R E S E A R C H
Open Access
Invariant curves for a delay differential
equation with a piecewise constant argument
Jinghua Liu
**Correspondence: [email protected] Department of Mathematics, Sichuan University, Chengdu, Sichuan 610064, P.R. China School of Mathematics and Computation Science, Lingnan Normal University, Zhanjiang, Guangdong 524048, P.R. China
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
The study of differential equations with piecewise constant argument (EPCA) initiated in [, ]. These equations represent a hybrid of continuous and discrete dynamical systems and combine the properties of both differential and difference equations, hence, they are of importance in control theory and in certain biomedical models []. In this paper the second order delay differential equation with a piecewise constant argument
x(t) +gx[t]= , t∈R,x∈R, ()
wherex(t) denotes the second order derivative ofx(t), [t] denotes the greatest integer less than or equal tot, andg:R→Ris a continuous or at least piecewise continuous function, is considered. In , Aftabizadehet al.discussed the oscillatory and periodic properties of the solutions of () in []. In , Gyori and Ladas investigated linearized oscillations of the solutions of () in []. Later, Wiener and Cooke considered oscillations of the solutions of systems of two differential equations with piecewise constant arguments in [].
The invariant curve [–] is another interesting problem in the study of dynamics be-cause it can be used to reduce a system to a -dimensional one. The problem of invariant curves is actually a part of the research on invariant manifolds. In , Ng and Zhang studied the nonlinearCinvariant curve of planar mappingG:R→R,
G(x,y) =
y, y–x–
g(y) +g(x), ()
derived from () in [] wheng is nonlinear and gave the conditions that Ghas linear invariant curves whengis linear. In , Yanget al.investigated nonlinearCinvariant
curves of () whengis nonlinear in []. So far, nonlinear invariant curves of () whengis linear have not been studied. So it is very interesting to look for nonlinear invariant curves
of () whengis linear. In this paper all the invariant curves of the planar mappingGare given including the linear and nonlinear ones whengis linear.
2 Main results
We discuss invariant curves of the formy=f(x) for the planar mapping (). Its invariant curves of the formy=f(x) satisfyf(y) = y–x–(g(y) +g(x)), which leads to the iterative functional equation
ff(x)= f(x) –x–
gf(x)+g(x), ∀x∈R. ()
Considering lineargandg(x) =ax+b, we compute that
ff(x)–
–a
f(x) +
+a
x= –b, ∀x∈R. ()
Thus, the invariant curves of planar mappingGwithg(x) =ax+b can be obtained by solving functional (). We mainly discuss the generic casesa∈ {/ –, }, but leave the special casesa= – anda= to the last part of this section. For generica∈ {/ –, }, () withb= is of the form discussed in [, ]. In order to apply the results of [], we let
r:=
( –a) – (a– a)
, r:=
( –a) + (a– a)
, ()
which are the roots of the characteristic polynomialP(r) :=r– ( –a
)r+ +
a
.
From () we see that the characteristic rootsr,rof () have the following possibilities:
(C) <r< <r, if and only if– <a< .
(C) r=r= , if and only ifa= .
(C) r< <r= andr= –r, if and only ifa< –.
(C) r=r< , if and only ifa= .
(C) r<r< –, if and only ifa> .
Note that the case r > r > is not listed because the case r >r > implies (–a)–(a–a)
> ,i.e., –a> (a
–a), which does not hold, and that the case <r
<r<
is not listed because <r<r< implies <(–a)+(a
–a)
< ,i.e., <a< , which
con-tradicts the requirement that=a– a≥, and that the case <a< is not listed because in this case () withb= has no continuous solutions, neitherrnorris real,
by []. Since we considera∈ {/ –, }, none of the caser= , the caser= , and the case
r= –r= is listed. Corresponding to the above list, we have the following results.
Theorem . (i)If– <a< ,then a continuous solutionsφof()with b= is either of the piecewise linear form that f(x) :=rix for x> ,or:= for x= ,or:=rjx for x< ,where
i,j= , ,or given by
f(x) :=
fn(x), x∈[xn,xn+),n= , , , . . . ,
f–
–n(x), x∈[x–n,x–n+),n= , , . . . ,
where xn = r
n
r–r(x –rx) + rn
r–r(–x +rx), n∈Z, with an arbitrarily chosen x ∈
(–∞, +∞)and x ∈[rx,rx], and fn(x) = (r+r)x–rrfn––(x)for all x∈[xn,xn+),
(r
Proof The proof is a simple application of well-known results in []. The result (i) is given by Theorem of [], where the characteristic rootsr,rsatisfyr> >r> as shown
in (C). We can deduce the result (ii) from Theorem of [], wherer=r= as shown
in (C). The proof is completed.
Theorem . (i)If a< –,then()with b= only has two continuous solutions f and
Proof Firstly, we consider (i). By Theorem in [], () withb= only has two continuous solutionsf andf(x) =rxorrx, where the characteristic rootsr,rsatisfyr< <r=
andr= –ras shown in (C). Next, we consider (ii). By Theorem in [], () withb=
just has a continuous solution f(x) = –x, where the characteristic rootsr,rsatisfy
r=r= – as shown in (C). Finally, we consider (iii). In order to piecewise construct all
solutions of () withb= we need a partition for the interval (–∞,∞). For this purpose we consider a homogeneous linear difference equation
xn+–
which has the same coefficients as () withb= correspondingly. Its characteristic equa-tion is
() and () can be, respectively, rewritten as
ff(x)– (r+r)f(x) +rrx= , ()
Iff is a solution of () withb= , we easily see thatf is invertible. In fact, iff(x) =f(x),
which is called the dual equation to () withb= . Solving the homogeneous linear differ-ence () with arbitrarily chosen real initial valuesxandx, we obtain
xn=
recursively define the homeomorphismsfn–: [x–n+,x–n+)→[x–n,x–n+),n= , , . . . ,
andfn: [x–n,x–n+)→[x–n+,x–n+),n= , , . . . , such that
fn–(x–n+) =x–n+, fn–(x–n+) =x–n, ()
fn(x–n) =x–n+, fn(x–n+) =x–n+, ()
r≤fn–(x,y)≤r, ∀x,y∈[x–n+,x–n+), ()
r≤fn(x,y)≤r, ∀x,y∈[x–n,x–n+). ()
In fact, forfndefined satisfying () and (), we let
fn+(x) = (r+r)x–rrf–n(x), ∀x∈[x–n+,x–n+).
Obviously,fn+(x–n+) =x–nandfn+(x–n+) =x–n–. Making use of (), we have r ≤
f–n(x)–f–n(y)
x–y ≤
r forx,y∈[x–n,x–n+). It is easy to deduce that
r≤fn+(x,y)≤r, ∀x,y∈[x–n+,x–n+).
Furthermore, we again let
fn+(x) = (r+r)x–rrf–n+(x), ∀x∈[x–n–,x–n).
By the same argument we can see that
fn+(x–n–) =x–n–, fn+(x–n) =x–n+, ()
r≤fn+(x,y)≤r, ∀x,y∈[x–n–,x–n). ()
By induction both fn– andfn are well defined. Similarly, we can also recursively
de-fine the homeomorphisms f–n+ : [xn–,xn) → [xn+,xn+), n = , , . . . , and f–n :
[xn+,xn+)→[xn,xn+),n= , , . . . . By the properties of the dual () we can obtain
f–n+(xn–) =xn+, f–n+(xn) =xn+,
f–n(xn+) =xn+, f–n(xn+) =xn,
r
≤f–n+(x,y)≤
r
, ∀x,y∈[xn–,xn),
r≤
f–n(x,y)≤
r
, ∀x,y∈[xn+,xn+).
Therefore,
f––n+(xn+) =xn–, f––n+(xn+) =xn,
f––n(xn+) =xn+, f––n(xn) =xn+,
r≤f––n+(x,y)≤r, ∀x,y∈[xn+,xn+),
Thus, we can define
f(x) :=
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
fn(x), x∈[x–n,x–n+),n= , , , . . . ,
fn+(x), x∈[x–n+,x–n+),n= , , , . . . ,
, x= ,
f–
–n(x), x∈[xn,xn+),n= , , . . . ,
f––n+(x), x∈[xn+,xn+),n= , , . . . .
f is continuous on R because fn(x–n) =x–n+ =fn+(x–n), fn+(x–n+) =x–n– =
fn+(x–n+), wheren= , , , . . . ,f–(x) =f–(x),f––n(xn+) =xn+=f––n–(xn+), and
f––n+(xn+) =xn+=f––n–(xn+), wheren= , , , . . . . we can easily check thatf defined
in Theorem . satisfies () withb= inR. In fact, ifx∈[x–n,x–n+), n= , , , . . . ,
f(x) =f
n+(fn(x)) = (r+r)fn(x) –rrx= (r+r)f(x) –rrx,i.e.,f(x) – (r+r)f(x) –
rrx= . Similarly, we can also check thatf satisfies () withb= forx∈[x–n+,x–n+),
x∈[xn+,xn+),x∈[xn,xn+) andx= , wheren= , , , . . . . The proof is completed.
Remark In the case thatb= , as indicated in [] for () therein, () can be reduced equivalently to the equation
˜
ff˜(x)–
–a
˜
f(x) +
+a
x= , ()
the same type of equation as the one considered in Theorems . and . with vanishing
b, by the replacement˜f(x) =f(x+ξ) –ξ, whereξ= (–r–b
)(–r), if its characteristic roots
r,rare both real but neither of them is equal to . In this case solutions can be found
from Theorems . and .. So () withb= can be reduced to () except for the case
a= . For the case ofa= andb= , () has no real continuous solutions. In fact, by induction and () we can obtainfn(x) =nf(x) – (n– )x–n(n+)
b,n∈Z. Furthermore, we
havefn+(x) –fn(x) =f(x) –x– (n+ )b,n∈Z. For an arbitraryx∈R,fn+(x) –fn(x) has the same sign whenntakes the valuesNand –N, whereNis a large positive integer, because
f is strictly monotonic. Butf(x) –x– (n+ )bhas not, which contradicts the requirement
fn+(x) –fn(x) =f(x) –x– (n+ )b,n∈Z.
In what follows, we consider the case that eithera= – ora= , which is not generic. Fora= –, () is of the formf(x) – f(x) = –b, from which we get with the replacement
y=f(x):f(x) = x–b. Fora= , () is of the form
f(x) = –x–b, ()
which is the problem of iterative roots of the linear functionF(x) := –x–b. By the theory of iterative roots, as shown in [], we know () has no real continuous solutions.
Competing interests
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.
Received: 12 September 2014 Accepted: 25 December 2014
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