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  • Open Access

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

Boundary Value Problems20142014:204

https://doi.org/10.1186/s13661-014-0204-5

  • Received: 18 August 2014
  • Accepted: 18 August 2014
  • Published:

The original article was published in Boundary Value Problems 2012 2012:109

Abstract

In this paper, we give a complementary proof on the paper ‘Existence and uniqueness of anti-periodic solutions for prescribed mean curvature Rayleigh equations’.

Keywords

  • complementary proof
  • prescribed mean curvature Rayleigh equations

1 Introduction

In [1], the authors were 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 is a constant.

The paper mentioned above obtained the main result by using Mawhin’s continuation theorem in the coincidence degree theory. Unfortunately, the proof of main result Theorem 3.1 (see [1]) has a serious problem: F μ ( x ) = μ L ( Q 1 ( t , x 1 , x 2 ) ) where Q 1 depends on ψ ( x 2 ) and ψ ( x ) = x 1 x 2 which is only defined for | x | < 1 and cannot be continuously extended; therefore, F μ should not be defined on Ω ¯ = { x X : x < M } since | x 2 ( t ) | > 1 can occur, where x = max { x 1 , x 2 } and M = 1 + max { D 1 , D 2 } .

In this paper, we shall give a complementary proof to correct the errors.

2 Complementary proof

Rewrite (1.1) in the equivalent form as follows:
{ 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 ) ,
(2.1)
where ψ ( x ) = x 1 x 2 . In [1], the authors embed (2.1) into a family of equations with one parameter λ ( 0 , 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 ) .
(2.2)
They have proved that there exists a constant D 1 > 0 such that
| x 1 | 2 D 1 , and | x 1 | D 1 ,
(2.3)

and there exists η [ 0 , T ] such that x 2 ( η ) = 0 .

In fact, to use the continuation theorem, it suffices to prove that there exists a positive constant 0 < ε 0 1 such that, for any possible solution ( x 1 ( t ) , x 2 ( t ) ) of (2.2), the following condition holds:
| x 2 ( t ) | < 1 ε 0 .
(2.4)

In what follows, we shall give a complementary proof for the main result in [1] by giving a proof of (2.4).

In [1], the authors assume that

(H1): ( 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 ;

(H2): 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 ;
(H3): there exists β, γ such that
γ lim inf | x | f ( t , x ) x lim sup | x | f ( t , x ) x β , uniformly in  t R ;
(H4): 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 ) .

Under the conditions mentioned above, we prove that (2.4) holds.

Since | x 1 | < D 1 and g, e are continuous, we find that there exists M 3 > 0 such that
M 3 < g ( t , x 1 ( t ) ) + e ( t ) < M 3 , t R .
(2.5)
By (H3), there exists a positive constant M 4 > 0 such that
f ( t , x ) γ x M 4 , x > 0  and  t R .
(2.6)
Next, we shall prove that
x ( t ) M 3 + M 4 ( M 3 + M 4 ) 2 + γ 2 , t R .
Assume by contradiction that there exist t 2 > t 1 > η such that
x 2 ( t 1 ) = M 3 + M 4 ( M 3 + M 4 ) 2 + γ 2 , x 2 ( t 2 ) > M 3 + M 4 ( M 3 + M 4 ) 2 + γ 2 ,
and
x 2 ( t ) > M 3 + M 4 ( M 3 + M 4 ) 2 + γ 2 , for  t ( t 1 , t 2 ) .
Noticing that λ ( 0 , 1 ] , we have, t ( t 1 , t 2 ) ,
x 2 ( t ) = λ ( f ( t , ψ ( x 2 ( t ) ) ) g ( t , x 1 ( t ) ) + e ( t ) ) < 0 ,

which is a contradiction.

By (H3), there exists a positive constant M 5 > 0 such that
f ( t , x ) β x + M 5 , x < 0  and  t R .
By using a similar argument, we can prove that
x 2 ( t ) M 3 + M 5 ( M 3 + M 5 ) 2 + β 2 , for  t R .
Therefore, we get from the continuity of x 2 ( t ) , for any solution ( x 1 ( t ) , x 2 ( t ) ) of (2.2),
M 3 + M 5 ( M 3 + M 5 ) 2 + β 2 x 2 ( t ) M 3 + M 4 ( M 3 + M 4 ) 2 + γ 2 , t R .

Consequently, (2.4) holds.

Putting
Ω = { x = ( x , x ) C T 0 , 1 2 ( R , R 2 ) = X : x < M , | x 2 ( t ) | < 1 ε 0 } ,

we can use Mawhin’s continuation theorem on Ω.

Notes

Declarations

Acknowledgements

The authors would like to thank Professor J Webb for pointing out the errors of the paper [1].

Authors’ Affiliations

(1)
School of Science, Jiujiang University, Jiujiang, 332005, China
(2)
School of Mathematical Sciences, Capital Normal University, Beijing, 100048, China

References

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

Copyright

© Li and Wang; licensee Springer 2014

This article is published under license to BioMed Central Ltd.Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

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