R E S E A R C H
Open Access
Neutral operator with variable parameter and
third-order neutral differential equation
Yun Xin
1*and Zhibo Cheng
2*Correspondence:
1College of Computer Science and
Technology, Henan Polytechnic University, Jiaozuo, 454000, China Full list of author information is available at the end of the article
Abstract
In this article, we discuss the properties of the neutral operator with variable parameter (Ax)(t) =x(t) –c(t)x(t–
δ
(t)) and by applying Green’s function of athird-order differential equation and a fixed point theorem in cones, we obtain some sufficient conditions for existence, nonexistence, multiplicity of positive periodic solutions for a generalized third-order neutral differential equation.
Keywords: neutral operator; variable parameter; positive solutions; third-order; Green’s function
1 Introduction
In [], Zhang discussed the properties of the neutral operator (Ax)(t) =x(t) –cx(t–δ), which became an effective tool for the research on differential equations with this pre-scribed neutral operator (see,e.g., [–]). Lu and Ge [] investigated an extension ofA, namely the neutral operator (Ax)(t) =x(t) –
n
i=cix(t–δi), and obtained the existence of
periodic solutions for the corresponding neutral differential equation. Afterwards, Duet al.[] studied the neutral operator (Ax)(t) =x(t) –c(t)x(t–δ), herec(t) isω-periodic func-tions. By means of Mawhin’s continuation theorem and the properties ofA, they obtained sufficient conditions for the existence of periodic solutions to a Liénard neutral differen-tial equation. Recently, in [], Renet al.investigated the neutral operator with variable delay (A)x(t) –cx(t–δ(t)). By applying coincidence degree theory, they obtained suffi-cient conditions for the existence of periodic solutions to a Rayleigh neutral differential equation.
Motivated by [, –], in this paper, we consider the neutral operator (Ax)(t) =x(t) –
c(t)x(t–δ(t)), here |c(t)| = , c,δ∈C(R,R) and δ is anω-periodic function for some ω> . Notice that here the neutral operatorAis a natural generalization of the famil-iar operatorAi, i= , , , . But Apossesses a more complicated nonlinearity thanAi,
i= , , , . For example, the neutral operatorAi,i= , , is homogeneous in the following sense (Aix)(t) = (Aix)(t),i= , , whereas the neutral operatorAin general is inhomo-geneous. As a consequence, many of the new results for differential equations with the neutral operatorAwill not be a direct extension of known theorems for neutral differen-tial equations.
The paper is organized as follows. In Section , we first analyze qualitative properties of the generalized neutral operatorAwhich will be helpful for further studies of differen-tial equations with this neutral operator; in Section , we consider a third-order neutral
differential equation as follows:
x(t) –c(t)xt–δ(t)=a(t)x(t) –λb(t)fxt–τ(t), (.) hereλis a positive parameter;δ(t) is said to be variable delay,c,δ∈C(R,R) andδis an ω-periodic function for someω> ,f∈C(R, [,∞)), andf(x) > forx> ;a∈C(R, (,∞)) withmax{a(t) :t∈[,ω]}<
√( π
ω),b∈C(R, (,∞)),τ∈C(R,R),a(t),b(t) andτ(t) are ω-periodic functions. By applying Green’s function of a third-order differential equation and a fixed point theorem in cones, we obtain sufficient conditions for the existence, mul-tiplicity and nonexistence of positive periodic solutions to the third-order neutral differ-ential equation. We will give an example to illustrate our results, and an example is also given in this section. Our results improve and extend the results in [–].
2 Analysis of the generalized neutral operator with variable parameter
Let
c∞= max
t∈[,ω]
c(t), c= min
t∈[,ω]
c(t).
LetX={x∈C(R,R) :x(t+ω) =x(t),t∈R}with the normx=maxt∈[,ω]|x(t)|, and letC+
ω={x∈C(R, (,∞)) :x(t+ω) =x(t)},Cω–={x∈C(R, (–∞, )) :x(t+ω) =x(t)}. Then (X, · ) is a Banach space. A coneKinXis defined byK={x∈X:x(t)≥αx,∀t∈R},
whereαis a fixed positive number withα< . Moreover, define operatorsA,B:Cω→Cω by
(Ax)(t) =x(t) –c(t)xt–δ(t), (Bx)(t) =c(t)xt–δ(t).
Lemma . If|c(t)| = ,then the operator A has a continuous inverse A–on Cω,satisfying
()
A–f(t) =
⎧ ⎪ ⎨ ⎪ ⎩
f(t) +∞j=ji=c(Di)x(t–ji=δ(Di)) for|c(t)|< ,∀f ∈Cω, –fc((tt++δδ((tt))))–∞j=f(t+δ(t)+
j i=δ(Di))
c(t+δ(t))ji=c(Di) for|c(t)|> ,∀f ∈Cω.
()
A–f(t)≤
⎧ ⎨ ⎩
f
–c∞ forc∞< ∀f ∈Cω,
f
c– forc> ∀f ∈Cω.
()
ω
A–f(t)dt≤ ⎧ ⎨ ⎩
–c∞
ω
|f(t)|dt forc∞< ∀f ∈Cω,
c–
ω
|f(t)|dt forc> ∀f ∈Cω.
Proof Case :|c(t)| ≤c∞< .
Lett=DandDj=t– j
i=δ(Di),j= , , . . . .
(Bx)(t) =c(t)xt–δ(t)=c(D)x
t–δ(D)
Case :|c(t)|>c> .
LetD=t,Dj=t+ji=δ(Di),j= , , . . . . And set
E:Cω→Cω, (Ex)(t) =x(t) –
c(t)x
t+δ(t),
B:Cω→Cω, (Bx)(t) =
c(t)x
t+δ(t).
By the definition of the linear operatorB, we have
Bjf(t) =j
i=c(Di) f
t+
j
i= δDi
,
hereDiis defined as in Case . Summing overjyields
∞
j=
Bjf(t) =f(t) +
∞
j=
j i=c(Di)
f
t+
j
i= δDi
.
SinceB< , we obtain that the operatorEhas a bounded inverseE–,
E–:Cω→Cω, E–= (I–B)–=I+
∞
j=
Bj,
and∀f ∈Cωwe get
E–f(t) =f(t) +
∞
j=
Bjf(t).
On the other hand, from (Ax)(t) =x(t) –c(t)x(t–δ(t)), we have
(Ax)(t) =x(t) –c(t)xt–δ(t)= –c(t)
xt–δ(t)–
c(t)x(t)
,
i.e.,
(Ax)(t) = –c(t)(Ex)t–δ(t).
Letf ∈Cωbe arbitrary. We are looking forxsuch that
(Ax)(t) =f(t),
i.e.,
–c(t)(Ex)t–δ(t)=f(t).
Therefore
(Ex)(t) = –f(t+δ(t))
and hence
proving thatA–exists and satisfies
Statements () and () are proved. From the above proof, () can easily be deduced.
3 Positive periodic solutions for third-order neutral equations
At first, we introduce the following Green’s functions and properties of Green’s functions, which can be found in [].
has a unique solution which is of the form
u(t) =
has a uniqueω-periodic solution
u(t) =
Now we present the properties of the Green’s functions for (.), (.).
Theorem . ωG(t,s)ds=ρ and if √
ρω<πholds,then <l<G(t,s)≤L for all
t∈[,ω]and s∈[,ω].
Theorem . ωG(t,s)ds=ρ and if √
ρω<πholds,then <l<G(t,s)≤L for all [,ω]and s∈[,ω].
Define the Banach spaceXas in Section . Denote
M=maxa(t) :t∈[,ω], m=mina(t) :t∈[,ω], ρ=M,
k=l(M+m) +σLM, k=
k–√k– σLlMm
σLM , α=
l[m– (M+m)c∞]
LM( –c∞) .
It is easy to see thatM,m,β,k,k> . Now we consider (.). First let
f=lim x→
f(x)
x , f∞=xlim→∞ f(x)
x , f=xlim→
f(x)
x , f∞=xlim→∞
f(x)
x ,
and denote
i= number of ’s in (f,f∞), i= number of ’s in (f,f∞);
i∞= number of∞’s in (f,f∞), i∞= number of∞’s in (f,f∞).
It is clear thati,i,i∞,i∞∈ {, , }. We will show that (.) hasiori∞positivew-periodic
solutions for sufficiently large or smallλ, respectively.
In what follows, we discuss (.) in two cases, namely the case wherec(t) < and –c∞>
–min{k,Mm+m}.
From –c∞> –Mm+m, we have α =
l[m–(M+m)c∞]
LM(–c∞) >
l(m–(M+m)· m M+m)
LM(–c∞) = . So, we getα > . Moreover, we consider the equation
σLMx–kx+lm= .
Then the equation has a solutionx=k=k–
√
k–σLlMm
σLM . Fromc∞<k, we can get
σLMc∞–kc∞+lm< . So, we have
σLMc∞–l(M+m) +σLMc∞+lm< ,
we get
σc∞>l[m– (M+m)c∞]
LM( –c∞) =α.
On the other hand, the case wherec> andc∞<min{Mm+m, LM–lm
(L–l)M–lm}(note thatc∞< m
M+m impliesα> ;c∞< LM–lm
(L–l)M–lm impliesα< ). Obviously, we havec∞< , which makes
LetK={x∈X:x(t)≥αx}denote the cone inXas defined in Section , whereαis just as defined above. We also useKr={x∈K:x<r}and∂Kr={x∈K:x=r}.
Lety(t) = (Ax)(t), then from Lemma . we havex(t) = (A–y)(t). Hence (.) can be trans-formed into
y(t) –a(t)A–y(t) = –λb(t)fA–yt–τ(t), (.) which can be further rewritten as
y(t) –a(t)y(t) +a(t)Hy(t)= –λb(t)fA–yt–τ(t), (.) whereH(y(t)) =y(t) – (A–y)(t) = –c(t)(A–y)(t–δ(t)).
Now we discuss the two cases separately.
3.1 Case I:c(t) < 0 and –c∞> – min{k1,Mm+m} Now we consider
y(t) –a(t)y(t) +a(t)Hy(t)=h(t), h∈C–ω, (.) and define the operatorsT,Hˆ :X→Xby
(Th)(t) =
t+ω
t
G(t,s)
–h(s)ds, (Hyˆ )(t) = –M+a(t)y(t) –a(t)Hy(t).
ClearlyT, Hˆ are completely continuous, (Th)(t) > forh(t) < and ˆH ≤(M–m+
M–c∞c∞). By Theorem ., the solution of (.) can be written in the form
y(t) = (Th))(t) + (THyˆ )(t). (.)
In view ofc(t) < and –c∞> –min{k,Mm+m}, we have
THˆ ≤ T ˆH ≤M–m+mc∞
M( –c∞) < , (.)
where we used the facttt+ωG(t,s)ds=M. Hence
y(t) = (I–THˆ)–(Th)(t).
Define an operatorP:X→Xby
(Ph)(t) = (I–THˆ)–(Th)(t).
Obviously, for anyh∈Cω–, ifmax{a(t) :t∈[,ω]}<√( π ω)
,y(t) = (Ph)(t) is the unique
positiveω-periodic solution of (.).
Lemma . P is completely continuous and
(Th)(t)≤(Ph)(t)≤ M( –c∞)
m– (M+m)c∞Th for all h∈C
–
Proof By the Neumann expansion ofP, we have
P= (I–THˆ)–T
=I+THˆ + (THˆ)+· · ·+ (THˆ)n+· · ·T
=T+THTˆ + (THˆ)T+· · ·+ (THˆ)nT+· · ·. (.)
SinceTandHˆ are completely continuous, so isP. Moreover, by (.), and recalling that THˆ ≤ M–m+mc∞
M(–c∞) < , we get
(Th)(t)≤(Ph)(t)≤ M( –c∞)
m– (M+m)c∞Th.
Define an operatorQ:X→Xby
Qy(t) =Pλb(t)fA–yt–τ(t). (.)
Lemma . Q(K)⊂K.
Proof From the definition ofQ, it is easy to verify thatQy(t+ω) =Qy(t). Fory∈K, we have from Lemma . that
Qy(t) =Pλb(t)fA–yt–τ(t)
≥Tλb(t)fA–yt–τ(t)
=λ
t+ω
t
G(t,s)b(s)f
A–ys–τ(s)ds
≥λl
ω
b(s)fA–ys–τ(s)ds.
On the other hand,
Qy(t) =Pλb(t)fA–yt–τ(t)
≤ M( –c∞)
m– (M+m)c∞
Tλb(t)fA–yt–τ(t)
=λ M( –c∞)
m– (M+m)c∞tmax∈[,ω]
t+ω
t
G(t,s)b(s)f
A–ys–τ(s)ds
≤λ M( –c∞)
m– (M+m)c∞L
ω
b(s)fA–ys–τ(s)ds.
Therefore
Qy(t)≥l[m– (M+m)c∞]
LM( –c∞) Qy=αQy,
i.e.,Q(K)⊂K.
the existence of positive periodic solutions for (.) is equivalent to the existence of fixed points ofQinK. Furthermore, ifQhas a fixed pointyinK, it means that (A–y)(t) is a positiveω-periodic solution of (.).
Lemma . If there existsη> such that
fA–yt–τ(t)≥A–yt–τ(t)η for t∈[,ω]and y∈K,
then
Qy ≥λlη
α –c
– c∞
–c
∞
ω
b(s)dsy, y∈K.
Proof By Lemma . and Lemma ., we have fory∈Kthat
Qy(t) =Pλb(t)fA–yt–τ(t)
≥Tλb(t)fA–yt–τ(t)
=λ
t+ω
t
G(t,s)b(s)f
A–ys–τ(s)ds
≥λlη
ω
b(s)A–ys–τ(s)ds
≥λlη
α –c
– c∞
–c
∞
ω
b(s)dsy.
Hence
Qy ≥λlη
α –c
– c∞
–c
∞
ω
b(s)dsy, y∈K.
Lemma . If there existsε> such that
fA–yt–τ(t)≤A–yt–τ(t)ε for t∈[,ω]and y∈K,
then
Qy ≤λε LM
ω b(s)ds
m– (M+m)c∞y, y∈K.
Proof By Lemma . and Lemma ., we have
Qy(t) ≤λ M( –c∞)
m– (M+m)c∞L
ω
b(s)fA–ys–τ(s)ds
≤λ M( –c∞)
m– (M+m)c∞Lε
ω
b(s)A–ys–τ(s)ds
≤λε LM
ω b(s)ds
m– (M+m)c∞y.
Define
F(r) =max
f(t) : ≤t≤ r
–c∞
f(r) =min
f(t) :
α –c
– c∞
–c
∞
r≤t≤ r
–c∞
.
Lemma . If y∈∂Kr,then
Qy ≥λlf(r)
ω
b(s)ds.
Proof By Lemma ., we obtain ( α –c –
c∞
–c∞)r≤(A
–y)(t–τ(t))≤ r
–c∞ fory∈∂Kr, which yieldsf((A–y)(t–τ(t)))≥f
(r). The lemma now follows analogous to the proof
of Lemma ..
Lemma . If y∈∂Kr,then
Qy ≤λLM( –c∞)F(r)
m– (M+m)c∞
ω
b(s)ds.
Proof By Lemma ., we can have ≤(A–y)(t–τ(t))≤ r
–c∞ fory∈∂Kr, which yields
f((A–y)(t–τ(t)))≤F(r). Similar to the proof of Lemma ., we get the conclusion.
We quote the fixed point theorem which our results will be based on.
Lemma .[] Let X be a Banach space and K be a cone in X.For r> ,define Kr= {u∈K:u<r}.Assume that T:Kr→K is completely continuous such that Tx=x for x∈∂Kr={u∈K:u=r}.
(i) IfTx ≥ xforx∈∂Kr,theni(T,Kr,K) = ; (ii) IfTx ≤ xforx∈∂Kr,theni(T,Kr,K) = .
Now we give our main results on positive periodic solutions for (.).
Theorem .
(a) Ifi= or,then(.)hasipositiveω-periodic solutions forλ>f ()l
ω
b(s)ds> ;
(b) Ifi∞= or,then(.)hasi∞positiveω-periodic solutions for
<λ<LM(–mc–(∞M)F+()m)cω∞
b(s)ds;
(c) Ifi∞= ori= ,then(.)has no positiveω-periodic solutions for sufficiently small
or sufficiently largeλ> ,respectively.
Proof (a) Chooser = . Take λ=f (r)l
ω
b(s)ds > , then for allλ>λ, we have from Lemma . that
Qy>y fory∈∂Kr. (.)
Case . Iff= , we can choose <¯r<r, so thatf(u)≤εufor ≤u≤ ¯r, where the constantε> satisfies
λε LM
ω b(s)ds
m– (M+m)c∞ < . (.)
Letr= ( –c∞)¯r, we havef((A–y)(t–τ(t)))≤ε(A–y)(t–τ(t)) fory∈Kr. By Lemma ., we have ≤(A–y)(t–τ(t))≤ y
have fory∈∂Kr that
Qy ≤λε LM
ω b(s)ds
m– (M+m)c∞y<y.
It follows from Lemma . and (.) that
i(Q,Kr,K) = , i(Q,Kr,K) = ,
thusi(Q,Kr\ ¯Kr,K) = – andQhas a fixed pointyinKr\ ¯Kr, which means that (A–y)(t) is a positiveω-positive solution of (.) forλ>λ.
Case . Iff∞= , there exists a constantH˜ > such thatf(u)≤εuforu≥ ˜H, where the constantε> satisfies
λε LM
ω b(s)ds
m– (M+m)c∞ < . (.)
Letr=max{r,
˜
H(–c )(–c∞)
α(–c∞)–c∞(–c)}, we havef((A
–y)(t–τ(t)))≤ε(A–y)(t–τ(t)) for y∈
Kr. By Lemma ., we have (A
–y)(t–τ(t))≥( α –c
– c∞
–c∞)y ≥ ˜Hfory∈∂Kr. Thus by Lemma . and (.), we have fory∈∂Kr that
Qy ≤λε LM
ω b(s)ds
m– (M+m)c∞y<y.
Recalling Lemma . and (.) that
i(Q,Kr,K) = , i(Q,Kr,K) = ,
theni(Q,Kr\ ¯Kr,K) = andQhas a fixed pointyinKr\ ¯Kr, which means that (A–y)(t) is a positiveω-positive solution of (.) forλ>λ.
Case . Iff=f∞= , from the above arguments, there exist <r<r<rsuch thatQ has a fixed pointy(t) inKr\ ¯Krand a fixed pointy(t) inKr\ ¯Kr. Consequently, (A–y)(t) and (A–y
)(t) are two positiveω-periodic solutions of (.) forλ>λ. (b) Letr= . Takeλ=LM(–mc∞–()MF+(rm)c∞
)ωb(s)ds > , then by Lemma . we know ifλ<λ then
Qy<y, y∈∂Kr. (.)
Case . Iff
=∞, we can choose <¯r<rso thatf(u)≥ηufor ≤u≤ ¯r, where the constantη> satisfies
λlη
α –c
– c∞
–c
∞
ω
b(s)ds> . (.)
Letr= ( –c∞)r¯, we havef((A–y)(t–τ(t)))≥η(A–y)(t–τ(t)) fory∈Kr. By Lemma ., we have ≤(A–y)(t–τ(t))≤ y
–c∞ ≤ ¯rfory∈∂Kr. Thus by Lemma . and (.),
Qy ≥λlη
α –c
– c∞
–c
∞
ω
It follows from Lemma . and (.) that
i(Q,Kr,K) = , i(Q,Kr,K) = ,
which impliesi(Q,Kr\ ¯Kr,K) = andQhas a fixed pointyinKr\ ¯Kr. Therefore (A–y)(t) is a positiveω-periodic solution of (.) for <λ<λ.
Case . Iff∞=∞, there exists a constantH˜ > such thatf(u)≥ηuforu≥ ˜H, where the constantη> satisfies
λlη
α –c
– c∞
–c
∞
ω
b(s)ds> . (.)
Letr=max{r,
˜
H(–c)(–c∞)
α(–c∞)–c∞(–c)}, we havef((A
–y)(t–τ(t)))≥η(A–y)(t–τ(t)) fory∈
Kr. By Lemma ., we have (A
–y)(t–τ(t))≥( α –c–
c∞
–c∞)y ≥ ˜Hfory∈∂Kr. Thus by Lemma . and (.), we have fory∈∂Kr that
Qy ≥λlη
α –c
– c∞
–c
∞
ω
b(s)dsy>y.
It follows from Lemma . and (.) that
i(Q,Kr,K) = , i(Q,Kr,K) = ,
i.e.,i(Q,Kr\ ¯Kr,K) = – andQhas a fixed pointyinKr\ ¯Kr. That means (A–y)(t) is a positiveω-periodic solution of (.) for <λ<λ.
Case . Iff
=f∞=∞, from the above arguments,Qhas a fixed pointy inKr\ ¯Kr and a fixed pointy inKr\ ¯Kr. Consequently, (A–y)(t) and (A–y)(t) are two positive ω-periodic solutions of (.) for <λ<λ.
(c) By Lemma ., ify∈K, then (A–y)(t–τ(t))≥( α –c–
c∞
–c∞)y> fort∈[,ω]. Case . Ifi= , we havef
> andf∞> . Letb=min{
f(u)
u ;u> }> , then we obtain
f(u)≥bu, u∈[, +∞).
Assume that y(t) is a positive ω-periodic solution of (.) for λ > λ, where λ = (–c)(–c∞)
lb[α(–c∞)–c∞(–c)]
ω
b(s)ds> . SinceQy(t) =y(t) fort∈[,ω], then by Lemma . if λ>λ,
we have
y=Qy ≥λlb
α –c
– c∞
–c
∞
ω
b(s)dsy>y,
which is a contradiction.
Case . Ifi∞= , we havef<∞andf∞<∞. Letb=max{f(uu) :u> }> , then we obtain
Assume that y(t) is a positive ω-periodic solution of (.) for <λ<λ, where λ=
m–(M+m)c∞
bLMωb(s)ds. SinceQy(t) =y(t) fort∈[,ω], it follows from Lemma . that
y=Qy ≤λb
LMωb(s)ds
m– (M+m)c∞y<y,
which is a contradiction.
Theorem .
(a) If there exists a constantb> such thatf(u)≥buforu∈[, +∞),then(.)has
no positiveω-periodic solution forλ> (–c)(–c∞)
lb[α(–c∞)–c∞(–c)]
ω
b(s)ds.
(b) If there exists a constantb> such thatf(u)≤buforu∈[, +∞),then(.)has
no positiveω-periodic solution for <λ<bm–(M+m)c∞ LMωb(s)ds.
Proof From the proof of (c) in Theorem ., we obtain this theorem immediately.
Theorem . Assume i=i=i∞=i∞= ,and that one of the following conditions holds:
() f≤f∞; () f
>f∞;
() f
≤f∞≤f≤f∞;
() f∞≤f
≤f∞≤f.
If
( –c
)( –c∞)
l[α( –c
∞) –c∞( –c)]
ω
b(s)dsmax{f,f,f∞,f∞}
<λ< m– (M+m)c∞
LMωb(s)dsmin{f
,f,f∞,f∞} ,
then(.)has one positiveω-periodic solution.
Proof Case . Iff≤f∞, then
( –c
)( –c∞)
l[α( –c
∞) –c∞( –c)]
ω b(s)ds
<λ<m– (M+m)c∞
LMωb(s)ds .
It is easy to see that there exists <ε<f∞such that
( –c
)( –c∞)
(f¯∞–ε)l[α( –c
∞) –c∞( –c)]ωb(s)ds<λ<
m– (M+m)c∞
(f
+ε)LM
ω b(s)ds
.
For the aboveε, we choose¯r> such thatf(u)≤(f+ε)ufor ≤u≤ ¯r. Letr= (–c∞)¯r, we havef((A–y)(t–τ(t)))≤(f
+ε)(A
–y)(t–τ(t)) fory∈Kr
. By Lemma ., we have ≤(A–y)(t–τ(t))≤ y
–c∞ ≤ ¯rforK∈∂Kr. Thus by Lemma . we have fory∈∂Krthat
Qy ≤λ(f
+ε)
LMωb(s)ds
On the other hand, there exists a constantH˜ > such thatf(u)≥(f∞–ε)uforu≥ ˜H. Letr=max{r,
˜
H(–c)(–c∞)
α(–c∞)–c∞(–c)}, we havef((A
–y)(t–τ(t)))≥(f
∞–ε)(A–y)(t–τ(t)) for
y∈Kr. By Lemma ., we have (A–y)(t–τ(t))≥( α –c–
c∞
–c∞)y ≥ ˜Hfory∈∂Kr. Thus by Lemma ., fory∈∂Kr,
Qy ≥λl(f∞–ε)
α –c–
c∞
–c
∞
ω
b(s)dsy>y.
It follows from Lemma . that
i(Q,Kr,K) = , i(Q,Kr,K) = ,
thusi(Q,Kr\ ¯Kr,K) = – andQhas a fixed pointyinKr\ ¯Kr. So (A
–y)(t) is a positive
ω-periodic solution of (.). Case . Iff
>f∞, in this case, we have ( –c)( –c
∞)
¯
fl[α( –c∞) –c∞( –c)]
ω b(s)ds
<λ< m– (M+m)c∞
f∞LMωb(s)ds.
It is easy to see that there exists <ε<fsuch that
( –c)( –c
∞)
(f¯–ε)l[α( –c∞) –c∞( –c)]
ω b(s)ds
<λ< m– (M+m)c∞ (f∞+ε)LMωb(s)ds.
For the aboveε, we chooser¯> such thatf(u)≥(f–ε)ufor ≤u≤ ¯r. Letr= (–c∞)¯r, we havef((A–y)(t–τ(t)))≥(f–ε)(A–y)(t–τ(t)) fory∈Kr. By Lemma ., we have ≤(A–y)(t–τ(t))≤ y
–c∞≤ ¯rfory∈∂Kr. Thus we have by Lemma . that fory∈∂Kr,
Qy ≥λl(f–ε)
α –c
– c∞
–c
∞
ω
b(s)dsy>y.
On the other hand, there exists a constantH˜ > such thatf(u)≤(f∞+ε)uforu≥ ˜H. Letr=max{r,
˜
H(–c)(–c∞)
α(–c∞)–c∞(–c)}, we havef((A
–y)(t–τ(t)))≤(f
∞+ε)(A–y)(t–τ(t)) for
y∈Kr. By Lemma . we have (A–y)(t–τ(t))≥( α –c –
c∞
–c∞)y ≥ ˜Hfory∈∂Kr. Thus by Lemma ., fory∈∂Kr,
Qy ≤λ(f∞+ε) LM
ω b(s)ds
m– (M+m)c∞y.
It follows from Lemma . that
i(Q,Kr,K) = , i(Q,Kr,K) = .
Thusi(Q,Kr\ ¯Kr,K) = – andQhas a fixed pointyinKr\ ¯Kr, proving that (A–y)(t) is a positiveω-periodic solution of (.).
Case .f
≤f∞≤f≤f∞. The proof is the same as in Case . Case .f∞≤f
3.2 Case II:c(t) > 0 andc∞< min{Mm+m,(LLM–l)M–lm–lm} Define
f(r) =min
f(t) : α –c
r≤t≤ r
–c∞
.
Similarly as in Section ., we get the following results.
Theorem .
(a) Ifi= or,then(.)hasipositiveω-periodic solutions forλ>f ()l
ω
b(s)ds> .
(b) Ifi∞= or,then(.)hasi∞positiveω-periodic solutions for
<λ<LM(–mc–(M+m)c∞ ∞)F()ωb(s)ds.
(c) Ifi∞= ori= ,then(.)has no positiveω-periodic solution for sufficiently small or largeλ> ,respectively.
Theorem .
(a) If there exists a constantb> such thatf(u)≥buforu∈[, +∞),then(.)has no
positiveω-periodic solution forλ> –c
lαbωb(s)ds.
(b) If there exists a constantb> such thatf(u)≤buforu∈[, +∞),then(.)has no
positiveω-periodic solution for <λ<bm–(M+m)c∞ LM
ω
b(s)ds.
Theorem . Assume that i=i=i∞=i∞= hold,and that one of the following con-ditions holds:
() f≤f∞; () f
>f∞;
() f
≤f∞≤f≤f∞;
() f∞≤f
≤f∞≤f.
If
–c
lαωb(s)dsmax{f
,f,f∞,f∞}
<λ< m– (M+m)c∞
LMωb(s)dsmin{f
,f,f∞,f∞} ,
then(.)has one positiveω-periodic solution.
Remark In a similar way, one can consider the third-order neutral functional differential equation (x(t) –c(t)x(t–δ(t)))+a(t)x(t) =λb(t)f(x(t–τ(t))).
We illustrate our results with an example.
Example . Consider the following third-order neutral differential equation:
u(t) +
– sint
ut–cost
–
– sin
t
u(t)
= –λ( –cost)ut–τ(t)au(t–τ(t)), (.)
whereλand <a< are two positive parameters,τ(t+π) =τ(t).
Comparing (.) to (.), we see thatδ(t) =cost,c(t) = – ( –sint),a(t) =( –
sin
t),b(t) = –cost,ω=π,f(u) =uau. Clearly,c
∞= ,c= ,M=
,m=
and we getρ=, noticing that
√
π <
π
holds.f= ,f∞= ,i= . By Theorem ., we easily get the following conclusion: equation (.) has two positiveπ-periodic solutions forλ>
πr, wherer=min{f(.),f( )}. In fact, by simple computations, we have
l=
ρ(exp(ρω) – )= ., L=
+ exp(–ρω ) ρ( –exp(–ρω
)) = .
k= ., k= ., α= .,
c∞= <min
k,
m M+m
= ., c∞=
< . =α,
and
f() =min
f(t) : .≈ α –c–
c∞
–c
∞ ≤
t≤
=min
f(.),f
=r,
f()l
π b(s)ds
= πr
.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
YX and ZBC worked together in the derivation of the mathematical results. Both authors read and approved the final manuscript.
Author details
1College of Computer Science and Technology, Henan Polytechnic University, Jiaozuo, 454000, China.2School of
Mathematics and Information Science, Henan Polytechnic University, Jiaozuo, 454000, China.
Acknowledgements
YX and ZBC would like to thank the referee for invaluable comments and insightful suggestions. This work was supported by NSFC project (Nos. 11326124, 11271339) and Education Department of Henan Province project (No. 14A110002).
Received: 16 June 2014 Accepted: 13 October 2014 Published: 24 October 2014
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doi:10.1186/1687-1847-2014-273