily π π choose K = 1, D > 2 and M = 2 such that (H1 )–(H3 ) hold. By Theorem 1, Eq. (10) has at lest one solution.
M. Zong, H. Liang / Applied Mathematics Letters 20 (2007) 43–47
47
2π Remark 1. If 0 e(t) dt = 0 and f (t, 0) = 0, the problem of existence of 2π-periodic solutions to Eq. (2) can be converted to the existence of 2π-periodic solutions to the equation (11) (ϕ p (x (t))) + f 1 (t, x (t − σ (t))) + g1 (t, x(t − τ (t))) = e1 (t) 2π 2π where f 1 (t, x) = f (t, x) − f (t, 0), g1 (t, x) = g(t, x) − 0 e(t) dt + f (t, 0) and e1 (t) = e(t) − 0 e(t) dt. Clearly, 2π 0 e1 (t) dt = 0 and f 1 (t, 0) = 0, Eq. (11) can be discussed by using Theorem 1 (or Theorem 2). Remark 2. When f (t, x) ≡ 0, Eq. (2) reduces to a Duffing type equation; the results also hold. Acknowledgement The authors are grateful to the referee for his or her suggestions on the first draft of the work. References [1] X. Huang, Z. Xiang, On the existence of 2π -periodic solution for delay Duffing equation x (t) + g(t, x(t − τ )) = p(t), Chinese Sci. Bull. 39 (3) (1994) 201–203. [2] Y. Li, Periodic solutions of the Li´enard equation with deviating arguments, J. Math. Res. Exposition 18 (4) (1998) 565–570 (in Chinese). [3] G.Q. Wang, S.S. Cheng, A priori bounds for periodic solutions of a delay Rayleigh equation, Appl. Math. Lett. 12 (1999) 41–44. [4] G. Wang, J. Yan, On existence of periodic solutions of the Rayleigh equation of retarded type, Int. J. Math. Math. Sci. 23 (1) (2000) 65–68. [5] S. Lu, W. Ge, Some new results on the existence of periodic solutions to a kind of Rayleigh equation with a deviating argument, Nonlinear Anal. 56 (2004) 501–514. [6] R.E. Gaines, J. Mawhin, Coincidence Degree, and Nonlinear Differential Equations, in: Lecture Notes in Mathematics, vol. 568, SpringerVerlag, Berlin, New York, 1977. [7] R. Man´asevich, J. Mawhin, Periodic solutions for nonlinear systems with p-Laplacian-like operators, J. Differential Equations 145 (1998) 367–393.