Existence and uniqueness of anti-periodic solutions for prescribed mean curvature Rayleigh equations

Boundary Value Problems20122012:109

DOI: 10.1186/1687-2770-2012-109

Received: 12 May 2012

Accepted: 22 September 2012

Published: 9 October 2012

Abstract

By means of the Leray-Schauder degree theory, we establish some sufficient conditions for the existence and uniqueness of anti-periodic solutions for prescribed mean curvature Rayleigh equations.

MSC: 34C25, 34D40.

Keywords

prescribed mean curvature Rayleigh equation anti-periodic solutions Leray-Schauder degree

1 Introduction

We are concerned with the existence and uniqueness of anti-periodic solutions of the following prescribed mean curvature Rayleigh equation:
( x 1 + x 2 ) + f ( t , x ( t ) ) + g ( t , x ( t ) ) = e ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ1_HTML.gif
(1.1)

where e C ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq1_HTML.gif is T-periodic, and f , g C ( R × R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq2_HTML.gif are T-periodic in the first argument, T > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq3_HTML.gif is a constant.

In recent years, the existence of periodic solutions and anti-periodic solutions for some types of second-order differential equations, especially for the Rayleigh ones, were widely studied (see [17]) and the references cited therein). For example, Liu [7] discussed the Rayleigh equation
x + f ( t , x ( t ) ) + g ( t , x ( t ) ) = e ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equa_HTML.gif
and established the existence and uniqueness of anti-periodic solutions. At the same time, a kind of prescribed mean curvature equations attracted many people’s attention (see [811] and the references cited therein). Feng [8] investigated the prescribed mean curvature Liénard equation
( x 1 + x 2 ) + f ( x ( t ) ) x ( t ) + g ( t , x ( t τ ( t ) ) ) = e ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equb_HTML.gif

and obtained some existence results on periodic solutions. However, to the best of our knowledge, the existence and uniqueness of anti-periodic solution for Eq. (1.1) have not been investigated till now. Motivated by [7, 8], we establish some sufficient conditions for the existence and uniqueness of anti-periodic solutions via the Leray-Schauder degree theory.

The rest of the paper is organized as follows. In Section 2, we shall state and prove some basic lemmas. In Section 3, we shall prove the main result. An example will be given to show the applications of our main result in the final section.

2 Preliminaries

We first give the definition of an anti-periodic function. Assume that N is a positive integer. Let u : R R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq4_HTML.gif be a continuous function. We call u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq5_HTML.gif an anti-periodic function on ℝ if u satisfies the following condition:
u ( t + T 2 ) = u ( t ) , for all  t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equc_HTML.gif

Obviously, a T 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq6_HTML.gif-anti-periodic function u is a T-periodic function.

Throughout this paper, we will adopt the following notations:
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equd_HTML.gif
which is a linear normal space endowed with the norm http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq7_HTML.gif defined by
x = max { | x | , | x | , , | x ( k ) | } , for all  x C T k , 1 2 ( R , R N ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Eque_HTML.gif

The following lemmas will be useful to prove our main results.

Lemma 2.1 [12]

If x C T 1 ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq8_HTML.gif and 0 T x ( t ) d t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq9_HTML.gif, then
0 T | x ( t ) | 2 d t ( T 2 / 4 π 2 ) 0 T | x ( t ) | 2 d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equf_HTML.gif
(Wirtinger inequality) and
| x | 2 ( T / 12 ) 0 T | x ( t ) | 2 d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equg_HTML.gif

(Sobolev inequality).

Lemma 2.2 Suppose that the following condition holds:

( H 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq10_HTML.gif ( g ( t , x 1 ) g ( t , x 2 ) ) ( x 1 x 2 ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq11_HTML.gif, for all t , x 1 , x 2 R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq12_HTML.gif and x 1 x 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq13_HTML.gif.

Then Eq. (1.1) has at most one T-periodic solution.

Proof Assume that x 1 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq14_HTML.gif and x 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq15_HTML.gif are two T-periodic solutions of Eq. (1.1). Then we obtain
( x i ( t ) 1 + x i 2 ( t ) ) + f ( t , x i ( t ) ) + g ( t , x i ( t ) ) = e ( t ) , i = 1 , 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ2_HTML.gif
(2.1)
It is easy to see that x i ( t ) C 2 [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq16_HTML.gif ( i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq17_HTML.gif). From (2.1), we know
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ3_HTML.gif
(2.2)
Set z ( t ) = x 1 ( t ) x 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq18_HTML.gif. Now, we prove
z ( t ) 0 , for all  t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equh_HTML.gif
Otherwise, we have
max t R z ( t ) = max t [ 0 , T ] z ( t ) > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equi_HTML.gif
Then there exists a t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq19_HTML.gif such that
x 1 ( t ) x 2 ( t ) = z ( t ) = max t R z ( t ) = max t [ 0 , T ] z ( t ) > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ4_HTML.gif
(2.3)
which implies that
z ( t ) = x 1 ( t ) x 2 ( t ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ5_HTML.gif
(2.4)
and
z ( t ) = x 1 ( t ) x 2 ( t ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ6_HTML.gif
(2.5)
It follows from (2.2), (2.4) and (2.5) that
g ( t , x 1 ( t ) ) g ( t , x 2 ( t ) ) = ( x 1 ( t ) ( 1 + x 1 2 ( t ) ) 3 x 2 ( t ) ( 1 + x 2 2 ( t ) ) 3 ) ( f ( t , x 1 ( t ) ) f ( t , x 2 ( t ) ) ) = 1 ( 1 + x 1 2 ( t ) ) 3 ( x 1 ( t ) x 2 ( t ) ) = 1 ( 1 + x 1 2 ( t ) ) 3 z ( t ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equj_HTML.gif
From ( H 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq10_HTML.gif, we get
x 1 ( t ) x 2 ( t ) 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equk_HTML.gif
which contradicts (2.3). Thus,
z ( t ) 0 , for all  t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equl_HTML.gif
By using a similar argument, we can also show
z ( t ) 0 , for all  t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equm_HTML.gif
Hence,
x 1 ( t ) x 2 ( t ) = z ( t ) = 0 , for all  t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equn_HTML.gif

Therefore, Eq. (1.1) has at most one T-periodic solution. The proof is completed. □

To prove the main result of this paper, we shall use a continuation theorem [13, 14] as follows.

Lemma 2.3 Let Ω be open bounded in a linear normal space X. Suppose that f ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq20_HTML.gif is a complete continuous field on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq21_HTML.gif. Moreover, assume that the Leray-Schauder degree
deg { f ˜ , Ω , p } 0 , for p X f ˜ ( Ω ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equo_HTML.gif

Then the equation f ˜ ( x ) = p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq22_HTML.gif has at least one solution in Ω.

3 Main result

In this section, we present and prove our main result concerning the existence and uniqueness of anti-periodic solutions of Eq. (1.1).

Theorem 3.1 Let ( H 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq10_HTML.gif hold. Moreover, assume that the following conditions hold:

( H 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq23_HTML.gif there exists l > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq24_HTML.gif such that
| g ( t , x 1 ) g ( t , x 2 ) | l | x 1 x 2 | , for all t , x 1 , x 2 R ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equp_HTML.gif
( H 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq25_HTML.gif there exists β , γ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq26_HTML.gif such that
γ lim inf | x | f ( t , x ) x lim sup | x | f ( t , x ) x β , uniformly in t R ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equq_HTML.gif
( H 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq27_HTML.gif for all t , x R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq28_HTML.gif,
f ( t + T 2 , x ) = f ( t , x ) , g ( t + T 2 , x ) = g ( t , x ) , e ( t + T 2 ) = e ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equr_HTML.gif

Then Eq. (1.1) has a unique anti-periodic solution for l T 2 π < γ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq29_HTML.gif.

Proof Rewrite Eq. (1.1) in the equivalent form:
{ x 1 ( t ) = ψ ( x 2 ( t ) ) = x 2 ( t ) 1 x 2 2 ( t ) , x 2 ( t ) = f ( t , ψ ( x 2 ( t ) ) ) g ( t , x 1 ( t ) ) + e ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ7_HTML.gif
(3.1)
where ψ ( x ) = x 1 x 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq30_HTML.gif. Now, we consider the auxiliary equation of (3.1),
{ x 1 ( t ) = λ x 2 ( t ) 1 x 2 2 ( t ) = λ ψ ( x 2 ( t ) ) , x 2 ( t ) = λ f ( t , ψ ( x 2 ( t ) ) ) λ g ( t , x 1 ( t ) ) + λ e ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ8_HTML.gif
(3.2)
where λ ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq31_HTML.gif is a parameter. Set
x ( t ) = ( x 1 ( t ) x 2 ( t ) ) , Q 1 ( t , x 1 ( t ) , x 2 ( t ) ) = ( ψ ( x 2 ( t ) ) f ( t , ψ ( x 2 ( t ) ) ) g ( t , x 1 ( t ) ) + e ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equs_HTML.gif
Then Eq. (3.2) can be reduced to the equation as follows:
x ( t ) = λ Q 1 ( t , x 1 ( t ) , x 2 ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equt_HTML.gif

By Lemma 2.2 and condition ( H 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq10_HTML.gif, it is easy to see that Eq. (1.1) has at most one anti-periodic solution. Thus, to prove Theorem 3.1, it suffices to show that Eq. (1.1) has at least one anti-periodic solution. To do this, we shall apply Lemma 2.3. Firstly, we will prove that the set of all possible anti-periodic solutions of Eq. (3.2) is bounded.

Let x ( t ) = ( x 1 ( t ) , x 2 ( t ) ) T C T 1 , 1 2 ( R , R 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq32_HTML.gif be an arbitrary possible anti-periodic solution of Eq. (3.2). Then x 1 ( t ) C T 1 , 1 2 ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq33_HTML.gif. Thus, we have
0 T x 1 ( t ) d t = 0 T 2 x 1 ( t ) d t + T 2 T x 1 ( t ) d t = 0 T 2 x 1 ( t ) d t + 0 T 2 x 1 ( t + T 2 ) d t = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equu_HTML.gif
It follows from Lemma 2.1 that
| x 1 | T 12 | x 1 | 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equv_HTML.gif
Obviously, Eq. (3.2) is equivalent to the following equation:
( 1 λ x 1 ( t ) 1 + 1 λ 2 x 1 2 ( t ) ) + λ f ( t , 1 λ x 1 ( t ) ) + λ g ( t , x 1 ( t ) ) = λ e ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ9_HTML.gif
(3.3)
Multiplying (3.3) by x 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq34_HTML.gif and integrating from 0 to T, we have
λ 0 T f ( t , 1 λ x 1 ( t ) ) x 1 ( t ) d t + λ 0 T g ( t , x 1 ( t ) ) x 1 ( t ) d t = λ 0 T e ( t ) x 1 ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ10_HTML.gif
(3.4)
Since l T 2 π < γ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq35_HTML.gif, there exists a constant ε > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq36_HTML.gif such that
l T 2 π < γ ε . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ11_HTML.gif
(3.5)
For such a ε > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq36_HTML.gif, in view of ( H 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq25_HTML.gif, there exists M 1 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq37_HTML.gif such that for all t , x R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq28_HTML.gif, x f ( t , x ) ( γ ε ) x 2 M 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq38_HTML.gif. Hence,
| λ 0 T f ( t , 1 λ x 1 ( t ) ) x 1 ( t ) d t | λ 2 0 T f ( t , 1 λ x 1 ( t ) ) x 1 ( t ) λ d t ( γ ε ) 0 T | x 1 ( t ) | 2 d t λ 2 M 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ12_HTML.gif
(3.6)
It follows from (3.4) and (3.6) that
( γ ε ) 0 T | x 1 ( t ) | 2 d t | λ 0 T f ( t , 1 λ x 1 ( t ) ) x 1 ( t ) d t | + λ 2 M 1 | 0 T g ( t , x 1 ( t ) ) x 1 ( t ) d t | + | 0 T e ( t ) x 1 ( t ) d t | + M 1 0 T | g ( t , x 1 ( t ) ) g ( t , 0 ) | | x 1 ( t ) | d t + 0 T ( | g ( t , 0 ) | + | e ( t ) | ) | x 1 ( t ) | d t + M 1 l 0 T | x 1 ( t ) | | x 1 ( t ) | d t + max t [ 0 , T ] { | g ( t , 0 ) | + | e ( t ) | } 0 T | x 1 ( t ) | d t + M 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equw_HTML.gif
For u , v C ( [ a , b ] , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq39_HTML.gif, we have the Schwarz inequality
a b | u ( x ) | | v ( x ) | d x ( a b | u ( x ) | 2 d x ) 1 2 ( a b | v ( x ) | 2 d x ) 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equx_HTML.gif
Hence,
( γ ε ) 0 T | x 1 ( t ) | 2 d t l ( 0 T | x 1 ( t ) | 2 d t ) 1 2 ( 0 T | x 1 ( t ) | 2 d t ) 1 2 + max t [ 0 , T ] { | g ( t , 0 ) | + | e ( t ) | } T ( 0 T | x 1 ( t ) | 2 d t ) 1 2 + M 1 = l | x 1 | 2 | x 1 | 2 + T | x 1 | 2 max t [ 0 , T ] { | g ( t , 0 ) | + | e ( t ) | } + M 1 l T 2 π | x 1 | 2 2 + T | x 1 | 2 max t [ 0 , T ] { | g ( t , 0 ) | + | e ( t ) | } + M 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ13_HTML.gif
(3.7)
From (3.5) and (3.7), we know that there exists a constant D 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq40_HTML.gif such that
| x 1 | 2 D 1 , and | x 1 | D 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ14_HTML.gif
(3.8)
By the first equation of (3.2), we have
0 T x 2 ( t ) 1 x 2 2 ( t ) d t = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equy_HTML.gif
Then there exists η [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq41_HTML.gif such that x 2 ( η ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq42_HTML.gif. It follows that x 2 ( t ) = x 2 ( η ) + η t x 2 ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq43_HTML.gif, and so
| x 2 | 0 T | x 2 ( t ) | d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equz_HTML.gif
According to ( H 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq25_HTML.gif, we know there exists M 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq44_HTML.gif such that for all t , x R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq28_HTML.gif,
| f ( t , x ) | ( β + 1 ) | x | + M 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equaa_HTML.gif
From the second equation of (3.2), we get
0 T | x 2 ( t ) | d t 0 T λ | f ( t , 1 λ x 1 ( t ) ) | d t + 0 T | g ( t , x 1 ( t ) ) | d t + 0 T | e ( t ) | d t ( β + 1 ) 0 T | x 1 ( t ) | d t + λ M 2 T + 0 T | g ( t , x 1 ( t ) ) | d t + 0 T | e ( t ) | d t ( β + 1 ) T | x 1 | 2 + 0 T | g ( t , x 1 ( t ) ) | d t + 0 T | e ( t ) | d t + M 2 T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equab_HTML.gif
From (3.8), we know that there exists a constant k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq45_HTML.gif such that
| g ( t , x 1 ( t ) | k , t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equac_HTML.gif
Thus,
0 T | x 2 ( t ) | d t ( β + 1 ) T D 1 + k T + T max t [ 0 , T ] | e ( t ) | + M 2 T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equad_HTML.gif
which implies that there exists a constant D 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq46_HTML.gif such that
| x 2 | D 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equae_HTML.gif
Let
M = max { D 1 , D 2 } + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ15_HTML.gif
(3.9)
Set
Ω = { x C T 0 , 1 2 ( R , R 2 ) = X : x < M } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equaf_HTML.gif

Then Eq. (3.2) has no anti-periodic solution on Ω for λ ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq31_HTML.gif.

Next, we consider the Fourier series expansions of two functions x j ( t ) C T k , 1 2 ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq47_HTML.gif ( j = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq48_HTML.gif). We have
x j ( t ) = i = 0 [ a 2 i + 1 j cos 2 π ( 2 i + 1 ) t T + b 2 i + 1 j sin 2 π ( 2 i + 1 ) t T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equag_HTML.gif
Define an operator L 1 : C T k , 1 2 ( R , R ) C T k + 1 , 1 2 ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq49_HTML.gif by setting
( L 1 x j ) ( t ) = 0 t x j ( s ) d s T 2 π i = 0 b 2 i + 1 j 2 i + 1 = T 2 π i = 0 [ a 2 i + 1 j 2 i + 1 sin 2 π ( 2 i + 1 ) t T b 2 i + 1 j 2 i + 1 cos 2 π ( 2 i + 1 ) t T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equah_HTML.gif
Then
d d t ( L 1 x j ) ( t ) = x j ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equai_HTML.gif
and
| ( L 1 x j ) ( t ) | 0 T | x j ( s ) | d s + T 2 π i = 0 | b 2 i + 1 j | 2 i + 1 T | x j | + T 2 π ( i = 0 ( b 2 i + 1 j ) 2 ) 1 2 ( i = 0 1 ( 2 i + 1 ) 2 ) 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equaj_HTML.gif
Since
( i = 0 1 ( 2 i + 1 ) 2 ) 1 2 = π 2 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equak_HTML.gif
and
0 T | x j ( s ) | 2 d s = T 2 i = 0 [ ( a 2 i + 1 j ) 2 + ( b 2 i + 1 j ) 2 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equal_HTML.gif
we obtain
| ( L 1 x j ) ( t ) | T | x j | + T 4 2 ( i = 0 [ ( a 2 i + 1 j ) 2 + ( b 2 i + 1 j ) 2 ] ) 1 2 T | x j | + T 4 2 ( 2 T 0 T | x j ( s ) | 2 d s ) 1 2 ( T + T 4 ) | x j | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equam_HTML.gif
Define L : C T k , 1 2 ( R , R 2 ) C T k + 1 , 1 2 ( R , R 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq50_HTML.gif by setting
( L x ) ( t ) = L ( x 1 ( t ) x 2 ( t ) ) = ( ( L 1 x 1 ) ( t ) ( L 1 x 2 ) ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equan_HTML.gif

Then | L x | ( T + T 4 ) | x | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq51_HTML.gif, and thus L is continuous.

For any x ( t ) C T 0 , 1 2 ( R , R 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq52_HTML.gif, we know from ( H 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq27_HTML.gif that
Q 1 ( t + T 2 , x 1 ( t + T 2 ) , x 2 ( t + T 2 ) ) = Q 1 ( t , x 1 ( t ) , x 2 ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equao_HTML.gif
Therefore, Q 1 ( t , x 1 ( t ) , x 2 ( t ) ) C T 0 , 1 2 ( R , R 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq53_HTML.gif. Define an operator F μ : Ω ¯ C T 1 , 1 2 ( R , R 2 ) X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq54_HTML.gif by setting
F μ ( x ) = μ L ( Q 1 ( t , x 1 ( t ) , x 2 ( t ) ) ) , μ [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equap_HTML.gif

It is easy to see that F μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq55_HTML.gif is a compact homotopy, and the fixed point of F 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq56_HTML.gif on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq21_HTML.gif is the anti-periodic of Eq. (3.1).

Define a homotopic field as follows:
H μ ( x ) : Ω ¯ × [ 0 , 1 ] C T 0 , 1 2 ( R , R 2 ) , H μ ( x ) = x F μ ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equaq_HTML.gif
From (3.9), we have
H μ ( Ω ) 0 , μ [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equar_HTML.gif
Using the homotopy invariance property of degree, we obtain
deg { x F 1 ( x ) , Ω , 0 } = deg { x , Ω , 0 } 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equas_HTML.gif

Till now, we have proved that Ω satisfies all the requirements in Lemma 2.3. Consequently, x F 1 ( x ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq57_HTML.gif has at least one solution in Ω, i.e., F 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq56_HTML.gif has a fixed point x ( t ) = ( x 1 ( t ) , x 2 ( t ) ) T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq58_HTML.gif on  Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq21_HTML.gif. Therefore, Eq. (1.1) has at least one anti-periodic solution x 1 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq14_HTML.gif. This completes the proof. □

4 An example

In this section, we shall construct an example to show the applications of Theorem 3.1.

Example 4.1 Let f ( t , x ) = ( 1 + 1 2 sin 2 t ) x 3 1 + x 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq59_HTML.gif, g ( t , x ) = ( 1 + sin 4 t ) x 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq60_HTML.gif. Then the prescribed mean curvature Rayleigh equation
( x 1 + x 2 ) + f ( t , x ( t ) ) + g ( t , x ( t ) ) = cos t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_Equ16_HTML.gif
(4.1)

has a unique anti-periodic solution with period 2π.

Proof Let T = 2 π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq61_HTML.gif. From the definitions of f ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq62_HTML.gif and g ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq63_HTML.gif, we can easily check that conditions ( H 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq10_HTML.gif and ( H 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq27_HTML.gif hold. Moreover, it is easy to see that ( H 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq23_HTML.gif holds for l = 2 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq64_HTML.gif and ( H 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq25_HTML.gif holds for γ = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq65_HTML.gif, β = 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq66_HTML.gif. Since l T 2 π < γ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-109/MediaObjects/13661_2012_Article_215_IEq67_HTML.gif, we know from Theorem 3.1 that Eq. (4.1) has a unique anti-periodic solution with period 2π. □

Declarations

Acknowledgements

The authors would like to express their thanks to the Editor of the journal and the anonymous referees for their carefully reading of the first draft of the manuscript and making many helpful comments and suggestions which improved the presentation of the paper. Research supported by National Science foundation of China, No. 10771145 and Beijing Natural Science Foundation (Existence and multiplicity of periodic solutions in nonlinear oscillations), No. 1112006.

Authors’ Affiliations

(1)
School of Mathematical Sciences, Capital Normal University
(2)
College of Science, Jiujiang University

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© Li and Wang; licensee Springer 2012

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