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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

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