# Antiperiodic Solutions for a Generalized High-Order(p,q)-Laplacian Neutral Differential System with Delays in  the Critical Case

Authors: Yongzhi Liao; Tianwei Zhang; Yongkun Li
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

By applying the method of coincidence degree, some criteria are established for the existence of antiperiodic solutions for a generalized high-order(p,q)-Laplacian neutral differential system with delays (φp((x(t)-cx(t-τ))(k)))(m-k)=F(t,xθ0(t),xθ1(t)′,…,xθk(t)(k),yϑ0(t),yϑ1(t)′,…,yϑl(t)(l)), (φq((y(t)-dy(t-σ))(l)))(n-l)=G(t,yμ0(t),yμ1(t)′,…,yμl(t)(l),xν0(t),xν1(t)′,…,xνk(t)(k)) in the critical case |c|=|d|=1. The results of this paper are completely new. Finally, an example is employed to illustrate our results.


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## Body

## 1. Introduction

During the last twenty years, there have been quite a few results on the existence of periodic solutions for delay differential equations and neutral functional differential equations. We can see [1–7]. For example, the authors of [8–11] investigated the existence of periodic solutions for the following types of neutral functional differential equations:
(1)(x(t)-cx(t-τ))′=g1(t,x(t))+g2(t,x(t-δ))+e(t),(x(t)-cx(t-τ))′′+g(x(t-δ))=e(t),(φp(x(t)-cx(t-τ))′)′+g(t,x(t-η(t)))=e(t),(φp(x′(t)-cx′(t-τ)))′=f(x(t))x′(t)+ξ(t)g(x(t-η(t)))+e(t),
respectively. But the condition of constant |c|≠1 is required. For example, under the assumption |c|≠1, they obtain that A:C2π:={x:x∈C(ℝ,ℝ),x(t+2π)≡x(t)}→C2π, (Ax)(t)=x(t)-cx(t-τ) has a unique inverse A-1:C2π→C2π defined by
(2)(A-1f)(t)={∑j≥0‍cjf(t-jτ),|c|<1,-∑j≥1‍c-jf(t+jτ),|c|>1,
and then
(3)∫02π‍|(A-1f)(s)|ds≤1|1-|c||∫02π‍|f(s)|ds,∀f∈C2π,
which was crucial to obtaining estimation of a priori bounds of periodic solutions in the noncritical case |c|≠1.Under the critical case|c|=1, the authors of [12–15] studied a first-order neutral differential equation
(4)(x(t)-cx(t-τ))′=g(t,x(t-δ(t)))+e(t),
a Duffing differential equation of neutral type
(5)(x(t)-cx(t-τ))′′=g(t,x(t-δ(t)))+e(t),
a Rayleigh differential equation of neutral type
(6)(x(t)-cx(t-τ))′′=f(x(t))x′(t)+g(t,x(t-δ(t)))+e(t),
and a p-Laplacian differential equation of neutral type
(7)(φp(x(t)-cx(t-τ))′)′=f(x(t))x′(t)+g(t,x(t-δ(t)))+e(t),
respectively.In the past thirty years, there has been a great deal of work on the problem of the periodic solutions of high-order nonlinear differential equations, especially for the third-order and fourth-order differential equations which have been used to describe nonlinear oscillations [16–20], and fluid mechanical and nonlinear elastic mechanical phenomena [21–27]. In [28], Jin and Lu discussed the existence of periodic solutions of third-order p-Laplacian equation with a deviating argument
(8)(φp(x′′(t)))′+f(t,x′(t),x′′(t))+g(t,x(t-η(t)))=e(t).Before continuing, by applying Mawhin’s continuation theorem of coincidence degree theory, the authors of [29] studied the existence of periodic solutions for a fourth-order p-Laplacian equation with a deviating argument
(9)(φp(x′′(t)))′′+f(x′′(t))+g(x(t-η(t)))=e(t).Arising from problems in applied sciences, it is well known that the existence of antiperiodic solutions plays a key role in characterizing the behavior of nonlinear differential equations as a special periodic solution and has been extensively studied by many authors during the past twenty years; see [30–44] and references therein. For example, antiperiodic trigonometric polynomials are important in the study of interpolation problems [45, 46], and antiperiodic wavelets are discussed in [47].However, to the best of our knowledge, due to the neutral term andp-Laplace operator term, the existence of antiperiodic solutions for (4)–(7) is very difficult to obtain by applying traditional researching methods. Therefore, to date, there are few papers to investigate the existence of antiperiodic solutions for (4)–(7).Motivated by above statements, in this paper, we will apply the method of coincidence degree to study the existence of antiperiodic solutions for a generalized high-order(p,q)-Laplacian neutral differential system with delays in the critical case
(10)(φp((x(t)-cx(t-τ))(k)))(m-k)=F(t,xθ0(t),xθ1(t)′,…,xθk(t)(k),yϑ0(t),yϑ1(t)′,…,yϑl(t)(l)),(φq((y(t)-dy(t-σ))(l)))(n-l)=G(t,yμ0(t),yμ1(t)′,…,yμl(t)(l),xν0(t),xν1(t)′,…,xνk(t)(k)),
where |c|=|d|=1, φp(s)=|s|p-2s, φq(s)=|s|q-2s, and s∈ℝ, p,q≥2; θi(t)(0≤i≤k), ϑi(t)(0≤i≤l), μi(t)(0≤i≤l), and νi(t)(0≤i≤k)∈C(ℝ,ℝ) are π-periodic functions; for any ρ(t)∈C(ℝ,ℝ), xρ(t) is defined by xρ(t)=x(t-ρ(t)); F,G∈C(ℝk+l+3,ℝ) are 2π-periodic in their first arguments; τ, σ are constants; m, n, k, and l are nonnegative integers, k<m, l<n.Throughout this paper, we will denote byℕ the set of nonnegative integers and by ℕ1 the set of odd positive integers.Letp=q, k=l, m=n, x=y, c=d, τ=σ, and F=G in system (10), then system (10) is reformulated as
(11)(φp((x(t)-cx(t-τ))(k)))(m-k)=F(t,xθ0(t),xθ1(t)′,…,xθk(t)(k),xϑ0(t),xϑ1(t)′,…,xϑl(t)(l)).
Furthermore, one can easily obtain the following.(a)
Ifp=q=2,x=y,|c|=|d|=1,k=l=0,m=n=1,θ0(t)=δ(t),
(12)F=F(t,xθ0(t))=g(t,x(t-δ(t)))+e(t),then system (10) reduces to (4).(b)
Ifp=q=2,x=y,|c|=|d|=1,k=l=0,m=n=2,θ0(t)=δ(t),
(13)F=F(t,xθ0(t))=g(t,x(t-δ(t)))+e(t),then system (10) reduces to (5).(c)
Ifp=q=2,x=y,|c|=|d|=1,k=l=0,m=n=2,θ0(t)=θ1(t)≡0,ϑ0(t)=δ(t),
(14)F=F(t,xθ0(t),xθ1(t)′,xϑ0(t))=f(x(t))x′(t)+g(t,x(t-δ(t)))+e(t),then system (10) reduces to (6).(d)
Ifp=q,x=y,|c|=|d|=1,k=l=1,m=n=2,θ0(t)=θ1(t)≡0,ϑ0(t)=δ(t),
(15)F=F(t,xθ0(t),xθ1(t)′,xϑ0(t))=f(x(t))x′(t)+g(t,x(t-δ(t)))+e(t),then system (10) reduces to (7).The main purpose of this paper is to establish sufficient conditions for the existence ofπ-antiperiodic solutions to system (10) by using the method of coincidence degree.The organization of this paper is as follows. In Section2, we make some preparations. In Section 3, by using the method of coincidence degree, we establish sufficient conditions for the existence of π-antiperiodic solutions to system (10). An illustrative example is given in Section 4.


## 2. Preliminaries

The following continuation theorem of coincidence degree is crucial in the arguments of our main results.Lemma 1 (see [48]).
Let𝕏,𝕐 be two Banach spaces; let Ω⊂𝕏 be open bounded and symmetric with 0∈Ω. Suppose that L:D(L)⊂𝕏→𝕐 is a linear Fredholm operator of index zero with D(L)∩Ω-≠∅ and N:Ω-→𝕐 is L-compact. Further, one also assumes that
(H)Lx-Nx≠λ(-Lx-N(-x)), for all x∈D(L)∩∂Ω,λ∈(0,1].
Then equationLx=Nx has at least one solution on D(L)∩Ω-.Definition 2.
Letu(t):ℝ→ℝ be continuous. u(t) is said to be T/2-antiperiodic on ℝ, if
(16)u(t+T)=u(t),u(t+T2)=-u(t),∀t∈ℝ.We will adopt the following notations:(17)C2πk:={u∈C(ℝ,ℝ):uis2π-periodic},k∈ℕ,|u|∞=maxt∈[0,2π]|u(t)|,
where u is a 2π-periodic function.For the sake of convenience, we introduce the following assumptions.(H1)
There exist nonnegative constantsα1,α2,…,αk+l+2,β1,β2,…,βk+l+2, such that
(18)|F(t,s1,s2,…,sk+l+2)-F(t,z1,z2,…,zk+l+2)|≤∑i=1k+l+2‍αi|si-zi|,|G(t,s1,s2,…,sk+l+2)-G(t,z1,z2,…,zk+l+2)|≤∑i=1k+l+2‍βi|si-zi|for any(t,s1,s2,…,sk+l+2), (t,z1,z2,…,zk+l+2)∈ℝk+l+3.(H2)
For all(t,s1,s2,…,sk+l+2)∈ℝk+l+3,
(19)F(t+π,-s1,-s2,…,-sk+l+2)=-F(t,s1,s2,…,sk+l+2),G(t+π,-s1,-s2,…,-sk+l+2)=-G(t,s1,s2,…,sk+l+2).In order to apply Lemma1 to study the existence of antiperiodic solutions for system (10), we set
(20)𝕏={x=(x1(t),x2(t),y1(t),y2(t))T∈C2πk×C2πm-k-1×C2πl×C2πn-l-1:x(t+π)=-x(t)(x1(t),x2(t),y1(t),y2(t))T},𝕐={x=(x1(t),x2(t),y1(t),y2(t))T∈C2π0×C2π0×C2π0×C2π0:x(t+π)=-x(t)(x1(t),x2(t),y1(t),y2(t))T}
are two Banach spaces with the norms
(21)∥x∥𝕏=∑i=0k‍|x1(i)|∞+∑i=0m-k-1‍|x2(i)|∞+∑i=0l‍|y1(i)|∞+∑i=0n-l-1‍|y2(i)|∞,∥x∥𝕐=∑j=12‍(|xj|∞+|yj|∞),
respectively. Define
(22)𝔻={x=(x1(t),x2(t),y1(t),y2(t))T∈C2πk×C2πm-k×C2πl×C2πn-l:x(t+π)=-x(t)(x1(t),x2(t),y1(t),y2(t))T}
and two difference operators A and B as follows:
(23)A:𝕐⟶𝕐,(Ax)(t)=x(t)-cx(t-τ),B:𝕐⟶𝕐,(By)(t)=y(t)-dy(t-σ).
Then system (10) reduces to
(24)(Ax1)(k)(t)=φp′(x2(t)),x2(m-k)(t)=F(t,x1θ0(t),x1θ1(t)′,…,x1θk(t)(k),y1ϑ0(t),y1ϑ1(t)′,…,yϑ1l(t)(l)),(By1)(l)(t)=φq′(y2(t)),y2(n-l)(t)=G(t,y1μ0(t),y1μ1(t)′,…,y1μl(t)(l),x1ν0(t),x1ν1(t)′,…,x1νk(t)(k)),
where 1/p+1/p′=1,1/q+1/q′=1,1<p′,q′≤2. Obviously, the existence of antiperiodic solutions to system (10) is equivalent to that of antiperiodic solutions to system (24). Thus, the problem of finding a π-antiperiodic solution for system (10) reduces to finding one for system (24).Define a linear operatorL:D(L)≡𝔻⊂𝕏→𝕐 by setting
(25)Lx=L(x1x2y1y2)=((Ax1)(k)x2(m-k)(By1)(l)y2(n-l)),∀x∈D(L)
and N:𝕏→𝕐 by setting
(26)Nx=N(x1x2y1y2)=(φp′(x2(t))F(t,x1θ0(t),x1θ1(t)′,…,x1θk(t)(k),y1ϑ0(t),y1ϑ1(t)′,…,y1ϑl(t)(l))φq′(y2(t))G(t,y1μ0(t),y1μ1(t)′,…,y1μl(t)(l),x1ν0(t),x1ν1(t)′,…,x1νk(t)(k))).
It is easy to see that
(27)KerL={0},ImL={x∈𝕐:∫02π‍(x1(s)x2(s)y1(s)y2(s))ds=(0000)}≡𝕐.
Thus, dimKerL=0=codimImL, and L is a linear Fredholm operator of index zero.Define the continuous projectorP:𝕏→KerL and the averaging projector Q:𝕐→𝕐 by
(28)P(x1x2y1y2)=Q(x1x2y1y2)=12π∫02π‍(x1(s)x2(s)y1(s)y2(s))ds≡(0000).
Hence ImP=KerL and Ker Q=ImL. Denoting by LP-1:ImL→D(L)∩KerP the inverse of L|D(L)∩KerP, we have
(29)(LP-1(x1x2y1y2))(t)=((A-1Kx1)(t)(Kx2)(t)(B-1Ky1)(t)(Ky2)(t)),
where
(30)(Kx1)(t)=∑i=0k-1‍1i!(Ah1)(i)(0)ti+1(k-1)!∫0t‍(t-s)k-1x1(s)ds,(Kx2)(t)=∑i=0m-k-1‍1i!h2(i)(0)ti+1(m-k-1)!∫0t‍(t-s)m-k-1x2(s)ds,(Ky1)(t)=∑i=0l-1‍1i!(Bh3)(i)(0)ti+1(l-1)!∫0t‍(t-s)l-1y1(s)ds,(Ky2)(t)=∑i=0n-l-1‍1i!h4(i)(0)ti+1(n-l-1)!∫0t‍(t-s)n-l-1y2(s)ds,
in which (Ah1)(i)(0)(i=0,1,…,k-1) are decided by E1Z1=B1, where
(31)E1=(200⋯00c120⋯00c2c12⋯00⋮⋮⋮⋮⋮ck-2ck-3ck-4⋯20ck-1ck-2ck-3⋯c12)k×k,Z1=((Ah1)(k-1)(0)(Ah1)(k-2)(0)(Ah1)(k-3)(0)⋮(Ah1)′(0)(Ah1)(0))k×1,B1=(b11,b12,…,b1k)T,b1i=-(1/i!)∫02π‍(π-s)ix1(s)ds,cj=(π)j/j!, and j=1,2,…,k-1;h2(i)(0)(i=0,1,…,m-k-1) are decided by E2Z2=B2, where
(32)E2=(20⋯00c12⋯00c2c1⋯00⋮⋮⋮⋮cm-k-2cm-k-3⋯20cm-k-1cm-k-2⋯c12)(m-k)×(m-k),Z2=(h2(m-k-1)(0)h2(m-k-2)(0)h2(m-k-3)(0)⋮h2′(0)h2(0))(m-k)×1,B2=(b21,b22,…,b2(m-k))T,b2i=-(1/i!)∫02π‍(π-s)ix2(s)ds,cj=(π)j/j!, and j=1,2,…,m-k-1;(Bh3)(i)(0)(i=0,1,…,l-1) are decided by E3Z3=B3, where
(33)E3=(200⋯00c120⋯00c2c12⋯00⋮⋮⋮⋮⋮cl-2cl-3cl-4⋯20cl-1cl-2cl-3⋯c12)l×l,Z3=((Bh3)(l-1)(0)(Bh3)(l-2)(0)(Bh3)(l-3)(0)⋮(Bh3)′(0)(Bh3)(0))l×1,B3=(b31,b32,…,b3l)T,b3i=-(1/i!)∫02π‍(π-s)iy1(s)ds,cj=(π)j/j!, and j=1,2,…,l-1;h4(i)(0)(i=0,1,…,n-l-1) are decided by E4Z4=B4, where
(34)E4=(20⋯00c12⋯00c2c1⋯00⋮⋮⋮⋮cn-l-2cn-l-3⋯20cn-l-1cn-l-2⋯c12)(n-l)×(n-l),Z4=(h4(n-l-1)(0)h4(n-l-2)(0)h4(n-l-3)(0)⋮h4′(0)h4(0))(n-l)×1,B4=(b41,b42,…,b4(n-l))T,b4i=-(1/i!)∫02π‍(π-s)iy2(s)ds,cj=(π)j/j!, and j=1,2,…,n-l-1.Clearly,QN and LP-1(I-Q)N are continuous. Using the Arzela-Ascoli theorem, it is not difficult to show that QN(Ω-),LP-1(I-Q)N(Ω-) are relatively compact for any open bounded set Ω⊂𝕏. Therefore, N is L-compact on Ω- for any open bounded set Ω⊂𝕏.Lemma 3 (see [12]).
Let|τ|=(b/a)π, where a and b are coprime positive integers. Then (1)
ifc=-1, b is odd and a is even, then
(35)ω1:=infk∈ℕ1|1-ce-kiτ|=infk∈ℕ1[2(1+coskτ)]1/2>0;(2)
ifc=1 and b is odd, then
(36)ω2:=infk∈ℕ1|1-ce-kiτ|=infk∈ℕ1[2(1-coskτ)]1/2>0;(3)
ifc=1 and a=b=1, then
(37)ω3:=infk∈ℕ1|1-ce-kiτ|=infk∈ℕ1[2(1-coskτ)]1/2=2>0.Lemma 4 (see [13]).
SupposeA:𝕐→𝕐, (Ax)(t)=x(t)-cx(t-τ) for all t∈[0,2π]. Then the following propositions are true. (1)
Ifc=-1, |τ|=(b/a)π, where a and b are coprime positive integers with b odd and a even, then A has a unique inverse A-1:𝕐→𝕐 satisfying ∥A-1∥≤(1/ω1).(2)
Ifc=1, |τ|=(b/a)π, where a and b are coprime positive integers with b odd, then A has a unique inverse A-1:𝕐→𝕐 satisfying ∥A-1∥≤(1/ω2).(3)
Ifc=1, |τ|=π, then A has a unique inverse A-1:𝕐→𝕐 satisfying ∥A-1∥≤(1/ω3).Lemma 5 (see [12]).
Let|σ|=(ν/μ)π, where μ and ν are coprime positive integers. Then (1)
ifd=-1, ν is odd and μ is even, then
(38)ω1:=infk∈ℕ1|1-de-kiτ|=infk∈ℕ1[2(1+coskτ)]1/2>0;(2)
ifd=1 and ν is odd, then
(39)ω2:=infk∈ℕ1|1-de-kiτ|=infk∈ℕ1[2(1-coskτ)]1/2>0;(3)
ifd=1 and μ=ν=1, then
(40)ω3:=infk∈ℕ1|1-de-kiτ|=infk∈ℕ1[2(1-coskτ)]1/2=2>0.Lemma 6 (see [13]).
SupposeB:𝕐→𝕐,(Bx)(t)=x(t)-dx(t-σ) for all t∈[0,2π]. Then the following propositions are true. (1)
Ifd=-1,|σ|=(ν/μ)π, where μ and ν are coprime positive integers with ν odd and μ even, then B has a unique inverse B-1:𝕐→𝕐 satisfying ∥B-1∥≤(1/ω1).(2)
Ifd=1,|σ|=(ν/μ)π, where μ and ν are coprime positive integers with ν odd, then B has a unique inverse B-1:𝕐→𝕐 satisfying ∥B-1∥≤(1/ω2).(3)
Ifd=1,|σ|=π, then B has a unique inverse B-1:𝕐→𝕐 satisfying ∥B-1∥≤(1/ω3).


## 3. Main Result

In this section, we will study the existence ofπ-antiperiodic solutions for system (10) in the critical case |c|=|d|=1.Theorem 7.
Assume that(H1)-(H2) hold. Suppose further that c=-1, |τ|=(b/a)π, where a and b are coprime positive integers with b odd and a even, then system (10) has at least one π-antiperiodic solution, if one of the following conditions holds. (1)
d=-1,|σ|=(ν/μ)π, where μ and ν are coprime positive integers with ν odd and μ even, and W1+W2<1, where
(41)W1=max{∑i=1k+1‍πm-i+1ω1αi,∑i=1k+1‍πn-l+k-i+1ω1βl+i+1},W2=max{∑i=1l+1‍πm-k+l-i+1ω1αk+i+1,∑i=1l+1‍πn-i+1ω1βi}.(2)
d=1,|σ|=(ν/μ)π, where μ and ν are coprime positive integers with ν odd, and W1+W3<1, where
(42)W3=max{∑i=1l+1‍πm-k+l-i+1ω2αk+i+1,∑i=1l+1‍πn-i+1ω2βi}.(3)
d=1,|σ|=π, and W1+W4<1, where
(43)W4=max{∑i=1l+1‍πm-k+l-i+1ω3αk+i+1,∑i=1l+1‍πn-i+1ω3βi},in whichω1,ω2, and ω3 are constants defined by Lemma 3 or Lemma 5.Proof.
As the proof of other cases works almost exactly as the proof of case (1), we will prove case (1) only. Consider the operator equation(44)Lx-Nx=λ(-Lx-N(-x)),λ∈(0,1].
Then we have
(45)(Ax1)(k)(t)=11+λφp′(x2(t))-λ1+λφp′(-x2(t)),x2(m-k)(t)=11+λF(t,x,y)-λ1+λF(t,-x,-y),(By1)(l)(t)=11+λφq′(y2(t))-λ1+λφq′(-y2(t)),y2(n-l)(t)=11+λG(t,y,x)-λ1+λG(t,-y,-x),
where
(46)F(t,x1,y1)=F(t,x1θ0(t),x1θ1(t)′,…,x1θk(t)(k),y1ϑ0(t),y1ϑ1(t)′,…,y1ϑl(t)(l)),G(t,y1,x1)=G(t,y1μ0(t),y1μ1(t)′,…,y1μl(t)(l),x1ν0(t),x1ν1(t)′,…,x1νk(t)(k)),F(t,-x1,-y1)=F(t,-x1θ0(t),-x1θ1(t)′,…,-x1θk(t)(k),-y1ϑ0(t),-y1ϑ1(t)′,…,-y1ϑl(t)(l)),G(t,-y1,-x1)=G(t,-y1μ0(t),-y1μ1(t)′,…,-y1μl(t)(l),-x1ν0(t),-x1ν1(t)′,…,-x1νk(t)(k){y1μl(t)(l)x1ν1(t)′}).
Suppose that x(t)=(x1(t),x2(t),y1(t),y2(t))T∈𝔻 is an arbitrary π-antiperiodic solution of system (45). Because x1(t)∈C2πk is T/2-antiperiodic, hence, we have
(47)∫02π‍x1(s)ds=∫0π‍x1(s)ds+∫π2π‍x1(s)ds=∫0π‍x1(s)ds+∫0π‍x1(s+π)ds=0.
Then there exists a constant ξ∈[0,2π] such that
(48)x1(ξ)=0.
Therefore, we have
(49)|x1(t)|=|x1(ξ)+∫ξt‍x1′(s)ds|≤∫ξt‍|x1′(s)|ds,|x1(t)|=|x1(t-2π)|=|x1(ξ)-∫t-2πξ‍x1′(s)ds|≤∫t-2πξ‍|x1′(s)|ds
for all t∈[ξ,ξ+2π]. Combining the above two inequalities, we can get
(50)|x1|∞=maxt∈[0,2π]|x1(t)|=maxt∈[ξ,ξ+2π]|x1(t)|≤maxt∈[ξ,ξ+2π]{12(∫ξt‍|x1′(s)|ds+∫t-2πξ‍|x1′(s)|ds)}≤12∫02π‍|x1′(s)|ds.
Similar to (50), one can easily get
(51)|x1′|∞≤12∫02π‍|x1′′(s)|ds,|x1′′|∞≤12∫02π‍|x1′′′(s)|ds,…,|x1(k-1)|∞≤12∫02π‍|x1(k)(s)|ds,
which yield
(52)|x1(i)|∞≤πk-i|x1(k)|∞fori=0,1,…,k-1.
By a parallel argument to (47)–(52), we can also obtain
(53)|y1(i)|∞≤πl-i|y1(l)|∞,fori=0,1,…,l-1,(54)|x2(j)|∞≤πm-k-j-1|x2(m-k-1)|∞,forj=0,1,…,m-k-2,(55)|y2(j)|∞≤πn-l-j-1|y2(n-l-1)|∞,forj=0,1,…,n-l-2.
Since x2(t)∈C2πm-k is π-antiperiodic, similar to (47), there exists a constant ζ such that x2(m-k-1)(ζ)=0. By a parallel argument to (50), we can obtain from (45), (H1), (52), and (53) that
(56)2|x2(m-k-1)|∞≤2|x2(m-k-1)(ζ)|+∫02π‍|x2(m-k)(s)|ds=∫02π‍|11+λF(s,x1,y1)-λ1+λF(s,-x1,-y1)|ds≤11+λ∫02π‍|F(s,x1,y1)|ds+λ1+λ∫02π‍|F(s,-x1,-y1)|ds≤11+λ∫02π‍|F(s,x1,y1)-F(s,0,0)|ds+λ1+λ∫02π‍|F(s,-x1,-y1)-F(s,0,0)|ds+11+λ∫02π‍|F(s,0,0)|ds+λ1+λ∫02π‍|F(s,0,0)|ds≤11+λ∫02π‍(α1|x1(s-θ0(s))|+α2|x1′(s-θ1(s))|+⋯+αk+1|x1(k)(s-θk(s))|+αk+2|y1(s-ϑ0(s))|+αk+3|y1′(s-ϑ1(s))|+⋯+αk+l+2|y1(l)(s-ϑl(s))|)ds+λ1+λ∫02π‍(y1(l)α1|x1(s-θ0(s))|+⋯+αk+1|x1(k)(s-θk(s))|+αk+2|y1(s-ϑ0(s))|+⋯+αk+l+2|y1(l)(s-ϑl(s))|)ds+∫02π‍|F(s,0,0)|ds≤∫02π‍(α1|x1(s-θ0(s))|+α2|x1′(s-θ1(s))|+⋯+αk+1|x1(k)(s-θk(s))|+αk+2|y1(s-ϑ0(s))|+αk+3|y1′(s-ϑ1(s))|+⋯+αk+l+2|y1(l)(s-ϑl(s))|)ds+∫02π‍|F(s,0,0)|ds≤∫02π‍(α1|x1|∞+α2|x1′|∞+⋯+αk|x1(k-1)|∞+αk+1|x1(k)|∞+αk+2|y1|∞+αk+3|y1′|∞+⋯+αk+l+1|y1(l-1)|∞+αk+l+2|y1(l)|∞)ds+∫02π‍|F(s,0,0)|ds≤2πα1|x1|∞+2πα2|x1′|∞+⋯+2παk|x1(k-1)|∞+2παk+1|x1(k)|∞+2παk+2|y1|∞+2παk+3|y1′|∞+⋯+2παk+l+1|y1(l-1)|∞+2παk+l+2|y1(l)|∞+2πmaxs∈[0,2π]|F(s,0,0)|≤2πα1(2π)k2k|x1(k)|∞+2πα2(2π)k-12k-1|x1(k)|∞+⋯+2παk2π2|x1(k)|∞+2παk+1|x1(k)|∞+2παk+2(2π)l2l|y1(l)|∞+2παk+3(2π)l-12l-1|y1(l)|∞+⋯+2παk+l+12π2|y1(l)|∞+2παk+l+2|y1(l)|∞+2πmaxs∈[0,2π]|F(s,0,0)|=2∑i=1k+1‍πk-i+2αi|x1(k)|∞+2∑i=1l+1‍πl-i+2αk+i+1|y1(l)|∞+2πmaxs∈[0,2π]|F(s,0,0)|.
Namely,
(57)|x2(m-k-1)|∞≤∑i=1k+1‍πk-i+2αi|x1(k)|∞+∑i=1l+1‍πl-i+2αk+i+1|y1(l)|∞+πmaxs∈[0,2π]|F(s,0,0)|.
Since y2(t)∈C2πn-l is π-antiperiodic, similar to (50), there exists a constant η such that y2(n-l-1)(η)=0. By a parallel argument to (50), we can obtain from (45), (H1), (52), and (53) that
(58)2|y2(n-l-1)|∞≤2|y2(n-l-1)(η)|+∫02π‍|y2(n-l)(s)|ds=∫02π‍|11+λG(s,y1,x1)-λ1+λG(s,-y1,-x1)|ds≤11+λ∫02π‍|G(s,y1,x1)|ds+λ1+λ∫02π‍|G(s,-y1,-x1)|ds≤11+λ∫02π‍|G(s,y1,x1)-G(s,0,0)|ds+λ1+λ∫02π‍|G(s,-y1,-x1)-G(s,0,0)|ds+11+λ∫02π‍|G(s,0,0)|ds+λ1+λ∫02π‍|G(s,0,0)|ds≤11+λ∫02π‍(x1(k)β1|y1(s-μ0(s))|+β2|y1′(s-μ1(s))|+⋯+βl+1|y1(l)(s-μl(s))|+βl+2|x1(s-ν0(s))|+βl+3|x1′(s-ν1(s))|+⋯+βl+k+2|x1(k)(s-νk(s))|)ds+λ1+λ∫02π‍(x1(k)β1|y1(s-μ0(s))|+β2|y1′(s-μ1(s))|+⋯+βl+1|y1(l)(s-μl(s))|+βl+2|x1(s-ν0(s))|+βl+3|x1′(s-ν1(s))|+⋯+βl+k+2|x1(k)(s-νk(s))|)ds+∫02π‍|G(s,0,0)|ds≤∫02π‍(β1|y1(s-μ0(s))|+β2|y1′(s-μ1(s))|+⋯+βl+1|y1(l)(s-μl(s))|+βl+2|x1(s-ν0(s))|+βl+3|x1′(s-ν1(s))|+⋯+βl+k+2|x1(k)(s-νk(s))|)ds+∫02π‍|G(s,0,0)|ds≤∫02π‍(β1|y1|∞+β2|y1′|∞+⋯+βl|y1(l-1)|∞+βl+1|y1(l)|∞+βl+2|x1|∞+βl+3|x1′|∞+⋯+βl+k+1|x1(k-1)|∞+βl+k+2|x1(k)|∞)ds+∫02π‍|G(s,0,0)|ds≤2πβ1|y1|∞+2πβ2|y1′|∞+⋯+2πβl|y1(l-1)|∞+2πβl+1|y1(l)|∞+2πβl+2|x1|∞+2πβl+3|x1′|∞+⋯+2πβl+k+1|x1(k-1)|∞+2πβl+k+2|x1(k)|∞+2πmaxs∈[0,2π]|G(s,0,0)|≤2πβ1(2π)l2l|y1(l)|∞+2πβ2(2π)l-12l-1|y1(l)|∞+⋯+2πβl2π2|y1(l)|∞+2πβl+1|y1(l)|∞+2πβl+2(2π)k2k|x1(k)|∞+2πβl+3(2π)k-12k-1|x1(k)|∞+⋯+2πβl+k+12π2|x1(k)|∞+2πβl+k+2|x1(k)|∞+2πmaxs∈[0,2π]|G(s,0,0)|=2∑i=1l+1‍πl-i+2βi|y1(l)|∞+2∑i=1k+1‍πk-i+2βl+i+1|x1(k)|∞+2πmaxs∈[0,2π]|G(s,0,0)|.
Namely,
(59)|y2(n-l-1)|∞≤∑i=1l+1‍πl-i+2βi|y1(l)|∞+∑i=1k+1‍πk-i+2βl+i+1|x1(k)|∞+πmaxs∈[0,2π]|G(s,0,0)|.
From (1) of Lemmas 4 and 6, one can obtain
(60)|x1(k)(t)|=|A-1Ax1(k)(t)|≤1ω1|Ax1(k)(t)|=1ω1|(Ax1)(k)(t)|≤1ω1φp′(|x2|∞),|y1(l)(t)|=|B-1By1(l)(t)|≤1ω1|Bx1(l)(t)|=1ω1|(By1)(l)(t)|≤1ω1φq′(|y2|∞).
That is,
(61)|x1(k)|∞≤1ω1φp′(|x2|∞),|y1(l)|∞≤1ω1φq′(|y2|∞).
From (54) and (55), we can get
(62)|x2|∞≤πm-k-1|x2(m-k-1)|∞,|y2|∞≤πn-l-1|y2(n-l-1)|∞.
With (57)–(62), we have
(63)|x2|∞≤∑i=1k+1‍πm-i+1ω1αi|x2|∞p′-1+∑i=1l+1‍πm-k+l-i+1ω1αk+i+1|y2|∞q′-1+T1,|y2|∞≤∑i=1k+1‍πn-l+k-i+1ω1βl+i+1|x2|∞p′-1+∑i=1l+1‍πn-i+1ω1βi|y2|∞q′-1+T2,
where
(64)T1=πm-kmaxs∈[0,2π]|F(s,0,0)|,T2=πn-lmaxs∈[0,2π]|G(s,0,0)|.
Let X0=max{|x2|∞,|y2|∞}. Then we have from (63) that
(65)X0≤W1X0p′-1+W2X0q′-1+max{T1,T2}
in which
(66)W1=max{∑i=1k+1‍πm-i+1ω1αi,∑i=1k+1‍πn-l+k-i+1ω1βl+i+1},W2=max{∑i=1l+1‍πm-k+l-i+1ω1αk+i+1,∑i=1l+1‍πn-i+1ω1βi}.
Since 1<p′,q′≤2, (65) implies from (1) that there exists a positive constant M0 such that X0≤M0; that is,
(67)|x2|∞≤M0,|y2|∞≤M0,
which implies from (61) that there exist positive constants M1 and M2 such that
(68)|x1(k)|∞≤M1,|y1(l)|∞≤M2,
which implies from (57) and (59) that
(69)|x2(m-k-1)|∞≤∑i=1k+1‍πk-i+2αiM1+∑i=1l+1‍πl-i+2αk+i+1M2+πmaxs∈[0,2π]|F(s,0,0)|≡M3,|y2(n-l-1)|∞≤∑i=1l+1‍πl-i+2βiM2+∑i=1k+1‍πk-i+2βl+i+1M1+πmaxs∈[0,2π]|G(s,0,0)|≡M4.
Substituting (68), (69) into (52), (53), (54), and (55), we get
(70)|x1(i)|∞≤Tk-i2k-i|x1(k)|∞≤Tk-i2k-iM1≡Nifori=0,1,…,k-1,|y1(i)|∞≤Tl-i2l-i|y1(l)|∞≤Tl-i2l-iM2≡Lifori=0,1,…,l-1,|x2(j)|∞≤Tm-k-j-12m-k-j-1|x2(m-k-1)|∞≤Tm-k-j-12m-k-j-1M3≡Hjforj=0,1,…,m-k-2,|y2(j)|∞≤Tn-l-j-12n-l-j-1|y2(n-l-1)|∞≤Tn-l-j-12n-l-j-1M4≡Kjforj=0,1,…,n-l-2.
Let
(71)M=∑i=0k-1‍Ni+∑i=0m-k-2‍Hi+∑i=0l-1‍Li+∑i=0n-l-2‍Ki+∑i=14‍Mi+1(Clearly,Misindependentofλ).
Take
(72)Ω={x∈𝕏:∥x∥𝕏<M}.
It is clear that Ω satisfies all the requirements in Lemma 1 and condition (H) is satisfied. In view of all the discussions above, we conclude from Lemma 1 that system (10) has at least one π-antiperiodic solution. This completes the proof.Similar to Theorem7, we can easily obtain the following results.Theorem 8.
Assume that(H1)-(H2) hold. Suppose further that c=1,|τ|=(b/a)π, where a and b are coprime positive integers with b odd, then system (10) has at least one π-antiperiodic solution, if one of the following conditions holds. (1)
d=-1,|σ|=(ν/μ)π, where μ and ν are coprime positive integers with ν odd and μ even, and W5+W2<1, where
(73)W5=max{∑i=1k+1‍πm-i+1ω2αi,∑i=1k+1‍πn-l+k-i+1ω2βl+i+1},W2=max{∑i=1l+1‍πm-k+l-i+1ω1αk+i+1,∑i=1l+1‍πn-i+1ω1βi}.(2)
d=1,|σ|=(ν/μ)π, where μ and ν are coprime positive integers with ν odd, and W5+W3<1, where
(74)W3=max{∑i=1l+1‍πm-k+l-i+1ω2αk+i+1,∑i=1l+1‍πn-i+1ω2βi}.(3)
d=1,|σ|=π, and W5+W4<1, where
(75)W4=max{∑i=1l+1‍πm-k+l-i+1ω3αk+i+1,∑i=1l+1‍πn-i+1ω3βi},in whichω1,ω2, and ω3 are constants defined by Lemma 3 or Lemma 5.Theorem 9.
Assume that(H1)-(H2) hold. Suppose further that c=1, |τ|=π, then system (10) has at least one π-antiperiodic solution, if one of the following conditions holds. (1)
d=-1,|σ|=(ν/μ)π, where μ and ν are coprime positive integers with ν odd and μ even, and W6+W2<1, where
(76)W6=max{∑i=1k+1‍πm-i+1ω3αi,∑i=1k+1‍πn-l+k-i+1ω3βl+i+1},W2=max{∑i=1l+1‍πm-k+l-i+1ω1αk+i+1,∑i=1l+1‍πn-i+1ω1βi}.(2)
d=1,|σ|=(ν/μ)π, where μ and ν are coprime positive integers with ν odd, and W6+W3<1, where
(77)W3=max{∑i=1l+1‍πm-k+l-i+1ω2αk+i+1,∑i=1l+1‍πn-i+1ω2βi}.(3)
d=1,|σ|=π, and W6+W4<1, where
(78)W4=max{∑i=1l+1‍πm-k+l-i+1ω3αk+i+1,∑i=1l+1‍πn-i+1ω3βi},in whichω1,ω2, and ω3 are constants defined by Lemma 3 or Lemma 5.


## 4. An Example

Example 1. Letp, q≥2. Then the following third-order (p,q)-Laplacian neutral differential system
(79)(φp((x(t)+x(t-π2))′))′′=sint+2π4x(t-1)+2π6y(t-1)+2π6y′(t-1),(φq((y(t)+y(t-π2))′′))′=cost+2π6y(t-1)+2π4x(t-1)
has at least one π-antiperiodic solution.Proof.
By calculation,ω1=2,α1=2/π4,α3=α4=2/π6,α2=α5=0,β1=2/π6,β4=2/π4, and β2=β3=β5=0. Therefore,
(80)W1=max{1π,1π2}=1π<12,W2=max{1π2+1π3,1π3}=1π2+1π3<12.
Hence,
(81)W1+W2<12+12=1,
which implies that case (1) in Theorem 7 holds. It is easy to verify that (H1)-(H2) hold and the result follows from Theorem 7. This completes the proof.



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*Source: 454619-2013-05-15.xml*