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

The Erratum to this article has been published in Boundary Value Problems 2014 2014:204

## 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.

## 1 Introduction

We are concerned with the existence and uniqueness of anti-periodic solutions of the following prescribed mean curvature Rayleigh equation:

$( x ′ 1 + x ′ 2 ) ′ +f ( t , x ′ ( t ) ) +g ( t , x ( t ) ) =e(t),$
(1.1)

where $e∈C(R,R)$ is T-periodic, and $f,g∈C(R×R,R)$ are T-periodic in the first argument, $T>0$ 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 ) and the references cited therein). For example, Liu  discussed the Rayleigh equation

$x ″ +f ( t , x ′ ( t ) ) +g ( t , x ( t ) ) =e(t),$

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  and the references cited therein). Feng  investigated the prescribed mean curvature Liénard equation

$( x ′ 1 + x ′ 2 ) ′ +f ( x ( t ) ) x ′ (t)+g ( t , x ( t − τ ( t ) ) ) =e(t)$

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 $u:R→ R N$ be a continuous function. We call $u(t)$ an anti-periodic function on if u satisfies the following condition:

Obviously, a $T 2$-anti-periodic function u is a T-periodic function.

Throughout this paper, we will adopt the following notations: which is a linear normal space endowed with the norm $∥⋅∥$ defined by

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

Lemma 2.1 

If $x∈ C T 1 (R,R)$ and $∫ 0 T x(t)dt=0$, then

$∫ 0 T | x ( t ) | 2 dt⩽ ( T 2 / 4 π 2 ) ∫ 0 T | x ′ ( t ) | 2 dt$

(Wirtinger inequality) and

$| x | ∞ 2 ⩽(T/12) ∫ 0 T | x ′ ( t ) | 2 dt$

(Sobolev inequality).

Lemma 2.2 Suppose that the following condition holds:

$( H 1 )$ $(g(t, x 1 )−g(t, x 2 ))( x 1 − x 2 )<0$, for all $t, x 1 , x 2 ∈R$ and $x 1 ≠ x 2$.

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

Proof Assume that $x 1 (t)$ and $x 2 (t)$ are two T-periodic solutions of Eq. (1.1). Then we obtain

$( x i ′ ( t ) 1 + x i ′ 2 ( t ) ) ′ +f ( t , x i ′ ( t ) ) +g ( t , x i ( t ) ) =e(t),i=1,2.$
(2.1)

It is easy to see that $x i (t)∈ C 2 [0,T]$ ($i=1,2$). From (2.1), we know (2.2)

Set $z(t)= x 1 (t)− x 2 (t)$. Now, we prove

Otherwise, we have

$max t ∈ R z(t)= max t ∈ [ 0 , T ] z(t)>0.$

Then there exists a $t ∗ ∈[0,T]$ such that

$x 1 ( t ∗ ) − x 2 ( t ∗ ) =z ( t ∗ ) = max t ∈ R z(t)= max t ∈ [ 0 , T ] z(t)>0,$
(2.3)

which implies that

$z ′ ( t ∗ ) = x 1 ′ ( t ∗ ) − x 2 ′ ( t ∗ ) =0,$
(2.4)

and

$z ″ ( t ∗ ) = x 1 ″ ( t ∗ ) − x 2 ″ ( t ∗ ) ⩽0.$
(2.5)

It follows from (2.2), (2.4) and (2.5) that

$g ( t ∗ , x 1 ( t ∗ ) ) − g ( t ∗ , x 2 ( t ∗ ) ) = − ( x 1 ″ ( t ∗ ) ( 1 + x 1 ′ 2 ( t ∗ ) ) 3 − x 2 ″ ( t ∗ ) ( 1 + x 2 ′ 2 ( t ∗ ) ) 3 ) − ( f ( t ∗ , x 1 ′ ( t ∗ ) ) − f ( t ∗ , x 2 ′ ( t ∗ ) ) ) = − 1 ( 1 + x 1 ′ 2 ( t ∗ ) ) 3 ( x 1 ″ ( t ∗ ) − x 2 ″ ( t ∗ ) ) = − 1 ( 1 + x 1 ′ 2 ( t ∗ ) ) 3 z ″ ( t ∗ ) ⩾ 0 .$

From $( H 1 )$, we get

$x 1 ( t ∗ ) − x 2 ( t ∗ ) ⩽0,$

By using a similar argument, we can also show

Hence,

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 $f ˜$ is a complete continuous field on $Ω ¯$. Moreover, assume that the Leray-Schauder degree

$deg{ f ˜ ,Ω,p}≠0, for p∈X∖ f ˜ (∂Ω).$

Then the equation $f ˜ (x)=p$ 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 $( H 1 )$ hold. Moreover, assume that the following conditions hold:

$( H 2 )$ there exists $l>0$ such that

$| g ( t , x 1 ) − g ( t , x 2 ) | ⩽l| x 1 − x 2 |, for all t, x 1 , x 2 ∈R;$

$( H 3 )$ there exists $β,γ>0$ such that

$γ⩽ lim inf | x | → ∞ f ( t , x ) x ⩽ lim sup | x | → ∞ f ( t , x ) x ⩽β, uniformly in t∈R;$

$( H 4 )$ for all $t,x∈R$,

$f ( t + T 2 , − x ) =−f(t,x),g ( t + T 2 , − x ) =−g(t,x),e ( t + T 2 ) =−e(t).$

Then Eq. (1.1) has a unique anti-periodic solution for $l⋅ T 2 π <γ$.

Proof Rewrite Eq. (1.1) in the equivalent form:

${ x 1 ′ ( t ) = ψ ( x 2 ( t ) ) = x 2 ( t ) 1 − x 2 2 ( t ) , x 2 ′ ( t ) = − f ( t , ψ ( x 2 ( t ) ) ) − g ( t , x 1 ( t ) ) + e ( t ) ,$
(3.1)

where $ψ(x)= x 1 − x 2$. Now, we consider the auxiliary equation of (3.1),

${ x 1 ′ ( t ) = λ x 2 ( t ) 1 − x 2 2 ( t ) = λ ψ ( x 2 ( t ) ) , x 2 ′ ( t ) = − λ f ( t , ψ ( x 2 ( t ) ) ) − λ g ( t , x 1 ( t ) ) + λ e ( t ) ,$
(3.2)

where $λ∈(0,1]$ is a parameter. Set

$x(t)=( x 1 ( t ) x 2 ( t ) ), Q 1 ( t , x 1 ( t ) , x 2 ( t ) ) =( ψ ( x 2 ( t ) ) − f ( t , ψ ( x 2 ( t ) ) ) − g ( t , x 1 ( t ) ) + e ( t ) ).$

Then Eq. (3.2) can be reduced to the equation as follows:

$x ′ (t)=λ Q 1 ( t , x 1 ( t ) , x 2 ( t ) ) .$

By Lemma 2.2 and condition $( H 1 )$, 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 $x(t)= ( x 1 ( t ) , x 2 ( t ) ) T ∈ C T 1 , 1 2 (R, R 2 )$ be an arbitrary possible anti-periodic solution of Eq. (3.2). Then $x 1 (t)∈ C T 1 , 1 2 (R,R)$. Thus, we have

$∫ 0 T x 1 ( t ) d t = ∫ 0 T 2 x 1 ( t ) d t + ∫ T 2 T x 1 ( t ) d t = ∫ 0 T 2 x 1 ( t ) d t + ∫ 0 T 2 x 1 ( t + T 2 ) d t = 0 .$

It follows from Lemma 2.1 that

$| x 1 | ∞ ⩽ T 12 | x 1 ′ | 2 .$

Obviously, Eq. (3.2) is equivalent to the following equation:

$( 1 λ x 1 ′ ( t ) 1 + 1 λ 2 x 1 ′ 2 ( t ) ) ′ +λf ( t , 1 λ x 1 ′ ( t ) ) +λg ( t , x 1 ( t ) ) =λe(t).$
(3.3)

Multiplying (3.3) by $x 1 ′$ and integrating from 0 to T, we have

$λ ∫ 0 T f ( t , 1 λ x 1 ′ ( t ) ) x 1 ′ (t)dt+λ ∫ 0 T g ( t , x 1 ( t ) ) x 1 ′ (t)dt=λ ∫ 0 T e(t) x 1 ′ (t)dt.$
(3.4)

Since $l⋅ T 2 π <γ$, there exists a constant $ε>0$ such that

$l⋅ T 2 π <γ−ε.$
(3.5)

For such a $ε>0$, in view of $( H 3 )$, there exists $M 1 ⩾0$ such that for all $t,x∈R$, $xf(t,x)⩾(γ−ε) x 2 − M 1$. Hence,

$| λ ∫ 0 T f ( t , 1 λ x 1 ′ ( t ) ) x 1 ′ ( t ) d t | ⩾ λ 2 ∫ 0 T f ( t , 1 λ x 1 ′ ( t ) ) x 1 ′ ( t ) λ d t ⩾ ( γ − ε ) ∫ 0 T | x 1 ′ ( t ) | 2 d t − λ 2 M 1 .$
(3.6)

It follows from (3.4) and (3.6) that

$( γ − ε ) ∫ 0 T | x 1 ′ ( t ) | 2 d t ⩽ | λ ∫ 0 T f ( t , 1 λ x 1 ′ ( t ) ) x 1 ′ ( t ) d t | + λ 2 M 1 ⩽ | ∫ 0 T g ( t , x 1 ( t ) ) x 1 ′ ( t ) d t | + | ∫ 0 T e ( t ) x 1 ′ ( t ) d t | + M 1 ⩽ ∫ 0 T | g ( t , x 1 ( t ) ) − g ( t , 0 ) | | x 1 ′ ( t ) | d t + ∫ 0 T ( | g ( t , 0 ) | + | e ( t ) | ) | x 1 ′ ( t ) | d t + M 1 ⩽ l ∫ 0 T | x 1 ( t ) | | x 1 ′ ( t ) | d t + max t ∈ [ 0 , T ] { | g ( t , 0 ) | + | e ( t ) | } ∫ 0 T | x 1 ′ ( t ) | d t + M 1 .$

For $u,v∈C([a,b],R)$, we have the Schwarz inequality

$∫ a b | u ( x ) | | v ( x ) | dx⩽ ( ∫ a b | u ( x ) | 2 d x ) 1 2 ( ∫ a b | v ( x ) | 2 d x ) 1 2 .$

Hence,

$( γ − ε ) ∫ 0 T | x 1 ′ ( t ) | 2 d t ⩽ l ( ∫ 0 T | x 1 ( t ) | 2 d t ) 1 2 ⋅ ( ∫ 0 T | x 1 ′ ( t ) | 2 d t ) 1 2 + max t ∈ [ 0 , T ] { | g ( t , 0 ) | + | e ( t ) | } T ( ∫ 0 T | x 1 ′ ( t ) | 2 d t ) 1 2 + M 1 = l | x 1 | 2 | x 1 ′ | 2 + T | x 1 ′ | 2 max t ∈ [ 0 , T ] { | g ( t , 0 ) | + | e ( t ) | } + M 1 ⩽ l ⋅ T 2 π | x 1 ′ | 2 2 + T | x 1 ′ | 2 max t ∈ [ 0 , T ] { | g ( t , 0 ) | + | e ( t ) | } + M 1 .$
(3.7)

From (3.5) and (3.7), we know that there exists a constant $D 1 >0$ such that

$| x 1 ′ | 2 ⩽ D 1 ,and | x 1 | ∞ ⩽ D 1 .$
(3.8)

By the first equation of (3.2), we have

$∫ 0 T x 2 ( t ) 1 − x 2 2 ( t ) dt=0.$

Then there exists $η∈[0,T]$ such that $x 2 (η)=0$. It follows that $x 2 (t)= x 2 (η)+ ∫ η t x 2 ′ (s)ds$, and so

$| x 2 | ∞ ⩽ ∫ 0 T | x 2 ′ ( t ) | dt.$

According to $( H 3 )$, we know there exists $M 2 ⩾0$ such that for all $t,x∈R$,

$| f ( t , x ) | ⩽(β+1)|x|+ M 2 .$

From the second equation of (3.2), we get

$∫ 0 T | x 2 ′ ( t ) | d t ⩽ ∫ 0 T λ | f ( t , 1 λ x 1 ′ ( t ) ) | d t + ∫ 0 T | g ( t , x 1 ( t ) ) | d t + ∫ 0 T | e ( t ) | d t ⩽ ( β + 1 ) ∫ 0 T | x 1 ′ ( t ) | d t + λ M 2 T + ∫ 0 T | g ( t , x 1 ( t ) ) | d t + ∫ 0 T | e ( t ) | d t ⩽ ( β + 1 ) T | x 1 ′ | 2 + ∫ 0 T | g ( t , x 1 ( t ) ) | d t + ∫ 0 T | e ( t ) | d t + M 2 T .$

From (3.8), we know that there exists a constant $k>0$ such that

$| g ( t , x 1 ( t ) | ⩽k,∀t∈[0,T].$

Thus,

$∫ 0 T | x 2 ′ ( t ) | dt⩽(β+1) T D 1 +kT+T max t ∈ [ 0 , T ] | e ( t ) | + M 2 T,$

which implies that there exists a constant $D 2 >0$ such that

$| x 2 | ∞ ⩽ D 2 .$

Let

$M=max{ D 1 , D 2 }+1.$
(3.9)

Set

$Ω= { x ∈ C T 0 , 1 2 ( R , R 2 ) = X : ∥ x ∥ < M } .$

Then Eq. (3.2) has no anti-periodic solution on Ω for $λ∈(0,1]$.

Next, we consider the Fourier series expansions of two functions $x j (t)∈ C T k , 1 2 (R,R)$ ($j=1,2$). We have

$x j (t)= ∑ i = 0 ∞ [ a 2 i + 1 j cos 2 π ( 2 i + 1 ) t T + b 2 i + 1 j sin 2 π ( 2 i + 1 ) t T ] .$

Define an operator $L 1 : C T k , 1 2 (R,R)→ C T k + 1 , 1 2 (R,R)$ by setting

$( L 1 x j ) ( t ) = ∫ 0 t x j ( s ) d s − T 2 π ∑ i = 0 ∞ b 2 i + 1 j 2 i + 1 = T 2 π ∑ i = 0 ∞ [ a 2 i + 1 j 2 i + 1 sin 2 π ( 2 i + 1 ) t T − b 2 i + 1 j 2 i + 1 cos 2 π ( 2 i + 1 ) t T ] .$

Then

$d d t ( L 1 x j )(t)= x j (t),$

and

$| ( L 1 x j ) ( t ) | ⩽ ∫ 0 T | x j ( s ) | d s + T 2 π ∑ i = 0 ∞ | b 2 i + 1 j | 2 i + 1 ⩽ T | x j | ∞ + T 2 π ( ∑ i = 0 ∞ ( b 2 i + 1 j ) 2 ) 1 2 ( ∑ i = 0 ∞ 1 ( 2 i + 1 ) 2 ) 1 2 .$

Since

$( ∑ i = 0 ∞ 1 ( 2 i + 1 ) 2 ) 1 2 = π 2 2 ,$

and

$∫ 0 T | x j ( s ) | 2 ds= T 2 ∑ i = 0 ∞ [ ( a 2 i + 1 j ) 2 + ( b 2 i + 1 j ) 2 ] ,$

we obtain

$| ( L 1 x j ) ( t ) | ⩽ T | x j | ∞ + T 4 2 ( ∑ i = 0 ∞ [ ( a 2 i + 1 j ) 2 + ( b 2 i + 1 j ) 2 ] ) 1 2 ⩽ T | x j | ∞ + T 4 2 ( 2 T ∫ 0 T | x j ( s ) | 2 d s ) 1 2 ⩽ ( T + T 4 ) | x j | ∞ .$

Define $L: C T k , 1 2 (R, R 2 )→ C T k + 1 , 1 2 (R, R 2 )$ by setting

$(Lx)(t)=L( x 1 ( t ) x 2 ( t ) )=( ( L 1 x 1 ) ( t ) ( L 1 x 2 ) ( t ) ).$

Then $| L x | ∞ ⩽(T+ T 4 ) | x | ∞$, and thus L is continuous.

For any $x(t)∈ C T 0 , 1 2 (R, R 2 )$, we know from $( H 4 )$ that

$Q 1 ( t + T 2 , x 1 ( t + T 2 ) , x 2 ( t + T 2 ) ) =− Q 1 ( t , x 1 ( t ) , x 2 ( t ) ) .$

Therefore, $Q 1 (t, x 1 (t), x 2 (t))∈ C T 0 , 1 2 (R, R 2 )$. Define an operator $F μ : Ω ¯ → C T 1 , 1 2 (R, R 2 )⊂X$ by setting

$F μ (x)=μL ( Q 1 ( t , x 1 ( t ) , x 2 ( t ) ) ) ,μ∈[0,1].$

It is easy to see that $F μ$ is a compact homotopy, and the fixed point of $F 1$ on $Ω ¯$ is the anti-periodic of Eq. (3.1).

Define a homotopic field as follows:

$H μ (x): Ω ¯ ×[0,1]→ C T 0 , 1 2 ( R , R 2 ) , H μ (x)=x− F μ (x).$

From (3.9), we have

$H μ (∂Ω)≠0,μ∈[0,1].$

Using the homotopy invariance property of degree, we obtain

$deg { x − F 1 ( x ) , Ω , 0 } =deg{x,Ω,0}≠0.$

Till now, we have proved that Ω satisfies all the requirements in Lemma 2.3. Consequently, $x− F 1 (x)=0$ has at least one solution in Ω, i.e., $F 1$ has a fixed point $x(t)= ( x 1 ( t ) , x 2 ( t ) ) T$ on $Ω ¯$. Therefore, Eq. (1.1) has at least one anti-periodic solution $x 1 (t)$. 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 $f(t,x)=(1+ 1 2 sin 2 t) x 3 1 + x 4$, $g(t,x)=−(1+ sin 4 t)⋅ x 3$. Then the prescribed mean curvature Rayleigh equation

$( x ′ 1 + x ′ 2 ) ′ +f ( t , x ′ ( t ) ) +g ( t , x ( t ) ) =cost$
(4.1)

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

Proof Let $T=2π$. From the definitions of $f(t,x)$ and $g(t,x)$, we can easily check that conditions $( H 1 )$ and $( H 4 )$ hold. Moreover, it is easy to see that $( H 2 )$ holds for $l= 2 3$ and $( H 3 )$ holds for $γ=1$, $β= 3 2$. Since $l⋅ T 2 π <γ$, we know from Theorem 3.1 that Eq. (4.1) has a unique anti-periodic solution with period 2π. □

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## 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.

## Author information

Correspondence to Jin Li.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

Both authors, AA and MHA, contributed to each part of this work equally and read and approved the final version of the manuscript.

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Li, J., Wang, Z. Existence and uniqueness of anti-periodic solutions for prescribed mean curvature Rayleigh equations. Bound Value Probl 2012, 109 (2012) doi:10.1186/1687-2770-2012-109 