# Periodic Solutions of a Type of Liénard Higher Order Delay Functional Differential Equation with Complex Deviating Argument

Authors: Haiqing Wang
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The author has studied the existence of periodic solutions of a type
of higher order delay functional differential equations with neutral type by using the theory of
coincidence degree, and some new sufficient conditions for the existence of periodic solutions have
been obtained.


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## Body

## 1. Introduction and Lemma

With the rapid development of modern science and technology, functional differential equation with time delay has been widely applied in many areas such as bioengineering, systems analysis, and dynamics. Functional differential equation with complex deviating argument is an important type of the above function. Because the property of the solution to this kind of equation is impossibly estimated, so the literature on the functional differential equation with complex argument is relatively rare [1]. In recent years, with the maturity of the theory of nonlinear functional analysis and algebraic topology, we have the powerful tools of the study on the functional differential equation with complex deviating argument, so it is possible to study the above equation. Furthermore, the study on the periodic solutions of functional differential equation is always one of the most important subject that people concerned for its widespread use. Many results of the study of Duffing-typed functional differential equation and Liénard-typed functional differential equation have been obtained, for example, the literatures [2–18]. Hitherto, the literature of the discussion of higher order functional differential equations has not been found a lot [19]. In this paper I have studied and derived some sufficient conditions that guarantee the existence of periodic solutions for a type of higher order functional differential equations with complex deviating argument as the following:
(*)∑i=1m‍aix(i)(t)+f(x(t))x˙(t)+β(t)g(x(x(t)))=p(t)(ai≠0),
and some new results have been obtained.In order to establish the existence ofT-periodic solutions of (*), we make some preparations.Definition 1.1.
LetX, Y are Banach spaces, and let Ω be an open and bounded subset in X, and let L:Dom(L)⊆X→Y be linear mapping; the mapping L will be called a Fredholm mapping of index zero if dimkerL=codimImL<+∝ and ImL is closed in Y.Definition 1.2.
LetP:X→ker(L), let Q:Y→Y/Im(L) be projectors, and let N:Ω-→Y be nonlinear mapping; the mapping N will be called L-compact on Ω- if QN:Ω-→Y/Im(L) and (L|ker(P))-1(I-Q)N:Ω-→X are compact.Lemma 1.3 (see [20]).
LetX, Y be Banach spaces; L:DL⊂X→Y is a Fredholm mapping of index zero P:X→X; Q:X→Y are continuous mapping projectors; Ω is an open bounded set in X; N:Ω-×[0,1]→Y is L-Compact on Ω-, furthermore suppose that:(a)
Lx≠λN(x,λ),forallx∈DL∩∂Ω,λ∈(0,1);(b)
QN(x,0)≠0,forallx∈ker(L)∩∂Ω;(c)
deg(QN(x,0),ker(L)∩Ω,0)≠0,then the equation Lx=N(x,1) has at least one solution on Ω-, where deg is Brouwer degree.


## 2. Main Results and Proof of Theorems

Theorem 2.1.
Suppose thatf, β, g, p are continuous for their variables, respectively, p(t+T)=p(t),  β(t+T)=β(t)>0, ∫0T‍p(t)dt=0, and furthermore suppose that(a)
∃A>0,forallx∈ℝ, when |x|>A, such that xg(x)>0;(b)
∃M>0,forallx∈ℝ, such that |g(x)|≤M;(c)
f1=supx∈ℝ|f(x)|<(am-k(Tm-1-Tm-2-⋯-T))/Tm-1,where k=max{|ai|}, i=1,2,…,m-1 and am>k(Tm-1+Tm-2+⋯+T), then (*) has at least one T-periodic solution.Proof of Theorem2.1.
In order to use continuation theorem to obtainT-periodic solution of (*), we firstly make some required preparations. Let
(2.1)X={x∈cm-1(ℝ,ℝ)∣x(t+T)=x(t)},Y={y∈C(ℝ,ℝ)∣y(t+T)=y(t)},
and the norm of X and Y is ∥x∥=max0≤i≤m-1{|x(i)|∞}, |x(i)|∞=maxt∈ℝ{|x(i)(t)|}, i=1,2,…,m-1, and ∥y∥=maxt∈ℝ{|y(t)|}, respectively; then the X and Y with this norm are Banach spaces.
Firstly, we study the priori bound ofT-periodic solution of following equation:
(2.2)∑i=1m‍aix(i)(t)+λf(x(t))x˙(t)+λβ(t)g(x(x(t)))=λ2p(t).
Suppose thatx=x(t)∈X is an arbitrary T-periodic solution of (2.2), put x(t) into, (2.2) and then integrate both sides of (2.2) on [0,T], so
(2.3)∫0T‍β(t)g(x(x(t)))dt=0.
For the continuity ofβ, g, x, there must exist a number t0∈[0,T] such that
(2.4)β(t0)g(x(x(t0)))=0,
that is,
(2.5)g(x(x(t0)))=0.
For the condition (a) of Theorem2.1, we have
(2.6)|x(x(t0))|≤A.
Let(2.7)x(t0)=nT-t1,n∈N,t1∈[0,T],
so
(2.8)|x(t1)|=|x(x(t0))|≤A.
In view of(2.9)∀t∈[0,T],x(t)=x(t1)+∫t1t‍x˙(s)ds,
we have
(2.10)|x(t)|=|x(t1)+∫t1t‍x˙(s)ds|≤A+∫t1t‍|x˙(s)|ds≤A+∫0T‍|x˙(t)|dt,
that is,
(2.11)|x(0)|∞=|x|∞≤A+∫0T‍|x˙(t)|dt.
Notingx(t)=x(t+T), so there must exist the number ξi∈[0,T] such that x(i)(ξi)=0, where i=1,2,3,…,m-1.
For allt∈[0,T],
(2.12)x(i)(t)=x(i)(ξi)+∫ξit‍x(i+1)(s)ds=∫ξit‍x(i+1)(s)ds,
we have(2.13)|x(i)(t)|=|∫ξit‍x(i+1)(s)ds|≤∫0T‍|x(i+1)(t)|dt≤T∫0T‍|x(i+2)(t)|dt≤T2·∫0T‍|x(i+3)(t)|dt≤⋯≤Tm-(i+1)∫0T‍|x(i+m-i)(t)|dt=Tm-(i+1)∫0T‍|x(m)(t)|dt,
that is,
(2.14)|x(i)|∞≤Tm-(i+1)∫0T‍|x(m)(t)|dt,i=1,2,…,m-1.
Combining (2.11), (2.14), we get
(2.15)|x(0)|∞=|x|∞≤A+Tm-1∫0T‍|x(m)(t)|dt.
By (2.2), we get
(2.16)∫0T‍|amxm(t)|dt≤∫0T‍|λf(x(t))x˙(t)|dt+∫0T‍|λβ(t)g(x(x(t)))|dt+∫0T‍|λ2p(t)|dt+∫0T‍|a1x˙(t)|dt+∫0T‍|a2x¨(t)|dt+⋯+∫0T‍|am-3x(m-3)(t)|dt+∫0T‍|am-2x(m-2)(t)|dt+⋯+∫0T‍|am-1x(m-1)(t)|dt,
where β1=maxt∈ℝβ(t),  p1=maxt∈ℝ{|p(t)|}, and k=max{|ai|},i=1,2,3,…,m-1.
Noting (2.14) and the conditions (b), (c) of Theorem 2.1, we have
(2.17)∫0T‍|amx(m)(t)|dt≤f1T·Tm-2∫0T‍|x(m)(t)|dt+β1TM+p1T+kT·Tm-(1+1)∫0T‍|x(m)(t)|dt+kT·Tm-(2+1)∫0T‍|x(m)(t)|dt+⋯+kT·Tm-(m-1+1)∫0T‍|x(m)(t)|dt,
so(2.18)am∫0T‍|x(m)(t)|dt≤(kTm-1+kTm-2+⋯+kT+f1Tm-1)∫0T‍|x(m)(t)|dt+β1TM+p1T,
where am>Tm-1∑i=1mfi+kTm-1+kTm-2+⋯+kT.
Let(2.19)β1TM+p1Ta0-(kTm-1+kTm-2+⋯+kT+f1Tm-1)≜A1,
that is,
(2.20)∫0T‍|x(m)(t)|dt≤A1.
Noting (2.14), (2.15), and (2.20), we have
(2.21)|x(0)|∞=|x|∞≤A+Tm-1A1≜ω0,|x(i)|∞≤Tm-(i+1)A1≜ωi,i=1,2,…,m-1.
Letω=max0≤i≤m{ωi+1}, and let Ω={x∣x∈X:∥x∥<ω}; then Ω is an open and bounded set in X.
Let(2.22)L:DL⊂X→Y:x→Lx=∑i=1m‍aix(i)(t),N:X×I→Y:x→N(x,λ)=-f(x(t))x˙(t)-β(t)g(x(x(t)))+λp(t);
then the corresponding equation of Lx=λN(x,λ) is (2.2).
Now, we define projection operators as follows;(2.23)P:X→ker(L):x→Px=1T∫0T‍x(t)dt,Q:Y→YIm(L):y→Qy=1T∫0T‍y(t)dt.
Obviously,P, Q are continuous operators, Im(P)=ℝ=ker(L), ker(Q)=Im(L), and it is easy to prove that L is a Fredholm mapping of index zero and is L-Compact on Ω-.
From the above discussion and the construction ofΩ, we know that for all x∈DL∩∂Ω, λ∈(0,1), Lx≠λN(x,λ), therefore the condition (a) of Lemma 1.3 holds.
For arbitraryx∈ker(L)∩∂Ω, ∥x∥=ω, by the definition of Q, N, we have
(2.24)QN(x,0)=1T∫0T‍[-f(x(t))x˙(t)-β(t)g(x(x(t)))]dt=-1T∫0T‍β(t)g(x(x(t)))dt,
so
(2.25)xQN(x,0)=-1Tx∫0T‍β(t)g(x(x(t)))dt=-1Txg(x)∫0T‍β(t)dt≠0,
therefore the condition (b) of Lemma 1.3 holds.
Making a transformation.(2.26)H(x,μ)=-μx+(1-μ)QN(x,0),∀x∈∂Ω∩ker(L),μ∈[0,1],
we have
(2.27)xH(x,μ)=-μx2+x(1-μ)QN(x,0)=-μx2-(1-μ)1Tg(x)x∫0T‍β(t)dt<0.
SoxH(x,μ)≠0, that is, H(x,μ)≠0 is a homotopy, deg(QN(x,0),ker(L)∩Ω,0) = deg(-I,ker(L)∩Ω,0)=deg(-I,ℝ∩Ω,0)≠0, where I is an identity mapping, and the condition (c) of Lemma 1.3 holds.
From above all, the requirements of Lemma1.3 are all satisfied, so (*) has at least one T-periodic solution under the condition of Theorem 2.1, so the proof of Theorem 2.1 is completed.Remark 2.2.
In Theorem2.1, if β(t)<0 and the condition (a) of Theorem 2.1 is when |x|>A, xg(x)<0, and the rest are unchangeable, then (*) has at least one T-periodic solution.If theg(x) is not a bounded function, we have the following theorem.Theorem 2.3.
Suppose thatf, β, g, p are continuous for their variables, respectively, p(t+T)=p(t), β(t+T)=β(t)>0, ∫0T‍p(t)dt=0, and furthermore suppose following:(a)
∃A>0,forallx∈ℝ, when |x|>A, such that xg(x)>0;(b)
∃M>0,forallx∈ℝ, such that |g(x)|≤M|x|+c;(c)
f1=supx∈ℝ|f(x)|<(am-kTm-1-kTm-2-⋯-kT-β1Tm)/Tm-1,where k=max{|ai|},i=1,2,…,m-1, and am>kTm-1+kTm-2+⋯+kT+β1Tm, then (*) has at least one T-periodic solution.Proof of Theorem2.3.
Banach spacesX,  Y and the mappings L, P, Q, and N are the same to Theorem 2.1, and their property are equal to Theorem 2.1, then the corresponding equation of Lx=λN(x,λ) is
(2.28)∑i=1m‍aix(i)(t)+λf(x(t))x˙(t)+λβ(t)g(x(x(t)))=λ2p(t).
It is similar to Theorem2.1, there must exist a number t1∈[0,T], such that
(2.29)|x(t1)|≤A,
and it is easy to obtain
(2.30)|x(i)|∞≤Tm-(i+1)∫0T‍|x(m)(t)|dt,i=1,2,…,m-1,|x(0)|∞=|x|∞≤A+Tm-1∫0T‍|x(m)(t)|dt.
Noting (2.28), (2.30) and the conditions (b), (c) of Theorem 2.3, we have
(2.31)∫0T‍|amx(m)|dt≤∫0T‍|λf(x(t))x˙(t)|dt+∫0T‍|λβ(t)g(x(x(t)))|dt+∫0T‍|λ2p(t)|dt+kT·Tm-(1+1)∫0T‍|xm(t)|dt+kT·Tm-(2+1)∫0T‍|xm(t)|dt+⋯+kT·Tm-(m-1+1)∫0T‍|xm(t)|dt≤f1T·Tm-2∫0T‍|xm(t)|dt+β1T[M|x(x(t))|+c]+p1T+kT·Tm-(1+1)∫0T‍|xm(t)|dt+kT·Tm-(2+1)∫0T‍|xm(t)|dt+⋯+kT·Tm-(m-1+1)∫0T‍|xm(t)|dt.
So(2.32)am∫0T‍|xm(t)|dt≤(kTm-1+kTm-2+⋯+kT+f1Tm-1)∫0T‍|xm(t)|dt+β1Tc+β1TM|x|∝+p1T≤(kTm-1+kTm-2+⋯+kT+f1Tm-1)∫0T‍|xm(t)|dt+β1Tc+β1TM(A+Tm-1∫0T‍|x(m)(t)|dt)+p1T≤(kTm-1+kTm-2+⋯+kT+f1Tm-1+β1Tm)∫0T‍|xm(t)|dt+β1Tc+β1TMA+p1T,
where k=max{|ai|},i=1,2,3,…,m-1, and am>kTm-1+kTm-2+⋯+kT+f1Tm-1+β1Tm.
Let(2.33)β1Tc+β1TMA+p1Tam-(kTm-1+kTm-2+⋯+kT+f1Tm-1+β1Tm)≜A1,
that is,
(2.34)∫0T‍|xm(t)|dt≤A1.
Noting (2.30) and (2.34), we have
(2.35)|x(0)|∞=|x|∞≤A+Tm-1A1≜ω0,|x(i)|∞≤Tm-(i+1)A1≜ωi,i=1,2,…,m-1.
Letω=max0≤i≤m{ωi+1}, and we take Ω={x∣x∈X:∥x∥<ω}; then Ω is an open and bounded set in X.
Similarly to Theorem2.1, we prove easily that L is a Fredholm mapping of index zero and N is L-compact on Ω- and the conditions (a), (b), and (c) of Lemma 1.3 hold.
From above all, the requirements of Lemma1.3 are all satisfied, so (*) has at least one T-periodic solution under the condition of Theorem 2.3, so far the proof of Theorem 2.3 is completed.Remark 2.4.
In Theorem2.3, if β(t)<0 and the condition (a) of Theorem 2.3 is when |x|>A, xg(x)<0, and the rest are unchangeable, then (*) has at least one T-periodic solution.If the∫0T‍p(t)dt≠0, we have the following theorem.Theorem 2.5.
Suppose thatf, β, g, p are continuous for their variables, respectively, β(t+T)=β(t)>0, and meet the condition (a) of Theorem 2.1 and furthermore suppose as follows:(a)
lim|x|→+∝|g(x)|=+∝;(b)
∃a,b,c>0, such that |g(x)|≤ag(x)+b|x|+c;(c)
f1=supx∈ℝ|f(x)|(am-kTm-1-kTm-2-⋯-kT-f1Tm-1-bβ1Tm)/Tm-1,where k=max{|ai|},i=1,2,…,m-1, and am>kTm-1+kTm-2+⋯+kT+f1Tm-1+bβ1Tm, then (*) has at least one T-periodic solution.Proof of Theorem2.5.
Banach spacesX,  Y and the mappings L, P, Q, and N are the same to Theorem 2.1, and their property are equal to Theorem 2.1, then the corresponding equation of Lx=λN(x,λ) is
(2.36)∑i=1m‍aix(i)(t)+λf(x(t))x˙(t)+λβ(t)g(x(x(t)))=λ2p(t).
Suppose thatx=x(t)∈X is an arbitrary T-periodic solution of (2.36), put x(t) into (2.36), and then integrate both sides of (2.36) on [0,T], so
(2.37)∫0T‍β(t)g(x(x(t)))dt=∫0T‍λp(t)dt.
For the continuity ofβ, g, x, there must exist a number t1∈[0,T], such that
(2.38)g(x(x(t1)))=λ∫0T‍p(t)dt∫0T‍β(t)dt.
Combing the condition (a) of Theorem2.5, there must exist A1>0, such that
(2.39)|x(x(t1))|≤A1.
Similarly to Theorem2.1, we have
(2.40)|x(i)|∞≤Tm-(i+1)∫0T‍|x(m)(t)|dt,i=1,2,…,m-1,(2.41)|x(0)|∞=|x|∞≤A1+Tm-1∫0T‍|x(m)(t)|dt.
By (2.36), (2.37), (2.39), and (2.41) and the conditions (b), (c) of Theorem 2.5, we have
(2.42)∫0T‍|amx(m)|dt≤∫0T‍|λf(x(t))x˙(t)|dt+∫0T‍|λβ(t)g(x(x(t)))|dt+∫0T‍|λ2p(t)|dt+kT·Tm-(1+1)∫0T‍|xm(t)|dt+kT·Tm-(2+1)∫0T‍|xm(t)|dt+⋯+kt·Tm-(m-1+1)∫0T‍|xm(t)|dt≤f1T|x˙|∝+(kTm-1+kTm-2+⋯+am-1T)∫0T‍|xm(t)|dt+∫0T‍aβ(t)g(x(x(t)))dt+∫0T‍bβ(t)[|x(x(t))|+c]dt+p1T≤f1TTm-2∫0T‍|x(m)(t)|dt+(kTm-1+kTm-2+⋯+kT)∫0T‍|xm(t)|dt+aβ1Tp1+bβ1T|x|∝+bβ1Tc+p1T≤f1Tm-1∫0T‍|x(m)(t)|dt+(kTm-1+kTm-2+⋯+kT)∫0T‍|xm(t)|dt+bβ1T(A1+Tm-1∫0T‍|x(m)(t)|dt)+aβ1Tp1+bβ1Tc+p1T≤(kTm-1+kTm-2+⋯+kT+f1Tm-1+bβ1Tm)∫0T‍|xm(t)|dt+bβ1TA1+aβ1Tp1+bβ1Tc+p1T.
So(2.43)[am-(kTm-1+kTm-2+⋯+kT+f1Tm-1+bβ1Tm)]∫0T‍|xm(t)|dt≤bβ1TA1+aβ1Tp1+bβ1Tc+p1T.
Let(2.44)bβ1TA1+aβ1Tp1+bβ1Tc+p1Tam-(kTm-1+kTm-2+⋯+kT+f1Tm-1+bβ1Tm)≜A2,
that is,
(2.45)∫0T‍|xm(t)|dt≤A2.
Noting (2.40), (2.41), and (2.45), we have
(2.46)|x(0)|∞=|x|∞≤A+Tm-1A2≜ω0,|x(i)|∞≤Tm-(i+1)A2≜ωi,i=1,2,…,m-1.
For condition (a), there existM0>0 and A0> 0, such that |x|>M0,|g(x)|>A0; let ω=max0≤i≤m{ωi+1,M0}, and we take Ω={x∣x∈X:∥x∥<ω}; then Ω is an open and bounded set in X.
Similarly to Theorem2.1, we prove easily that L is a Fredholm mapping of index zero and N is L-compact on Ω- and the conditions (a), (b), and (c) of Lemma 1.3 hold.
From above all, the requirements of Lemma1.3 are all satisfied, so (*) has at least one T-periodic solution under the condition of Theorem 2.5, so the proof of Theorem 2.5 is completed.Remark 2.6.
In Theorem2.5, if β(t)<0 and the condition (a) of Theorem 2.1 is when |x|>A, xg(x)<0, and the rest are unchangeable, then (*) has at least one T-periodic solution.



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