R E S E A R C H
Open Access
Singularity problems to fourth-order
Rayleigh equation with time-dependent
deviating argument
Yun Xin
1and Hongmin Liu
1**Correspondence:
[email protected] 1College of Computer Science and
Technology, Henan Polytechnic University, Jiaozuo, China
Abstract
The paper is devoted to an investigation of the existence of a positive periodic solution for a kind of fourth-order singular Rayleigh equation with time-dependent deviating argument, where the nonlinear termgsatisfies singularities of attractive and repulsive type at the origin and has time-dependent deviating argument. By applications of coincidence degree theory, we prove that this equation has at least one positive periodic solution. At last, two examples are given to show applications of theorems.
MSC: 34B16; 34C25
Keywords: Positive periodic solution; Fourth-order Rayleigh equation; Singularities of attractive and repulsive type; Time-dependent deviating argument
1 Introduction
In this paper, we consider the following fourth-orderp-Laplacian singular Rayleigh equa-tion with time-dependent deviating argument:
φp
x(t)+ft,x(t)+gt,xt–δ(t)=e(t), (1.1)
whereφp:R→Ris given byφp(s) =|s|p–2s, andp> 1 is a constant;f :R×R→Ris a continuousT-periodic function abouttandf(t, 0) = 0;e:R→Ris a continuous periodic functions withe(t+T)≡e(t);δ∈C1(R,R) is aT-periodic function;g:R×(0, +∞)→R is aL2-Carathéodory function defined onR2, andg(t,·) =g(t+T,·); it is said that Eq. (1.1) is singularity of attractive type (resp. repulsive type) ifg(t,x)→+∞(resp.g(t,x)→–∞) asx→0+fort∈R.
As is well known, the Rayleigh equation can be derived from many fields, such as en-gineering technique, physics and mechanics fields, and an important question is whether we have periodic solutions to the Rayleigh equation. Gaines and Mawhin [1] in 1977 intro-duced continuation theorems and applied this theorem to prove the existence of a periodic solutions for the Rayleigh equation ([1], p. 99)
x(t) +fx(t)+gt,x(t)= 0. (1.2)
Gaines and Mawhin’s work has attracted the attention of many scholars to the Rayleigh equation. More recently, the method of lower and upper solutions [2], Mawhin’s continu-ous theorem [3–5], topological degree and time maps [6,7], and the Manásevich–Mawhin continuation theorem [8–10] have been employed to investigate the existence of a periodic solution of Rayleigh equations.
On the other hand, the existence of a periodic solutions for a singular differential equa-tion was extensively studied (see [11–20]). Among these, there are some results on the Liénard equation with a singularity of repulsive type [14,15,19,20]. Zhang [19] in 1996 introduced the problem of periodic solutions of the following Liénard equation with a repulsive singularity:
x(t) +fx(t)x(t) +gt,x(t)= 0, (1.3)
where the nonlinear termghas a singularity of repulsive type at the origin and satisfies semilinear condition atx=∞, the author proved that Eq. (1.3) has at least one positive
T-periodic solution by using Mawhin’s continuous theorem. Wang [20] in 2014 improved Eq. (1.3) and proved the existence of a positive periodic solution of the following Liénard equation with a singularity of repulsive type and a deviating argument:
x(t) +fx(t)x(t) +gt,x(t–δ)= 0, (1.4) whereδis a constant andδ∈[0,T].
Nowadays, a good deal of work has been performed on the existence of a positive pe-riodic solution of the Rayleigh equation with a singularity [21–24]. Wang and Ma [24] in 2015 discussed a kind of singular Rayleigh equation as follows:
x(t) +ft,x(t)+gx(t)=p(t), (1.5)
where the nonlinear termghas a singularity of repulsive type at the origin and satisfies semilinear condition. The authors obtained the existence of a positive periodic solution for Eq. (1.5) by applications of the limit properties of time map. Lu and Chen [21] in 2017 studied the following Rayleigh equation with singularity of repulsive type:
x+ft,x(t)+ϕ(t)x(t) – 1
xγ(t)=p(t), (1.6)
whereγ ≥1. By using topological degree theory, the authors proved Eq. (1.6) has at least one positiveT-periodic solution.
Inspired by the above paper [1,19–21,24], in this paper, we further consider the ex-istence of a positiveT-periodic solution for Eq. (1.1) with singularities of attractive and repulsive type. By applications of coincidence degree theory, we obtain the following con-clusions.
Theorem 1.1 Assume that the following conditions hold: (H1) There exists a positive constantNsuch that
(H2) There exist two positive constantsD1,D2withD1<D2such thatg(t,x) –e(t) < –N for all(t,x)∈[0,T]×(0,D1),andg(t,x) –e(t) >Nfor all(t,x)∈[0,T]×(D2, +∞).
(H3) There exist positive constantsa,bsuch that
g(t,x)≤axp–1+b, for all(t,x)∈[0,T]×(0, +∞).
(H4) g(t,x) =g0(x) +g1(t,x),whereg0∈C((0,∞);R)andg1: [0,T]×[0,∞)→Ris an
L2-Carathéodory function. (H5) (Singularity of repulsive type)
1
0
g0(x)dx= –∞.
Then Eq. (1.1)has at least one positiveT-periodic solution if
0 < aT 1 –δ
T
πp
2p–1 < 1,
whereδ=:maxt∈[0,T]|δ(t)|,and1≤p<∞,πp= 2
(p–1)/p 0
ds (1–psp–1)1/p
=2pπsin(p–1)(π/1/p)p.
Remark1.2 The friction termf(x(t))x(t) in Eqs. (1.3) and (1.4) satisfy0Tf(x(t))x(t)dt= 0, which is crucial to estimate a priori boundsof a positiveT-periodic solution for these equations. However, in this paper, the friction termf(t,x) may not satisfy0Tf(t,x(t))dt= 0. For example, let
ft,x=cos2(2t) + 100sinx(t).
Obviously,0T(cos2(2t) + 100)sinx(t)= 0. This implies that our methods to estimatea priori boundsof a positiveT-periodic solution for Eq. (1.1) is more complex than Eqs. (1.3) and (1.4).
Remark1.3 From Eqs. (1.3), (1.4), (1.5), (1.6) in [19–21,24], the nonlinear termg has a deviating argument (i.e.,δis a positive constant and 0≤δ<T). But, in this paper, the nonlinear termgsatisfies a time-dependent deviating argument. For example, let
δ(t) = 1 4sin2t.
Obviously, the work on estimatinga lower boundsof a positiveT-periodic solution for Eq. (1.1) is more difficult than the corresponding work on Eqs. (1.3), (1.4), (1.5), (1.6). Therefore, we have to find another way to get over the difficulty.
Theorem 1.5 Assume that conditions(H1)and(H4)hold.Suppose the following condi-tions are satisfied:
(H6) There exist two positive constantsD3,D4withD3<D4such thatg(t,x) –e(t) >Nfor all(t,x)∈[0,T]×(0,D3),andg(t,x) –e(t) < –Nfor all(t,x)∈[0,T]×(D4, +∞).
(H7) There exist positive constantsa,bsuch that
–g(t,x)≤axp–1+b, for all(t,x)∈[0,T]×(0, +∞).
(H8) (Singularity of attractive type)
1
0
g0(x)dx= +∞.
Then Eq. (1.1)has at least one positiveT-periodic solution if
0 < a
T
1 –δ
T
πp
2p–1 < 1.
2 Preparation
Lemma 2.1(Gaines and Mawhin [1]) Suppose that X and Y are two Banach spaces,and L:D(L)⊂X→Y is a Fredholm operator with index zero.Let⊂X be an open bounded set and N:→Y be L-compact on.Assume that the following conditions hold:
(1) Lx =λNx,∀x∈∂∩D(L),λ∈(0, 1); (2) Nx∈/ImL,∀x∈∂∩KerL;
(3) deg{JQN,∩KerL, 0} = 0,whereJ:ImQ→KerLis an isomorphism. Then the equation Lx=Nx has a solution in∩D(L).
Lemma 2.2([25]) Ifω∈C1(R,R)andω(0) =ω(T) = 0,then
T
0
ω(t)pdt 1
p
≤
T
πp
T
0
ω(t)pdt 1
p
.
In order to apply topological degree theorem to study the existence of a positive periodic solution for Eq. (1.1), we rewrite Eq. (1.1) in the form:
⎧ ⎨ ⎩
x1(t) = (φq(x2(t)),
x2(t) = –f(t,x1(t)) –g(t,x1(t–δ(t))) +e(t), (2.1) where 1
p+ 1
q= 1. Clearly, ifx(t) = (x1(t),x2(t))is aT-periodic solution to Eq. (2.1), then
x1(t) must be aT-periodic solution to Eq. (1.1). Thus, the problem of finding aT-periodic solution for Eq. (1.1) reduces to finding one for Eq. (2.1).
Let
X:=x=x1(t),x2(t)∈C2R,R2:x(t+T) –x(t)≡0
with the normx:=max{x1,x2};
with the normx∞:=max{x,x}. Clearly,XandY are both Banach spaces.
Mean-while, define
L:D(L)⊂X→Y, by (Lx)(t) =
x1(t)
x2(t)
, (2.2)
where D(L) ={x= (x1,x2)∈C2(R,R2) :x(t+T) –x(t)≡0,t∈R}. Define a nonlinear operatorN:X→Yas follows:
(Nx)(t) =
φq(x2(t))
–f(t,x1(t)) –g(t,x1(t–δ(t))) +e(t)
. (2.3)
Then Eq. (2.1) can be converted to the abstract equationLx=Nx. From the definition ofL, one can easily see that
KerL∼=R2, ImL=
y∈Y:
T
0
y1(s)
y2(s)
ds=
0 0
.
SoLis a Fredholm operator with index zero. LetP:X→KerLandQ:Y→ImQ⊂R2be defined by
Px:=
(x1)(0)
x2(0)
; Qy:= 1
T
T
0
y1(s)
y2(s)
ds,
thenImP=KerL,KerQ=ImL. letKdenote the inverse ofL|Kerp∩D(L). It is easy to see that KerL=ImQ=R2and
[Ky](t) =col
T
0
G1(t,s)y1(s)ds,
T
0
G2(t,s)y2(s)ds
,
where
Gi(t,s) =
⎧ ⎨ ⎩
–s(T–t)
T , 0≤s≤t≤T, –t(T–s)
T , 0≤t<s≤T,
i= 1, 2. (2.4)
3 Main results
In the section, we first consider the existence of a positiveT-periodic solution for Eq. (1.1) with singularity of repulsive type.
Proof of Theorem1.1 Consider the operator equation
Lx=λNx, λ∈(0, 1),
whereLandNare defined by Eqs. (2.2) and (2.4). Set
1=
Ifx(t) = (x1(t),x2(t))∈1, then
⎧ ⎨ ⎩
x1(t) =λφq(x2(t)),
x2(t) = –λf(t,x1(t)) –λg(t,x1(t–δ(t))) +λe(t). (3.1)
Substitutingx2(t) = λp1–1(φp(x1)(t)) into the second equation of (3.1)
φp
x1(t)(t) +λpft,x1(t)+λpgt,x1
t–δ(t)=λpe(t). (3.2) Integrating both side of Eq. (3.2) over [0,T], we have
T
0
ft,x1(t)+gt,x1
t–δ(t)–e(t)dt= 0, (3.3)
since0T(φp(Ax1)(t))= 0. From Eq. (3.3) and condition (H1), we deduce
–NT≤
T
0
gt,x1
t–δ(t)–e(t)dt≤NT.
Then, by condition (H2), we know that there are two pointsξ,η∈(0,T) such that
x1
ξ–δ(ξ)≥D1, x1(η)≤D2. (3.4)
Then, from Eq. (3.4), we have
x(t) = 1 2
x(t) +x(t–T)
=1 2
x(η) +
t
η
x(s)ds+x(η) –
η
t–T
x(s)ds
=x(η) +1 2
t
t–T
x(s)ds
≤D2+ 1 2
T
0
x(t)dt. (3.5)
Multiplying both sides of Eq. (3.2) byx1(t) and integrating over the interval [0,T], we get
T
0
φp
x1(t)x1(t)dt+λp
T
0
ft,x1(t)x1(t)dt+λp
T
0
gt,x1
t–δ(t)x1(t)dt
=λp
T
0
e(t)x1(t)dt. (3.6)
Substituting0Tφp(x1(t))x1(t)dt=
T
0 |x1(t)|pdtinto Eq. (3.6), we arrive at
T
0
x
1(t) p
dt= –λp
T
0
ft,x1(t)x1(t)dt–λp
T
0
gt,x1
t–δ(t)x1(t)dt
+λp
T
0
Therefore, from condition (H1), we deduce
by using the Hölder inequality, we see that
Letω(t) =x1(t+η) –x1(η), herex1(η)≤D2, and thenω(0) =ω(T) = 0. By Lemma2.2 and the Minkowski inequality [26], it is clear that
T
Substituting Eqs. (3.11) into (3.10), we obtain
T
Furthermore, substituting Eqs. (3.11) and (3.12) into (3.9), we see that
Next, we introduce a classical inequality, there exists ak(p) > 0 which is dependent onp
only,
(1 +x)p≤1 + (1 +p)x, forx∈0,k(p). (3.14)
Then, we consider the following two cases:
Case 1.If D2T
From Eqs. (3.5) and (3.11), by using the Hölder inequality, we deduce
it is easy to see that there exists a positive constantM1such that
T
0
x1(t)pdt≤M1.
From Eq. (3.5) and Lemma2.2, we have
x1(t)≤D2+
On the other hand, it follows from Eq. (3.2) and condition (H4) that
Letξ ∈[0,T] be as in Eq. (3.4), for anyt∈[ξ,T]. Multiplying both sides of Eq. (3.21) by
Furthermore, by Eqs. (3.16), (3.17), (3.18) and (3.19), we have
λp
gular condition (H5), we know that there exists a positive constantM5such that
x1
t–δ(t)≥M5, ∀t∈[ξ,T]. (3.24)
From Eqs. (3.16), (3.17), (3.18), (3.19) and (3.24), we let
=x= (x1,x2):E1≤x1(t)≤E2,x1≤E3,x2 ≤E4andx2≤E5,∀t∈[0,T]
,
where 0 <E1<min(M5,D1),E2>max(M1,D2),E3>M2,E4>M4andE5>M3.2={x:x∈
∂∩KerL}then∀x∈∂∩KerL
QNx= 1
T
T
0
φq(x2(t)) –f(t,x1) –g(t,x1) +e(t)
dt.
IfQNx= 0, thenx2(t) = 0,x1=E2orE1. But ifx1(t) =E2, we know
0 =
T
0
g(t,E2) –e(t)
dt.
From condition (H2), we havex1(t)≤D2≤E2, which yields a contradiction. Similarly if
x1=E1. We also haveQNx = 0, i.e.,∀x∈∂∩KerL,x∈/ImL, so assumptions (1) and (2) of Lemma2.1are both satisfied. Define the isomorphismJ:ImQ→KerLas follows:
J(x1,x2)= (x2, –x1).
LetH(μ,x) = –μx+ (1 –μ)JQNx, (μ,x)∈[0, 1]×, then∀(μ,x)∈(0, 1)×(∂∩KerL),
H(μ,x) =
–μx1–1–Tμ
T
0 (g(t,x1) –e(t))dt –μx2– (1 –μ)φq(x2)
.
From condition (H2), we getxH(μ,x)= 0,∀(μ,x)∈(0, 1)×(∂∩KerL). Hence
deg{JQN,∩KerL, 0}=degH(0,x),∩KerL, 0 =degH(1,x),∩KerL, 0 =deg{I,∩KerL, 0} = 0.
So assumption (3) of Lemma2.1is satisfied. By applying Lemma2.1, we conclude that the equationLx=Nxhas a solutionx= (x1,x2)on¯ ∩D(L), i.e., Eq. (2.1) has aT-periodic
solutionx1(t).
Next, we investigate the existence of a positiveT-periodic solution for Eq. (1.1) with a singularity of attractive type.
Proof of Theorem1.5 We follow the same strategy and notation as in the proof of The-orem1.1. From Eq. (3.3) and condition (H6), we know that there are pointsτ,ν∈(0,T) such that
x1
Next, we consider0T|g(t,x1(t–δ(t)))|dt. From Eq. (3.11) and conditions (H1), (H7), we
The proof further is the same as Theorem1.1.
4 Examples
In this section, we present two examples to illustrate Theorems1.1and1.5.
Example4.1 Consider the fourth-order Rayleigh equation with singularity of repulsive type and time-dependent deviating argument:
Example4.2 Consider the fourth-order Rayleigh equation with singularity of attractive type:
x(4)(t) –cos2(2t) + 100sinx(t) – 1
3π(sin4t+ 5)x
t–1
8cos4t
+ 5
xκ(t–1 8cos4t)
=esin2(2t), (4.2)
whereκ≥1 andp= 2. It is clear that T = π
2,f(t,x) = –(cos2(2t) + 100)sinx,g(t,x) = – 1
3π(sin4t+ 5)x+
5 xκ,
δ=18cos4t,δ=12,π2=2π(p–1)
1
p
psin(π/p) = 2π(2–1)12
4×12 =π. TakeN= 6,a
= 2
π,b= 1. It is obvious
that (H1), (H4), (H6)–(H8) hold. Now we consider
aT
1 –δ
T
πp
2p–1 = 4
π × π
2 ×
1 2
3 =1
4< 1.
Therefore, applying Theorem1.5, we know that Eq. (4.2) has at least one positive π
2 -periodic solution.
5 Conclusions
In this paper, we introduce the existence of a positiveT-periodic solution for the fourth-orderp-Laplacian singular Rayleigh equation with time-dependent deviating argument. The nonlinear function has a time-dependent deviating argument. This implies that the work on estimatinglower boundsof periodic solutions for Eq. (1.1) is more difficult than the corresponding work on Eq. (1.4) in [21]. Secondly, attractive conditions (H6), (H7) and (H8) are in contradiction with the repulsive conditions (H2), (H3) and (H5), and the methods of singularity of repulsive type are no longer applicable to the proof of a periodic solution for Eq. (1.1) with singularity of attractive type. In this paper, by using coincidence degree theory and conditions (H1)–(H5), we prove the existence of a positiveT-periodic solution for Eq. (1.1) with a singularity of repulsive type; applying conditions (H1), (H4), (H6)–(H8), we find that Eq. (1.1) with singularity of attractive type has at least one positive
T-periodic solution.
Acknowledgements
YX and HML are grateful to anonymous referees for their constructive comments and suggestions, which have greatly improved this paper.
Funding
This work was supported by Education Department of Henan Province project (No. 16B110006) and Henan Polytechnic University Outstanding Youth Fund (J2016-03).
Abbreviations Not applicable.
Availability of data and materials Not applicable.
Ethics approval and consent to participate
YX and HML contributed to each part of this study equally and declare that they have no competing interests.
Competing interests
YX and HML declare that they have no competing interests.
Consent for publication
Authors’ contributions
YX and HML contributed equally and significantly in writing this article. Both authors read and approved the final manuscript.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Received: 20 June 2018 Accepted: 10 September 2018
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