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Invariant curves for a delay differential equation with a piecewise constant argument
- Jinghua Liu^{1, 2}Email author
https://doi.org/10.1186/s13662-014-0349-7
© Liu; licensee Springer 2015
- Received: 12 September 2014
- Accepted: 25 December 2014
- Published: 30 January 2015
Abstract
In order to understand the dynamics of a second order delay differential equation with a piecewise constant argument, we investigate invariant curves of the derived planar mapping from the equation. All invariant curves are given in this paper.
Keywords
- difference equation
- invariant curve
- piecewise construction
- characteristic root
- dual equation
1 Introduction
2 Main results
- (C1)
\(0< r_{1}<1<r_{2}\), if and only if \(-2< a<0\).
- (C2)
\(r_{1}=r_{2}=1\), if and only if \(a=0\).
- (C3)
\(r_{1}<0<r_{2}\neq1\) and \(r_{1}\ne-r_{2}\), if and only if \(a<-2\).
- (C4)
\(r_{1}=r_{2}<0 \), if and only if \(a=16\).
- (C5)
\(r_{1}< r_{2}<-1\), if and only if \(a>16\).
Note that the case \(r_{2}>r_{1}>1\) is not listed because the case \(r_{2}>r_{1}>1\) implies \(\frac{(4-a)-(a^{2}-16a)^{\frac{1}{2}}}{4}>1\), i.e., \(-a>(a^{2}-16a)^{\frac{1}{2}}\), which does not hold, and that the case \(0< r_{1}<r_{2}<1\) is not listed because \(0< r_{1}<r_{2}<1\) implies \(0<\frac{(4-a)+(a^{2}-16a)^{\frac{1}{2}}}{4}<1\), i.e., \(0< a<4\), which contradicts the requirement that \(\Delta=a^{2}-16a\geq0\), and that the case \(0< a<16\) is not listed because in this case (4) with \(b=0\) has no continuous solutions, neither \(r_{1}\) nor \(r_{2}\) is real, by [14]. Since we consider \(a\notin\{-2, 4\}\), none of the case \(r_{1}=0\), the case \(r_{2}=0\), and the case \(r_{1}=-r_{2}\ne0\) is listed. Corresponding to the above list, we have the following results.
Theorem 2.1
Proof
The proof is a simple application of well-known results in [14]. The result (i) is given by Theorem 2 of [14], where the characteristic roots \(r_{1}\), \(r_{2}\) satisfy \(r_{2}>1>r_{1}>0\) as shown in (C1). We can deduce the result (ii) from Theorem 8 of [14], where \(r_{1}=r_{2}=1\) as shown in (C2). The proof is completed. □
Theorem 2.2
Proof
Remark
In what follows, we consider the case that either \(a=-2\) or \(a=4\), which is not generic.
For \(a=-2\), (4) is of the form \(f^{2}(x)-3f(x)=-b\), from which we get with the replacement \(y=f(x)\): \(f(x)=3x-b\).
Declarations
Acknowledgements
The author would like to thank the referees for their valuable comments and suggestions, which have helped to improve the quality of this paper. This work was supported by the general item (L0801) of Zhanjiang Normal University.
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.
Authors’ Affiliations
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