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Periodic solutions for a kind of Rayleigh equation with a deviating argument
Computers and Mathematics with Applications 53 (2007) 825–830 www.elsevier.com/locate/camwa
Periodic solutions for a kind of Rayleigh equation with a deviating argumentI Yinggao Zhou ∗ , Xianhua Tang School of Mathematical Science and Computing Technology, Central South University, Changsha, Hunan 410083, PR China Received 20 January 2006; received in revised form 21 October 2006; accepted 12 December 2006
Abstract Existence of periodic solutions for a kind of Rayleigh equation with a deviating argument x 00 (t) + f (x 0 (t)) + g(x(t − τ (t))) = p(t) is studied, and some new results are obtained. Our work generalizes and improves the known results in the literature. c 2007 Published by Elsevier Ltd
Keywords: Rayleigh equations; Periodic solution; A priori bound; Continuation theorem; Deviating argument
1. Introduction Consider the Rayleigh equation with a deviating argument x 00 (t) + f (x 0 (t)) + g(x(t − τ (t))) = p(t),
(1.1)
and its auxiliary equation x 00 (t) + λ f (x 0 (t)) + λg(x(t − τ (t))) = λp(t),
λ ∈ (0, 1)
(1.2)
where f, g, p and τ are real continuous functions defined on R, τ and p are periodic with period 2π . In recent years, the existence of periodic solutions for a kind of Rayleigh equations was studied by some researchers (see [1–9]). In [1,2], continuation theorems are introduced and applied to the existence of solutions of differential equations. In particular, a specific example is given in [1, p. 99] (see also [2, p. 175]) on how periodic solutions can be obtained by means of these theorems. According to these theory, in the course of derivations, it is realized that once appropriate a priori bounds for the 2π-periodic solutions of the auxiliary equation (1.2) are known for each λ ∈ (0, 1), then standard procedures will allow these theorems to imply existence of periodic solutions to Eq. (1.1). Applying these approaches, Wang [3] established a priori bounds for 2π -periodic solutions of Eq. (1.2) in the case of f (0) = 0 I This work is partially supported by the NNSF of China (No: 10471153). ∗ Corresponding author.
E-mail address: [email protected] (Y. Zhou). c 2007 Published by Elsevier Ltd 0898-1221/$ - see front matter doi:10.1016/j.camwa.2006.12.003
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Y. Zhou, X. Tang / Computers and Mathematics with Applications 53 (2007) 825–830
R 2π and 0 p(t)dt = 0, these bounds implied that Eq. (1.1) had a periodic solution. Lu [4,5] also studied the existence of 2π-periodic solutions of Eq. (1.1) and generalized the results in [3]. In [6], the researchers continued to discuss the above equation and got some new results on the existence of T -periodic solutions under the assumption of f (0) = 0. In this paper, we will also discuss the existence of 2π-periodic solutions to Eq. (1.1) without the restriction of R 2π f (0) = 0 and 0 p(t)dt = 0. Using the theory in [1,2] and some improved prior estimate, we obtain some new results which generalize and improve the known results in [3–6]. For the sake of convenience, we denote by C2π the space of continuous 2π -periodic functions, endowed with the norm kxk0 = maxt∈[0,2π] |x(t)|. 2. Main results Lemma 2.1. Let x(t) be continuous differentiable T -periodic function (T > 0). Then for any t∗ ∈ (−∞, ∞) Z 1 T 0 |x (s)|ds. max |x(t)| ≤ |x(t∗ )| + t∈[t∗ ,t∗ +T ] 2 0
(2.1)
Proof. Choose t ∗ ∈ [t∗ , t∗ + T ] such that |x(t ∗ )| = maxt∈[t∗ ,t∗ +T ] |x(t)|. Then Z t∗ Z t∗ 0 ∗ |x 0 (s)|ds, x (s)ds ≤ |x(t∗ )| + |x(t )| = x(t∗ ) + t∗ t∗ and Z t∗ |x(t )| = |x(t − T )| = x(t∗ ) − ∗ ∗
∗
t −T
Z t∗ x (s)ds ≤ |x(t∗ )| + ∗ 0
|x 0 (s)|ds.
t −T
Combining the above two inequalities, we have Z ∗ Z 1 T 0 1 t |x (s)|ds. |x 0 (s)|ds = |x(t∗ )| + |x(t ∗ )| ≤ |x(t∗ )| + 2 t ∗ −T 2 0 The proof is complete.
Lemma 2.2 (Wirtinger Inequality [10]). Let x(t) be twice continuous differentiable 2π -periodic function. Then Z 2π Z 2π |x 0 (s)|2 ds ≤ |x 00 (s)|2 ds. (2.2) 0
0
Theorem 2.1. Assume that there exist constants r1 , r2 ≥ 0, d > 0, K > 0 and M > 0 such that (H1) | f (x)| ≤ r1 |x| + K , for x ∈ R; (H2) xg(x) > 0 and |g(x)| > L for |x| > d; (H3) g(x) ≥ r2 x − M for x ≤ −d, where L = maxt∈[0,2π] | p(t) − f (0)|. If 2π(r1 + πr2 ) < 1,
(2.3)
then Eq. (1.1) has at least a 2π-periodic solution. Proof. From the results (degree theory) in [1,2] (see also [3]), it is sufficient to show that there are positive constants M0 and M1 , independent of λ, such that if x(t) is a 2π-periodic solution of Eq. (1.2), then kxk0 < M0 and kx 0 k0 < M1 . Now, let x = x(t) be any 2π-periodic solution of Eq. (1.2). Then there exist t1 , t2 ∈ [0, 2π] such that x(t1 ) = min x(t), [0,2π]
x(t2 ) = max x(t). [0,2π]
It follows that x 0 (t1 ) = x 0 (t2 ) = 0,
x 00 (t1 ) ≥ 0,
x 00 (t2 ) ≤ 0.
(2.4)
Y. Zhou, X. Tang / Computers and Mathematics with Applications 53 (2007) 825–830
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By (1.2) and (2.4), we have g(x(t1 − τ (t1 ))) ≤ p(t1 ) − f (0),
and
g(x(t2 − τ (t2 ))) ≥ p(t2 ) − f (0),
which yields g(x(t1 − τ (t1 ))) ≤ L ,
and
g(x(t2 − τ (t2 ))) ≥ −L .
(2.5)
If g(x(t1 − τ (t1 ))) ≥ −L, then |g(x(t1 − τ (t1 )))| ≤ L, it follows from (H2) that |x(t1 − τ (t1 ))| ≤ d.
(2.6)
If g(x(t1 − τ (t1 ))) < −L, then by the second inequality in (2.5) and the continuity of the function g(x(t − τ (t))), there exists t3 ∈ [0, 2π] such that g(x(t3 − τ (t3 ))) = −L, and so |g(x(t3 − τ (t3 )))| = L, it follows from (H2) that |x(t3 − τ (t3 ))| ≤ d.
(2.7)
According to the above discussion, it can be seen that there exists a t ∗ ∈ [0, 2π] such that |x(t ∗ )| ≤ d.
(2.8)
By Lemma 2.1, we have 1 kxk0 ≤ |x(t )| + 2 ∗
2π
Z
|x 0 (s)|ds
(2.9)
0
≤ d + πkx 0 k0 .
(2.10)
Let E 1 = {t : t ∈ [0, 2π ], x(t − τ (t)) > d}, E 2 = {t : t ∈ [0, 2π ], x(t − τ (t)) < −d}, E 3 = {t : t ∈ [0, 2π], |x(t − τ (t))| ≤ d}. Integrating both sides of (1.2) on [0, 2π], we obtain Z Z Z 2π Z |g(x(s − τ (s)))|ds ≤ | f (x 0 (s))|ds + + |g(x(s − τ (s)))|ds + 2π k pk0 . 0
E1
E2
E3
Thus, 1 kx k0 ≤ 2 0
2π
Z
|x 00 (s)|ds
0
1 ≤ 2
"Z
1 ≤ 2 Z ≤
"Z
2π
| f (x (s))|ds + 0
0
|g(x(s − τ (s)))|ds +
2π
| f (x 0 (s))|ds +
| f (x (s))|ds +
0
Z
Z
Z +
Z
+ E2
Z + E2
| p(s)|ds 0
E1 0
# |g(x(s − τ (s)))|ds + 2π k pk0
E3
|g(x(s − τ (s)))|ds + 2π k pk0
E3
≤ 2π(r1 kx 0 k0 + r2 kxk0 ) + 2π(K + M + gd + k pk0 ) ≤ 2π(r1 + πr2 )kx 0 k0 + 2π(K + M + gd + r2 d + k pk0 ), where gd = max|x|≤d |g(x)|. Therefore, kx 0 k0 ≤
2π(K + M + gd + r2 d + k pk0 ) , M1 . 1 − 2π(r1 + r2 π )
It follows from (2.10) that kxk0 ≤ d + M1 π , M0 . This completes the proof.
#
2π
Z
0
0 2π
2π
Z
Similarly, we can get the following Theorem 2.2.
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Y. Zhou, X. Tang / Computers and Mathematics with Applications 53 (2007) 825–830
Theorem 2.2. Assume that there exist constants r1 , r2 ≥ 0, d > 0, K > 0 and M > 0 such that (H1) | f (x)| ≤ r1 |x| + K , for x ∈ R; (H2) xg(x) > 0 and |g(x)| > L for |x| > d; (H4) g(x) ≤ r2 x + M for x ≥ d, where L is defined as in Theorem 2.1. If (2.3) holds, then Eq. (1.1) has at least a 2π -periodic solution. Remark 2.1. Note that the condition 2π(r1 + πr2 ) < 1 is more weaker than 4π(r R 2π 1 + (2π + 1)r2 ) < 1 and 8π(r1 + πr1r2 ) < 1. Therefore, even if under the assumption of f (0) = 0 and 0 p(s)ds = 0, Theorems 2.1 and 2.2 still improve Theorem 1 and 2 in [4], and Theorem 1 and 2 in [5] as well. Theorem 2.3. Assume that there exist constants r ≥ 0, d > 0 and K > 0 such that (H5) |g(x)| > L for |x| > d; (H6) |g(x)| ≤ r |x| + K , for x ∈ R, where L is defined as in Theorem 2.1. If πr < 1, then Eq. (1.1) has at least a 2π -periodic solution. Proof. According to the proof of Theorem 2.1, there exists a t ∗ ∈ [0, 2π] such that |x(t ∗ )| ≤ d, and (2.9) and (2.10) hold. Applying Schwarz inequality and Lemma 2.2, we have " Z # Z 2π Z 2π Z 2π 2π |x 00 (s)|2 ds = λ − f (x 0 (s))x 00 (s)ds − g(x(s − τ (s)))x 00 (s)ds + p(s)x 00 (s)ds 0
0 2π
Z
0
0
|g(x(s − τ (s)))||x 00 (s)|ds + k pk0
≤
2π
Z
0
|x 00 (s)|ds
0
≤ (r kxk0 + k pk0 + K )
2π
Z
|x 00 (s)|ds
0
r 2
≤
2π
Z
|x 0 (s)|ds + r d + k pk0 + K
0
!Z
2π
|x 00 (s)|ds
0
√
s s Z 2π Z 2π 2πr 0 2 |x (s)| ds + r d + k pk0 + K |x 00 (s)|2 ds ≤ 2π 2 0 0 √ s s Z 2π Z 2π √ 2πr 00 2 ≤ 2π |x (s)| ds + r d + k pk0 + K |x 00 (s)|2 ds 2 0 0 s Z 2π Z 2π √ 00 2 ≤ πr |x (s)| ds + 2π (r d + k pk0 + K ) |x 00 (s)|2 ds. √
0
0
And so, we have #2 "√ Z 2π 2π (r d + k pk + K ) 0 |x 00 (s)|2 ds ≤ . 1 − rπ 0 By Lemma 2.1, we get 1 kx k0 ≤ 2 0
2π
Z 0
1√ |x (s)|ds ≤ 2π 2 00
It follows that kxk0 ≤ d + M1 π. This completes the proof.
s
2π
Z 0
|x 00 (s)|2 ds =
π(r d + k pk0 + K ) , M1 . 1 − rπ
Y. Zhou, X. Tang / Computers and Mathematics with Applications 53 (2007) 825–830
829
Remark 2.2. Even if in the case of f (0) = 0, the condition (H6) improves (B2) of Theorem 3 in [5], especially, πr < 1 of Theorem 2.3 is more weaker than 4π 2r3 < 1 of Theorem 3 in [5]. Theorem 2.4. Assume that there exist constants n ≥ 1, σ > 0, r ≥ 0, d > 0 and K > 0 such that (H5) holds and that (H7) x f (x) ≥ σ |x|n+1 , for x ∈ R or x f (x) ≤ −σ |x|n+1 , for x ∈ R; (H8) |g(x)| ≤ r |x|n + K , for x ∈ R. If π n r < σ , then Eq. (1.1) has at least a 2π-periodic solution. Proof. According to the proof of Theorem 2.1, there exists a t ∗ ∈ [0, 2π] such that |x(t ∗ )| ≤ d, and (2.9) and (2.10) hold. By Holder inequality and (H7) and (H8), we have Z Z 2π 2π 0 n+1 0 0 |x (s)| ds ≤ f (x (s))x (s)ds σ 0 0 Z 2π Z 2π ≤ |g(x(s − τ (s)))||x 0 (s)|ds + | p(s)||x 0 (s)|ds 0
0
Z
2π
≤ (r kxkn0 + k pk0 + K ) |x 0 (s)|ds 0 !n #Z " Z 2π 1 2π 0 |x (s)|ds + k pk0 + K |x 0 (s)|ds ≤ r d+ 2 0 0 !1/n !(n+1)/n n Z 2π Z 2π 1 = r d |x 0 (s)|ds + |x 0 (s)|ds 2 0 0 Z 2π + (k pk0 + K ) |x 0 (s)|ds 0 !1/n(n+1) !1/n n Z 2π Z 2π |x 0 (s)|n+1 ds +π |x 0 (s)|n+1 ds ≤ r d(2π )1/(n+1) 0
0
+ (k pk0 + K )(2π )
n/(n+1)
2π
Z
!1/(n+1) |x (s)| 0
n+1
ds
0
This implies that there exists a positive D such that Z 2π |x 0 (s)|n+1 ds ≤ D. 0
And so, we have Z 2π n 1 |x 0 (s)|ds ≤ (2π ) n+1 D n+1 . 0
From (2.10), we get n 1 1 kxk0 ≤ d + (2π ) n+1 D n+1 , M0 . 2 By Lemma 2.1, we obtain Z 1 2π 00 0 kx k0 ≤ |x (s)|ds 2 0 "Z # Z 2π Z 2π 2π 1 0 ≤ | f (x (s))|ds + |g(x(s − τ (s)))|ds + | p(s)|ds 2 0 0 0
.
830
Y. Zhou, X. Tang / Computers and Mathematics with Applications 53 (2007) 825–830 2π
Z ≤
|g(x(s − τ (s)))|ds +
2π
Z
| p(s)|ds 0
0
≤ 2π(r kxkn0 + k pk0 + K ) ≤ 2π(r M0n + k pk0 + K ) , M1 . This completes the proof.
By a similar proof of Theorem 2.4, we can get the following Theorem 2.5. Theorem 2.5. In Theorem 2.4, if (H7) is replaced by (H9) f (x) ≥ σ |x|n , for x ∈ R or f (x) ≤ −σ |x|n , for x ∈ R, then the conclusion still holds. Remark 2.3. If p(t) and τ (t) are T -periodic, then the existence of T -periodic solutions for Eq. (1.1) can be investigated by transformation s = 2π ˆ , p(t) are 2π-periodic functions, T t, which implies τˆ (s) , τ (t) and p(s) and some corresponding results can also be obtained directly applying the above theorems. As an example, we give a corollary related to Theorem 2.4. Corollary 2.1. Assume that p(t) and τ (t) are T -periodic and that there exist constants n ≥ 1, σ > 0, r ≥ 0, d > 0 n and K > 0 such that (H5), (H7) and (H8) hold for L = maxs∈[0,T ] | p(s) − f (0)|. If T2 r < σ , then Eq. (1.1) has at least a T -periodic solution. n Remark 2.4. Note that the condition T2 r < σ is weaker than T n r < σ . Therefore, even if in the case of f (0) = 0, Corollary 2.1 still improves Theorem 3.2 in [6]. References [1] R.E. Gains, J.L. Mawhin, Coincidence Degree and Nonlinear Differential Equations, in: Lecture Notes in Math., vol. 568, Springer-Verlag, 1977. [2] K. Deimling, Nonlinear Functional Analysis, Springer-Verlag, 1985. [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] S.P. Lu, W.G. Ge, Z.X. Zheng, Periodic solutions for a kind of Rayleigh equation with a deviating argument, Acta Math. Sinica 47 (2) (2004) 299–304 (in Chinese). [5] S.P. Lu, W.G. Ge, Z.X. Zheng, Periodic solutions for a kind of Rayleigh equation with a deviating argument, Appl. Math. Lett. 17 (2004) 443–449. [6] S.P. Lu, W.G. 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. [7] F. Liu, On the existence of the periodic solutions of Rayleigh equation, Acta Math. Sinica 37 (5) (1994) 639–644 (in Chinese). [8] G.Q. Wang, J.R. Yan, Existence theorem of periodic positive solutions for the Rayleigh equation of retarded type, Port. Math. 57 (2) (2000) 153–160. [9] G.Q. Wang, J.R. Yan, On existence of periodic solutions of the Rayleigh equation of retarded type, Int. J. Math. Math. Sci. 23 (1) (2000) 65–68. [10] J.C. Kuang, Applied Inequalities, 3rd ed., Shangdong Science and Technology Press, 2004 (in Chinese).
Periodic solutions for a kind of Rayleigh equation with a deviating argumentI Yinggao Zhou ∗ , Xianhua Tang School of Mathematical Science and Computing Technology, Central South University, Changsha, Hunan 410083, PR China Received 20 January 2006; received in revised form 21 October 2006; accepted 12 December 2006
Abstract Existence of periodic solutions for a kind of Rayleigh equation with a deviating argument x 00 (t) + f (x 0 (t)) + g(x(t − τ (t))) = p(t) is studied, and some new results are obtained. Our work generalizes and improves the known results in the literature. c 2007 Published by Elsevier Ltd
Keywords: Rayleigh equations; Periodic solution; A priori bound; Continuation theorem; Deviating argument
1. Introduction Consider the Rayleigh equation with a deviating argument x 00 (t) + f (x 0 (t)) + g(x(t − τ (t))) = p(t),
(1.1)
and its auxiliary equation x 00 (t) + λ f (x 0 (t)) + λg(x(t − τ (t))) = λp(t),
λ ∈ (0, 1)
(1.2)
where f, g, p and τ are real continuous functions defined on R, τ and p are periodic with period 2π . In recent years, the existence of periodic solutions for a kind of Rayleigh equations was studied by some researchers (see [1–9]). In [1,2], continuation theorems are introduced and applied to the existence of solutions of differential equations. In particular, a specific example is given in [1, p. 99] (see also [2, p. 175]) on how periodic solutions can be obtained by means of these theorems. According to these theory, in the course of derivations, it is realized that once appropriate a priori bounds for the 2π-periodic solutions of the auxiliary equation (1.2) are known for each λ ∈ (0, 1), then standard procedures will allow these theorems to imply existence of periodic solutions to Eq. (1.1). Applying these approaches, Wang [3] established a priori bounds for 2π -periodic solutions of Eq. (1.2) in the case of f (0) = 0 I This work is partially supported by the NNSF of China (No: 10471153). ∗ Corresponding author.
E-mail address: [email protected] (Y. Zhou). c 2007 Published by Elsevier Ltd 0898-1221/$ - see front matter doi:10.1016/j.camwa.2006.12.003
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Y. Zhou, X. Tang / Computers and Mathematics with Applications 53 (2007) 825–830
R 2π and 0 p(t)dt = 0, these bounds implied that Eq. (1.1) had a periodic solution. Lu [4,5] also studied the existence of 2π-periodic solutions of Eq. (1.1) and generalized the results in [3]. In [6], the researchers continued to discuss the above equation and got some new results on the existence of T -periodic solutions under the assumption of f (0) = 0. In this paper, we will also discuss the existence of 2π-periodic solutions to Eq. (1.1) without the restriction of R 2π f (0) = 0 and 0 p(t)dt = 0. Using the theory in [1,2] and some improved prior estimate, we obtain some new results which generalize and improve the known results in [3–6]. For the sake of convenience, we denote by C2π the space of continuous 2π -periodic functions, endowed with the norm kxk0 = maxt∈[0,2π] |x(t)|. 2. Main results Lemma 2.1. Let x(t) be continuous differentiable T -periodic function (T > 0). Then for any t∗ ∈ (−∞, ∞) Z 1 T 0 |x (s)|ds. max |x(t)| ≤ |x(t∗ )| + t∈[t∗ ,t∗ +T ] 2 0
(2.1)
Proof. Choose t ∗ ∈ [t∗ , t∗ + T ] such that |x(t ∗ )| = maxt∈[t∗ ,t∗ +T ] |x(t)|. Then Z t∗ Z t∗ 0 ∗ |x 0 (s)|ds, x (s)ds ≤ |x(t∗ )| + |x(t )| = x(t∗ ) + t∗ t∗ and Z t∗ |x(t )| = |x(t − T )| = x(t∗ ) − ∗ ∗
∗
t −T
Z t∗ x (s)ds ≤ |x(t∗ )| + ∗ 0
|x 0 (s)|ds.
t −T
Combining the above two inequalities, we have Z ∗ Z 1 T 0 1 t |x (s)|ds. |x 0 (s)|ds = |x(t∗ )| + |x(t ∗ )| ≤ |x(t∗ )| + 2 t ∗ −T 2 0 The proof is complete.
Lemma 2.2 (Wirtinger Inequality [10]). Let x(t) be twice continuous differentiable 2π -periodic function. Then Z 2π Z 2π |x 0 (s)|2 ds ≤ |x 00 (s)|2 ds. (2.2) 0
0
Theorem 2.1. Assume that there exist constants r1 , r2 ≥ 0, d > 0, K > 0 and M > 0 such that (H1) | f (x)| ≤ r1 |x| + K , for x ∈ R; (H2) xg(x) > 0 and |g(x)| > L for |x| > d; (H3) g(x) ≥ r2 x − M for x ≤ −d, where L = maxt∈[0,2π] | p(t) − f (0)|. If 2π(r1 + πr2 ) < 1,
(2.3)
then Eq. (1.1) has at least a 2π-periodic solution. Proof. From the results (degree theory) in [1,2] (see also [3]), it is sufficient to show that there are positive constants M0 and M1 , independent of λ, such that if x(t) is a 2π-periodic solution of Eq. (1.2), then kxk0 < M0 and kx 0 k0 < M1 . Now, let x = x(t) be any 2π-periodic solution of Eq. (1.2). Then there exist t1 , t2 ∈ [0, 2π] such that x(t1 ) = min x(t), [0,2π]
x(t2 ) = max x(t). [0,2π]
It follows that x 0 (t1 ) = x 0 (t2 ) = 0,
x 00 (t1 ) ≥ 0,
x 00 (t2 ) ≤ 0.
(2.4)
Y. Zhou, X. Tang / Computers and Mathematics with Applications 53 (2007) 825–830
827
By (1.2) and (2.4), we have g(x(t1 − τ (t1 ))) ≤ p(t1 ) − f (0),
and
g(x(t2 − τ (t2 ))) ≥ p(t2 ) − f (0),
which yields g(x(t1 − τ (t1 ))) ≤ L ,
and
g(x(t2 − τ (t2 ))) ≥ −L .
(2.5)
If g(x(t1 − τ (t1 ))) ≥ −L, then |g(x(t1 − τ (t1 )))| ≤ L, it follows from (H2) that |x(t1 − τ (t1 ))| ≤ d.
(2.6)
If g(x(t1 − τ (t1 ))) < −L, then by the second inequality in (2.5) and the continuity of the function g(x(t − τ (t))), there exists t3 ∈ [0, 2π] such that g(x(t3 − τ (t3 ))) = −L, and so |g(x(t3 − τ (t3 )))| = L, it follows from (H2) that |x(t3 − τ (t3 ))| ≤ d.
(2.7)
According to the above discussion, it can be seen that there exists a t ∗ ∈ [0, 2π] such that |x(t ∗ )| ≤ d.
(2.8)
By Lemma 2.1, we have 1 kxk0 ≤ |x(t )| + 2 ∗
2π
Z
|x 0 (s)|ds
(2.9)
0
≤ d + πkx 0 k0 .
(2.10)
Let E 1 = {t : t ∈ [0, 2π ], x(t − τ (t)) > d}, E 2 = {t : t ∈ [0, 2π ], x(t − τ (t)) < −d}, E 3 = {t : t ∈ [0, 2π], |x(t − τ (t))| ≤ d}. Integrating both sides of (1.2) on [0, 2π], we obtain Z Z Z 2π Z |g(x(s − τ (s)))|ds ≤ | f (x 0 (s))|ds + + |g(x(s − τ (s)))|ds + 2π k pk0 . 0
E1
E2
E3
Thus, 1 kx k0 ≤ 2 0
2π
Z
|x 00 (s)|ds
0
1 ≤ 2
"Z
1 ≤ 2 Z ≤
"Z
2π
| f (x (s))|ds + 0
0
|g(x(s − τ (s)))|ds +
2π
| f (x 0 (s))|ds +
| f (x (s))|ds +
0
Z
Z
Z +
Z
+ E2
Z + E2
| p(s)|ds 0
E1 0
# |g(x(s − τ (s)))|ds + 2π k pk0
E3
|g(x(s − τ (s)))|ds + 2π k pk0
E3
≤ 2π(r1 kx 0 k0 + r2 kxk0 ) + 2π(K + M + gd + k pk0 ) ≤ 2π(r1 + πr2 )kx 0 k0 + 2π(K + M + gd + r2 d + k pk0 ), where gd = max|x|≤d |g(x)|. Therefore, kx 0 k0 ≤
2π(K + M + gd + r2 d + k pk0 ) , M1 . 1 − 2π(r1 + r2 π )
It follows from (2.10) that kxk0 ≤ d + M1 π , M0 . This completes the proof.
#
2π
Z
0
0 2π
2π
Z
Similarly, we can get the following Theorem 2.2.
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Y. Zhou, X. Tang / Computers and Mathematics with Applications 53 (2007) 825–830
Theorem 2.2. Assume that there exist constants r1 , r2 ≥ 0, d > 0, K > 0 and M > 0 such that (H1) | f (x)| ≤ r1 |x| + K , for x ∈ R; (H2) xg(x) > 0 and |g(x)| > L for |x| > d; (H4) g(x) ≤ r2 x + M for x ≥ d, where L is defined as in Theorem 2.1. If (2.3) holds, then Eq. (1.1) has at least a 2π -periodic solution. Remark 2.1. Note that the condition 2π(r1 + πr2 ) < 1 is more weaker than 4π(r R 2π 1 + (2π + 1)r2 ) < 1 and 8π(r1 + πr1r2 ) < 1. Therefore, even if under the assumption of f (0) = 0 and 0 p(s)ds = 0, Theorems 2.1 and 2.2 still improve Theorem 1 and 2 in [4], and Theorem 1 and 2 in [5] as well. Theorem 2.3. Assume that there exist constants r ≥ 0, d > 0 and K > 0 such that (H5) |g(x)| > L for |x| > d; (H6) |g(x)| ≤ r |x| + K , for x ∈ R, where L is defined as in Theorem 2.1. If πr < 1, then Eq. (1.1) has at least a 2π -periodic solution. Proof. According to the proof of Theorem 2.1, there exists a t ∗ ∈ [0, 2π] such that |x(t ∗ )| ≤ d, and (2.9) and (2.10) hold. Applying Schwarz inequality and Lemma 2.2, we have " Z # Z 2π Z 2π Z 2π 2π |x 00 (s)|2 ds = λ − f (x 0 (s))x 00 (s)ds − g(x(s − τ (s)))x 00 (s)ds + p(s)x 00 (s)ds 0
0 2π
Z
0
0
|g(x(s − τ (s)))||x 00 (s)|ds + k pk0
≤
2π
Z
0
|x 00 (s)|ds
0
≤ (r kxk0 + k pk0 + K )
2π
Z
|x 00 (s)|ds
0
r 2
≤
2π
Z
|x 0 (s)|ds + r d + k pk0 + K
0
!Z
2π
|x 00 (s)|ds
0
√
s s Z 2π Z 2π 2πr 0 2 |x (s)| ds + r d + k pk0 + K |x 00 (s)|2 ds ≤ 2π 2 0 0 √ s s Z 2π Z 2π √ 2πr 00 2 ≤ 2π |x (s)| ds + r d + k pk0 + K |x 00 (s)|2 ds 2 0 0 s Z 2π Z 2π √ 00 2 ≤ πr |x (s)| ds + 2π (r d + k pk0 + K ) |x 00 (s)|2 ds. √
0
0
And so, we have #2 "√ Z 2π 2π (r d + k pk + K ) 0 |x 00 (s)|2 ds ≤ . 1 − rπ 0 By Lemma 2.1, we get 1 kx k0 ≤ 2 0
2π
Z 0
1√ |x (s)|ds ≤ 2π 2 00
It follows that kxk0 ≤ d + M1 π. This completes the proof.
s
2π
Z 0
|x 00 (s)|2 ds =
π(r d + k pk0 + K ) , M1 . 1 − rπ
Y. Zhou, X. Tang / Computers and Mathematics with Applications 53 (2007) 825–830
829
Remark 2.2. Even if in the case of f (0) = 0, the condition (H6) improves (B2) of Theorem 3 in [5], especially, πr < 1 of Theorem 2.3 is more weaker than 4π 2r3 < 1 of Theorem 3 in [5]. Theorem 2.4. Assume that there exist constants n ≥ 1, σ > 0, r ≥ 0, d > 0 and K > 0 such that (H5) holds and that (H7) x f (x) ≥ σ |x|n+1 , for x ∈ R or x f (x) ≤ −σ |x|n+1 , for x ∈ R; (H8) |g(x)| ≤ r |x|n + K , for x ∈ R. If π n r < σ , then Eq. (1.1) has at least a 2π-periodic solution. Proof. According to the proof of Theorem 2.1, there exists a t ∗ ∈ [0, 2π] such that |x(t ∗ )| ≤ d, and (2.9) and (2.10) hold. By Holder inequality and (H7) and (H8), we have Z Z 2π 2π 0 n+1 0 0 |x (s)| ds ≤ f (x (s))x (s)ds σ 0 0 Z 2π Z 2π ≤ |g(x(s − τ (s)))||x 0 (s)|ds + | p(s)||x 0 (s)|ds 0
0
Z
2π
≤ (r kxkn0 + k pk0 + K ) |x 0 (s)|ds 0 !n #Z " Z 2π 1 2π 0 |x (s)|ds + k pk0 + K |x 0 (s)|ds ≤ r d+ 2 0 0 !1/n !(n+1)/n n Z 2π Z 2π 1 = r d |x 0 (s)|ds + |x 0 (s)|ds 2 0 0 Z 2π + (k pk0 + K ) |x 0 (s)|ds 0 !1/n(n+1) !1/n n Z 2π Z 2π |x 0 (s)|n+1 ds +π |x 0 (s)|n+1 ds ≤ r d(2π )1/(n+1) 0
0
+ (k pk0 + K )(2π )
n/(n+1)
2π
Z
!1/(n+1) |x (s)| 0
n+1
ds
0
This implies that there exists a positive D such that Z 2π |x 0 (s)|n+1 ds ≤ D. 0
And so, we have Z 2π n 1 |x 0 (s)|ds ≤ (2π ) n+1 D n+1 . 0
From (2.10), we get n 1 1 kxk0 ≤ d + (2π ) n+1 D n+1 , M0 . 2 By Lemma 2.1, we obtain Z 1 2π 00 0 kx k0 ≤ |x (s)|ds 2 0 "Z # Z 2π Z 2π 2π 1 0 ≤ | f (x (s))|ds + |g(x(s − τ (s)))|ds + | p(s)|ds 2 0 0 0
.
830
Y. Zhou, X. Tang / Computers and Mathematics with Applications 53 (2007) 825–830 2π
Z ≤
|g(x(s − τ (s)))|ds +
2π
Z
| p(s)|ds 0
0
≤ 2π(r kxkn0 + k pk0 + K ) ≤ 2π(r M0n + k pk0 + K ) , M1 . This completes the proof.
By a similar proof of Theorem 2.4, we can get the following Theorem 2.5. Theorem 2.5. In Theorem 2.4, if (H7) is replaced by (H9) f (x) ≥ σ |x|n , for x ∈ R or f (x) ≤ −σ |x|n , for x ∈ R, then the conclusion still holds. Remark 2.3. If p(t) and τ (t) are T -periodic, then the existence of T -periodic solutions for Eq. (1.1) can be investigated by transformation s = 2π ˆ , p(t) are 2π-periodic functions, T t, which implies τˆ (s) , τ (t) and p(s) and some corresponding results can also be obtained directly applying the above theorems. As an example, we give a corollary related to Theorem 2.4. Corollary 2.1. Assume that p(t) and τ (t) are T -periodic and that there exist constants n ≥ 1, σ > 0, r ≥ 0, d > 0 n and K > 0 such that (H5), (H7) and (H8) hold for L = maxs∈[0,T ] | p(s) − f (0)|. If T2 r < σ , then Eq. (1.1) has at least a T -periodic solution. n Remark 2.4. Note that the condition T2 r < σ is weaker than T n r < σ . Therefore, even if in the case of f (0) = 0, Corollary 2.1 still improves Theorem 3.2 in [6]. References [1] R.E. Gains, J.L. Mawhin, Coincidence Degree and Nonlinear Differential Equations, in: Lecture Notes in Math., vol. 568, Springer-Verlag, 1977. [2] K. Deimling, Nonlinear Functional Analysis, Springer-Verlag, 1985. [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] S.P. Lu, W.G. Ge, Z.X. Zheng, Periodic solutions for a kind of Rayleigh equation with a deviating argument, Acta Math. Sinica 47 (2) (2004) 299–304 (in Chinese). [5] S.P. Lu, W.G. Ge, Z.X. Zheng, Periodic solutions for a kind of Rayleigh equation with a deviating argument, Appl. Math. Lett. 17 (2004) 443–449. [6] S.P. Lu, W.G. 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. [7] F. Liu, On the existence of the periodic solutions of Rayleigh equation, Acta Math. Sinica 37 (5) (1994) 639–644 (in Chinese). [8] G.Q. Wang, J.R. Yan, Existence theorem of periodic positive solutions for the Rayleigh equation of retarded type, Port. Math. 57 (2) (2000) 153–160. [9] G.Q. Wang, J.R. Yan, On existence of periodic solutions of the Rayleigh equation of retarded type, Int. J. Math. Math. Sci. 23 (1) (2000) 65–68. [10] J.C. Kuang, Applied Inequalities, 3rd ed., Shangdong Science and Technology Press, 2004 (in Chinese).